Looking for that blessed hope, and the glorious appearing (epiphany) of the great God and our Saviour Jesus Christ;  Titus 2:13 511

course, appears three times, since, as shown above, there were 20 boards on the south side, 20 on the north and a total of 20 full boards on the three sides of the most holy, each having two sockets, or pedestals. The 8 boards on the west side, like all the other boards, had each two sockets, hence 16 in all. Applying to these 16 our three key numbers we find the following: 16 x 18 x 10 x 10 x 5 = 144,000— the same truth again hidden in the tabernacle. The second veil had for its supports four pillars, each of which had a socket (vs. 35, 36). Thus there were four sockets for these four pillars. As before, let us apply to this number 4 our key numbers, as follows: 4 x 18 x 10 x 10 x 10 x10÷5 = 144,000. This result occurs twice, because the pillars, as well as their sockets, were four in number. The same remark as in the other cases applies here as to the prefigured number of the Christ's members. The first veil had five pillars and five sockets (v. 38). Please note the following result of applying to this number, 5, our key numbers: 5 x 18 x 10 x 10 x 10 x 10 x 10 x 10 ÷ (5 x 5 x 5 x 5) = 144,000. Here, because of there being five pillars and five sockets at the first veil God twice hid the thought of the number of members in the Christ class. Also combinedly the pillars show the same truth: They total 9. To this we apply the key numbers, 10 and 5, as follows: 9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 ÷ (5 x 5 x 5 x 5) = 144,000. The same result is also found in using the combined number of pillar sockets—9. There was a set of five bars on each of the three boarded sides of the tabernacle (vs. 31, 32). In each case the middle bar extended the whole length of each wall (v. 33). The top and the bottom bars in each of the three boarded walls were half the length of each wall. Applying to the five bars of each of the three boarded walls our key numbers, the following results: 5 x 18 x 10 x 10 x 10 x 10 x 10 x 10 ÷ (5 x 5 x 5 x 5) = 144,000, the same symbolization as before on the number of the Christ's members. This appears three times, once for each of the three walls. This is also shown in the combined number of bars together with the combined number

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of boards that they held together; for there were 15 bars in all and 48 boards in all. Now 15 x 48 = 720. Applying our key numbers as follows we have: 720 x 10 x 10 x 10 ÷ 5 = 144,000. Also the number of the small bars (12) shows it: 12 x 12 x 10 x 10 x 10 = 144,000. So, too, 12 x 4 (the number of short bars on each side) x 3 (the number of long bars) x 10 x 10 x 10=144,000. Incidentally the key number 18 is indicated in the combined number of holy and most holy pillars and their sockets (9 x 2 = 18); also in the number of full bars (3) multiplied by the number of half length bars (6) considered as full length bars (3 x 6 = 18).

(36) From vs. 21, 23, 25, 27, 28 we find that the tabernacle's length, height and width were respectively: 30, 10 and 10 cubits. Their sum was 50 cubits. Applying to this number our key numbers as follows, we find the same result as often before was found on the number of the Christ's members: 50 x 18 x 10 x 10 x 10 x 10 x 10 ÷ (5 x 5 x 5 x 5) =144,000. The sum of the tabernacle's length and height was (30+10) 40 cubits. Combining this with our key numbers we get the following: 40 x 18 x 10 x 10 x 10 ÷ 5 = 144,000. The same sum, 40, results from adding the length and width of the tabernacle; hence in a second way the previous figures are used, with the same result. Again, the height and width of the tabernacle, being respectively 10 cubits, total 20 cubits; and by applying our key numbers we reach the same secret hidden here. Thus: 20 x 18 x 10 x 10 x 10 x 10 ÷ (5 x 5) = 144,000. The diagonal of each of the long sides of the tabernacle is 31.6227766 cubits, which, being the hypotenuse of a right-angled triangle with sides of 30 and 10 cubits, is obtained by squaring each of these two sides, whose sum equals the square of that hypotenuse. The diagonal of the short, the west, side of the tabernacle is 14.1421356 cubits, which is obtained in the same way as the other hypotenuse, here using 10 cubits for each side. If the first of these hypotenuses or diagonals is squared the result is 1,000 square cubits; and if the second is squared the result is 200 square cubits.

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Their sum equals 1,200 square cubits. Now 1,200 x 1,200 ÷ 10 =144,000, the same hidden symbolism often before gotten, here lying in the tabernacle's diagonals, a fact brought out four times, since there are four long sides in the tabernacle, counting also the ceiling and floor such sides. The tabernacle had four length lines, two at the top and two at the bottom, these four lines of 30 cubits each, which added together give us the sum of 120 cubits. Applying to the square of 120 the key number 10 we have: 120 x 120 x 10 = 144,000. Again, the same secret is the four length lines of the tabernacle.

(37) There are also four height lines, of 10 cubits each, in the tabernacle: two in the front and two in the rear. Their sum is 40 cubits. Applying to this number our key numbers, we get as follows: 40 x 18 x 10 x 10 x 10 ÷ 5=144,000. The same secret, this time hidden in the four height lines. There are also four width lines, of 10 cubits each, in the tabernacle: two in the front top and bottom, and two in the rear top and bottom. Their sum is 40 cubits. Applying to this sum our key numbers as follows, we get the same hidden truth, this time in the width lines of the tabernacle, thus: 40 x 18 x 10 x 10 x 10 ÷ 5=144,000. If we add all the cubit length, height and width lines (4 x 30+ 4 x 10 + 4 x 10 = 120 + 40 + 40 = 200) we obtain 200 cubits. Applying to these 200 cubits our key numbers we will again get the same secret, this time hidden in the sum of all the cubit length, height and width lines. Thus, 200 x 18 x 10 x 10 x 10 ÷ (5 x 5) =144,000. There are eight long diagonals in the tabernacle's four sides, counting the floor and the ceiling also as sides of it, each side, of course, having two diagonals. If we square the long diagonal, which we saw above makes 1,000 square cubits and multiply it by 8, the number of long diagonals (8 x 1,000), we get 8,000 square cubits as the sum of the squares of all the long diagonals of the four surfaces of the tabernacle. If we multiply these 8,000 square cubits by the main tabernacle key number (8,000 x 18 =144,000), we again get the same secret, this time from the sum of the

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squares of all the tabernacle surfaces' long diagonals. We get the same secret hidden in the short diagonals. There are four of such: two in the front and two in the rear of the tabernacle. Above we saw that the square of the short diagonal is 200 square cubits, which would make the sum of the squares of all four 800 square cubits (4 x 200 = 800). Applying two of our key numbers to this number we have the same secret hidden in the short diagonals. Thus 800 x 18 x 10 = 144,000.

(38) The Lord has been careful to put this same secret hiddenly in the lengthwise circuit of the tabernacle. Its lengthwise circuit is 80 cubits, its two sides being 30 cubits each, and its front and rear 10 cubits each (30+30+10+10=80). Applying to these 80 cubits two of our key numbers we get 80 x 18 x 10 x 10= 144,000. Thus is hidden the same secret in the lengthwise circuit of the tabernacle. The same secret lies hidden in the crosswise circuit of the tabernacle. Its crosswise circuit was 40 cubits, as the height of each side was 10 cubits and its floor and ceiling were each 10 cubits (10+10+10+10=40). Applying to this our three key numbers we get this secret hidden in the crosswise circuit of the tabernacle: 40 x 18 x 10 x 10 x 10 ÷ 5 = 144,000. In still another way God has embedded this secret in the tabernacle's dimensions. The area of each one of its four long sides (30 x 10 = 300) is 300 square cubits. The area of its four long sides would be four times 300 square cubits, which is 1,200 square cubits. Applying the key number 10 to the square of 1,200 cubits we get 144,000. Thus: 1,200 x 1,200 = 1,444,000; and this divided by 10 is 144,000. Thus God has given us this secret hidden in the area of the floor, ceiling and the two long sides of the tabernacle. Finally, He has given us this secret in a sixteenth way, i.e., in the areas of both ends of the tabernacle. The area of each of these is 100 square cubits; for each of these ends is 10 cubits high and 10 cubits wide. Thus 10 x 10 = 100. The sum of the areas of the two ends was (2 x 100) 200 square cubits. Applying our key numbers to these 200

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square cubits we get 144,000. Thus: 200 x 18 x 10 x 10 x 10 ÷ (5x5) = 144,000. Thus in the tabernacle's end walls God has embedded the truth that the Christ class would consist of 144,000 members. This can be shown from the 4 cubical diagonals; also from the total of diagonals (4 end, 8 lengthwise and 4 cubical, or 16): 16 x 18 x 10 x 10 x 5=144,000. This truth has been embedded in the dimensions of the tabernacle at least 19 times.

(39) But in the tabernacle as a whole there were other symbolizations, nine of them pointing out the Millennium and five of them pointing out God's name. Above we saw that the square of the long diagonal (31.6227766 x 31.6227766 = 1,000) was 1,000 square cubits, indicating 1,000 years. Such a long diagonal we find occurs eight times: two times each in the roof, floor and each of the side walls. Hence by the long diagonals of the tabernacle God has indicated eight times that the Millennium will last 1,000 years. The square of the tabernacle's length (30 x 30=900) is 900 square cubits; the square of the height (10 x 10) is 100 square cubits; and the sum of 900 and 100 is 1,000. The width being the same as the height, the sum of the squares of the length and of the width is also 1,000. The length of the tabernacle in inches (30 x 25) was 750 inches. The height in inches (10 x 25) was 250 inches. And the sum of these two numbers (750+250) is 1,000. The width being the same as the height, the sum of the length and width in inches is also 1,000 inches. The cross circuit, as we saw above, was 40 cubits, which reduced to inches (40 x 25) make 1,000 inches. We also saw above that the area of each end wall (10 x 10) was 100 square cubits. Applying to this area the key number 10 we have (100 x 10), again 1,000. This holds true for each end. Also the entire perimeter: 4 x 30 (length) +4 x 10 (one end perimeter) + 4 x 10 (the other end perimeter) = 200, which multiplied by the key number 5 = 1,000. Thus in nine ways in the tabernacle as a whole God indicates that the Millennium is a period of 1,000 years. Now there will be pointed out the fact that in five ways

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God has pointed out that He was the tabernacle's Architect. Above we pointed out the four consonants that spell God's name: Yod (10), He (5), Vav (6) and He (5). The tabernacle was 10 cubits high = Yod. The sum of its length, height and width divided by its height equals 5=He. Thus: 30+10+10 = 50, and 50 ÷ 10 = 5 =He. Its length multiplied by its height (30 x 10) is 300. And 300 divided by the sum of its length, height and width (30+10+10 = 50) gives us 6. Thus 300 ÷ 50 = 6 =Vav. The height multiplied by the width (10 x 10 = 100) and this product (100) divided by the sum of the height and width (10+10 = 20) gives us the quotient 5. Thus 100 ÷ 20 = 5 =He. Thus we have gotten God's name spelled in English: YHVH. But this is also obtained in three other ways. It will be noted that in getting the first He we divided the sum of the length, height and width (30+10+10 = 50) by the height (10); but the division of this sum can be made by the width (10), which is equal to the height. This change, using the same methods of getting the other three letters, gives us a second way of finding God's name in the tabernacle as a whole. It will be also noted that in getting the letter Vav (6) we divided the product of the length and height (30 x 10 = 300) by the sum of the length, height and width (30+10+10 = 50); for 300 ÷ 50 = 6. But since the width is the same as the height (10 cubits) we could have taken the product of the length and width (30 x 10 = 300) and divided it by the sum of the length, height and width (30+10+10 = 50) and thus have gotten the quotient 6 =Vav. Similarly, by changing the dimension names but not values of the dimensions in the fourth process above we get He (5). This change, using the other three sets of figures as before will give us a third way of reaching the name of God in the tabernacle as a whole. Finally, by using the second way by which we found the first He and the second way by which we found the Vav (6), we furthermore reach a fourth way of getting God's name in the tabernacle as a whole. The six sides of the

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tabernacle also give us Vav. Combining this with the other ways used above for getting the other three letters of God's name yields us another way of finding God's name in the tabernacle as a whole—five in all.

(40) We will now present some things hidden in the first and second veils (vs. 35, 37). We will first show how God has pointed out the Christ in these two veils, one typing the death of the Christ's will, the other the death of the Christ's body. These veils were 10 cubits high and 10 cubits wide. The sum of the height and width=20 cubits. Applying to this number of the second veil our key numbers we have the following: 20 x 18 x 10 x 10 x 10 x 10 ÷ (5 x 5) =144,000. Applying them to the sum of the first veil's height and width gives the same results, because both were of the same size. The sum of the two heights of each veil, being 20 cubits, by the same operation produces the same results twice again. The sum of the two widths of each veil, being also 20 cubits, by the same operation gives us the same results twice. There were four height lines in both veils (4 x 10 = 40). Applying our key numbers to 40 we have the following: 40 x 18 x 10 x 10 x 10 ÷ 5 = 144,000. The four width lines in the two veils, subjected to the same operation, give us again the same result. The perimeter of the first veil (10+10+10+10 = 40) was 40 cubits. Subjected to the same process, it gives us the same result. The same is true of the perimeter of the second veil. The sum of the perimeters of both veils (40+40) was 80 cubits. Subjecting this number to our key numbers 18 and 10 we have: 80 x 18 x 10 x 10 =144,000. The sum of the areas of both veils (2 x 10 x 10) =200 square cubits. Subjecting this number to our key numbers, we have the following: 200 x 18 x 10 x 10 x 10 ÷ (5 x 5)=144,000. As shown in the second veil alone: The area facing the most holy was (10 x 10) 100 square cubits. Its area facing the holy was also 100 square cubits. This sum, 200 square cubits, 200 x 40 (this veil's perimeter, 4 x10) = 8000. 8000 x 18 (key number) = 144,000. The same appears in the first veil. Thus in many ways in

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the two veils the Christ is set forth as consisting of 144,000 members. In six ways in connection with the two veils the Millennium is set forth as a period of 1,000 years, in which the Christ will bless the world: (1) The area of the first veil (10 x 10) =100 square cubits. Applying to this number the key number 10 we have the following: 100 x 10 = 1,000. (2) The second veil having the same dimensions, the same process produces the same result. (3) The sum of the areas of both veils (100+100) = 200. Subjected to the key number 5 it yields the following result: 200 x 5 =1,000. (4) As shown above, the perimeter of the first veil is 40 cubits, which (40 x 25) is 1,000 inches. (5) The perimeter of the second veil, being the same as that of the first veil, subjected to the same operation yields the same result. (6) The four diagonals of the two veils are each 14.1421356 cubits long, whose square is 200. The sum of their squares is 800 square cubits, adding to which that of the veils (100+100) yields 1000.

(41) There are four ways in which God embedded His name in the two veils: (1) In the first veil the height is 10 cubits. 10=Yod. The product of its height and width (100) divided by their sum (20) is 5=He. The sum of its three sides (two heights and one width) not touching the floor (10+ 10+ 10) = 30. This sum multiplied by the sum of the height and width (10+10 = 20) and divided by the area of the first veil (10 x 10 =100) gives us 6 =Vav. Thus 30 x 20 = 600; and 600 ÷ 100 = 6. Finally, the product of the height and the width (10 x 10 = 100) divided by the sum of the height and width (10+10 = 20) yields a quotient of 5=He. Thus, 100 ÷ 20=5. Accordingly, these four features give us the Hebrew numbers that are the consonants of Jehovah's name, whose English equivalents are YHVH. (2) The second veil, having the same dimensions as the first, subjected to the same four processes as the first veil was above subjected to, yields the same four results. This is also shown two more times—once in each veil: The main dimensions of each was 10 (Yod); the

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product of the heights and width of each (10x10=100) divided by the sum of each (10+10 = 20) = 5 (He); each, being quite thick, had 6 (Vav) surfaces (2 sides and 4 edges). The perimeter of each (4 x 10 = 40) plus the height of each (10) divided by the width of each (10) was 5 (He). Thus we have found in the two veils 12 ways of pointing out that the Little Flock will consist of 144,000 members, 6 ways of pointing out that it will be during a 1,000-year period, the Millennium, when the Little Flock will bless the world, and 6 ways in which God has put His name in the veils, 24 in all, a multiple of 12, the Little Flock's number.

(42) From the sizes of the boards given in vs. 20-23; 27-30, especially of those in the most holy, we deduce the tabernacle dimensions as a whole and in the holy and most holy; and these we will now study. That there were 6⅔ boards on the north side and the same number of boards on the south side of the most holy proves that these two sides were each 10 cubits long. That there were six full boards on the west side and a third of another board (which is a half-cubit width) in each corner of the west side of the holy of holies, proves that it also was 10 cubits wide. Thus the holy of holies was a perfect cube, each of whose six sides was 10 cubits square. Since these two corner boards were sawn in halves from top to bottom, each of these halves was ¾ of a cubit wide, and since the parts of these ¾ that were visible in the most holy were in each case a half-cubit wide, the parts of them that ended even with the north and south corner boards must have been ¼ of a cubit wide, which proves that the 48 boards of the tabernacle were ¼ of a cubit, i.e., 6¼ inches, thick. The pillars in the most holy fitting flush against the eastmost boards in the most holy's north and south sides so touched them as to make the two ⅔ in the most holy and ⅓ in the holy, to yield the right sum. The 13⅓ boards in the north side, and the 13⅓ boards in the south side of the holy, prove that the holy was 20 cubits long; and its width being the

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same as that of the most holy, it was 10 cubits wide. From now on, instead of making the remark about "adjusting," "subjecting," "applying," etc., the various numbers to the key numbers, we will by the process indicate that such is done. We will now bring out some of the secrets that God hid in the holy on the 144,000, as the number of the Christ's members, on the Millennium as being the 1,000 years when the Christ will bless the world and on the name Jehovah as designating the tabernacle's Architect: (1) The length of the holy was 20 cubits. 20x18x10x10x10x10 ÷ (5 x 5) =144,000. (2) The sum of its length, height and width was 40 cubits (20+10+10 = 40). 40x18x10x10x10 ÷ 5 = 144,000. (3) Its side length diagonals were each 22.36068 cubits long. As we know, the square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides, we must square the length of the holy and add to this the square of its height, or width, then extract the square root of this sum, which will give us the diagonals of the four length sides, i.e., the north, south, ceiling and floor length sides (20x20=400; and 10x10=100; and their sum is 500, whose square root is 22.36068). Its square is, therefore, 500. 500x18x10x10x10x10÷ (5x5x5x5) = 144,000. Since each side has two diagonals, the eight show this eight times. Hence the next point is (11).

(43) (11) The end diagonals measured 14.1421356 cubits each, gotten by the same process as was shown in point (3). The square of this diagonal is 200. 200x 18x10x10x10÷5x5)=144,000. As the holy had two ends and each of these ends had two diagonals, we have here this secret given in four ways. Accordingly, we will number our next general point (15). (15) As seen under (3), the square of each long side diagonal is 500 square cubits, and as there are eight of them the sum of the squares of all eight is 4,000 square cubits (8x500=4,000). 4,000x18x10÷5=144,000. (16) As the square of each of the end diagonals is 200 square cubits [see (11) above], and as there are four

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of such diagonals, we get (4x200) as their sum 800 square cubits. 800x18x10 =144,000. (17) The square of each of the four solid diagonals of the holy = 600, which, as the hypotenuse of the right-angled triangle, is the sum of the square of the side diagonal and the square of the height, the two short sides of this right-angled triangle, = 500+100 = 600, whose square root is 24.4949. There are eight long side diagonals, four end, or short side, diagonals and four solid diagonals to the holy. The square root of 500, the square of each long side diagonal, is [see (3) above] 22.36068, and the square root of 200, the square of each end diagonal, is 14.1421356. Since there are eight side, four end, and four solid diagonals, the sum of the squares of the eight side diagonals = 4,000 (8x500); the sum of the squares of the four end diagonals = 800 (4x200) and the sum of the squares of the four solid diagonals (4x600 = 2,400, the sum of the squares of all 16 diagonals (4,000+800+2,400) = 7,200 square cubits. 7,200 x 10 x 10 ÷ 5 = 144,000. (18) The area of the side wall or floor or ceiling (20 x 10) was 200 square cubits. 200 x 18 x 10 x 10 x 10 ÷ (5x5) = 144,000. Since this is the area of the two sides, the ceiling and the floor, this secret appears four times. Hence we will number our next point (22). (22) The sum of the areas of the two sides, the ceiling and the floor was (4 x 200) 800 square cubits [see (18)]. 800 x 18 x 10 = 144,000. (23) The sum of the length lines of the two long sides, ceiling and floor is (4 x 20) 80 cubits. 80 x 18 x 10 x 10 = 144,000. (24) Take the sum of the squares of all the length lines (4 x 20 x 10 = 1,600), of the height lines (4 x 10 x 10 = 400), of the width lines (4 x 10 x 10 = 400), of the four sides' diagonals (8 x 500 = 4,000) and of the four end diagonals (4 x 200 = 800), and this totals 7,200 square cubits. Thus: 1,600+400+400+4,000+800 = 7,200. And 7,200 x 10 x 10 ÷ 5 = 144,000. (25) The holy's lengthwise circuit (20+10+20+10) was 60 cubits. If this length is taken as a side of a square, the square would be (60 x 60) 3,600 square cubits. And 3,600 x 10 x 10 x 10

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÷ (5x5) = 144,000. (26) The holy's crosswise circuit (10+10+10+10) was 40 cubits. 40x18x10x10x10 ÷ 5 = 144,000. (27) If the length of the crosswise circuit is taken as the side of a square, this square (40x40 = 1,600) would be 1,600 square cubits. And 1,600x18x5 = 144;000. (28) If we take the sum of the two sides' perimeters (60+60 = 120) and consider this sum as the side of a square, the square cubits (120x120 = 14,400) are 14,400 square cubits. 14,400x10 = 144,000. (29) If we add the perimeters of both end walls (40 + 40) we obtain 80 cubits. 80x18x10x10 = 144,000. (30) The volume of the holy (20x10x10) was 2,000 cubic cubits. 2,000x18x10x10 ÷ (5x5) = 144,000. Thus in the holy as a structure we find that God has at least 30 times shown the Christ to have 144,000 members.

(44) We will now study some symbolisms of the holy that hide the thought, now revealed, without and with the key numbers, that in the holy the Millennium as a period of 1,000 years will be the time when as God's Vicegerent the Christ will bless the world: First we will point out several features gotten without the use of key numbers. (1) The sum of the two side walls' lengths (20+20) was 40 cubits. 40 cubits are 1,000 inches (40x25 = 1,000). (2) This occurs twice, the second time on the ceiling and floor. (3) The sum of the heights of the two veils (10+10 = 20) and of the length of the ceiling (20) is 40 cubits, or 1,000 inches (40x25 = 1,000). (4) The same is true of the sum of these two heights and the length of the floor line; (5) also of the sum of the two end widths and the length of each long side. Thus this occurs four times; hence this makes 8 points so far indicated without the use of key numbers, that the Christ in the Millennium will bless the world. But we will point out seven others of such features. (9) Its south side was 20 cubits, or 500 inches long (20x25 = 500). Its north side was the same; and 500+500 = 1,000. (10) This is also true of the sum of the ceiling's and floor's lengths. (11) The height was 10 cubits, or 250 inches (10x25 = 250);

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and any two of its four lengths are 500 inches; and the width was 10 cubits, or 250 inches (10x25= 250); and 250 + 500 + 250 = 1,000. This occurs two times, hence we number our next point (13). (13) The sum of all the holy's areas is 1,000 square cubits. Thus the area of each (20 x 10 = 200) of its four sides is 4 x 200=800 square cubits; and of each (10 x 10 = 100) of its two ends is 100 x 2 = 200 square cubits. The sum of these two sets of square cubits is 1,000. (14) The sum of the squares of both diagonals of anyone of the two long sides, the ceiling or the floor, is 1,000 (500+500; see point (3), at end of par. 42). Since there are eight of such diagonals God has pictured the 1,000 years four times in this feature; hence we will number our next point (18). The same truth is brought out by the holy's dimensions and key numbers: (18) Five (key number) times the square of the length of any diagonal in the end walls equals 1,000. The square of each of these diagonals is 200 [see par. 63, (11)]. 5x200= 1,000. Four of such exist; our next point is (22).

(45) (22) The crosswise circuit of the holy (4x10) was 40 cubits. Its lengthwise circuit was (20+10+20+10) 60 cubits. The sum of these two circuits is (40+60) 100 cubits. 100x10 (key number) =1,000. (23) The area of each of the two long sides, of the ceiling and of the floor of the holy is 200 square cubits (20x10=200). 200x5 (key number)=1,000. This being true for each of the four involved areas, we number our next point (27). (27) The area of each of the holy's end walls is 100 square cubits (10x10= 100). 100x10=1,000. This being true of both ends makes, at least, a total of 28 ways in which the holy symbolizes that the Christ will bless the world in the Millennium. There are several ways in which the name, Jehovah, is embedded in the holy: (1) The lengthwise circuit is 1,500 inches (60 x 25=1,500). 1,500= 10(Yod) x 5(He) x 6(Vav) x 5(He). (2) God's name, Jehovah, is embedded in the holy's main dimensions. Thus: The height of the holy was 10 (Yod) cubits.

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The product of its height and width (10x10= 100) divided by its length (100 ÷ 20) is 5 (He). The sum of the squares of its length, height and width (20x20 + 10x10 + 10x10=400+100+100 = 600) divided by the square of its width or height (600 ÷ 100 =6) gives us 6 (Vav). The length multiplied by the height, or width, divided by the sum of the length, height and width [(20x10) 200 ÷ (20+ 10+ 10) 40=5] gives us 5 (He). As there are in the process of finding Vav and in the second way of finding He in this way of finding the name Jehovah an alternative use of height and width, and still another as follows: Vav (6) is derived from the number of the holy's surfaces, which were six, there are resultantly five ways of finding the name Jehovah in the holy—(3), (4) and (5). Thus we have found 30 times the 144,000; 28 times the Millennium and 5 times the name Jehovah as Architect symbolized in the holy; but 30+28+5=63 (= 9x7, a multiple of seven, the Divine number) symbolizations in the dimensions of the holy.

(46) There now remains of Ex. 36 the study of the truths hidden in the measurements of the most holy. As we know, the most holy was a perfect cube, whose length, height and width were each 10 cubits: (1) Its six sides were, therefore, bounded by 12 lines, each of the same length—10 cubits. 12 x 10 =120. 120 x 120 x 10 =144,000. (2) Each one of its six sides had two diagonals, hence 12 in all, each one of which was 14.1421356 cubits long. 12 x 12 x 10 x 10 x 10 = 144,000. (3) If we add any two of its dimensions (height, 10; and length, 10; or width, 10) the sum would be 20 cubits. 20x18x10x10x10 ÷ (5 x 5)=144,000. There are three ways of using these dimensions—length and height, length and width, and height and width, with the same result as in (3). Hence we will number our next point (6). (6) As we have seen, each of the most holy's 12 surface diagonals was 14.1421356, whose square is 200 square cubits. 200 x 18 x 10 x 10 x 10 ÷ (5 x 5) =144,000. The 12 diagonals give us this point 12 times;

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hence the next point will be numbered (18). (18) Since the sum of the squares of 10 and any of the most holy's side diagonals gives us the square of the solid diagonal, the solid diagonal is thus found to be 17.32051 [10 x 10 (100) + 14.1421356 x 14.1421356 (200) = 300, whose square root is 17.32051]. There are four of such solid diagonals in the most holy. 4 x 300 = 1,200 square cubits. 1,200 x 1,200 ÷ 10 = 144,000. (19) Each of its circuits was 40 cubits. 40 x 18 x 10 x 10 x 10 ÷ 5 = 144,000. Since there are three circuits in a cube the next point will be called (22). (22) Based upon the same fact that its circuit is 40 cubits, these three circuits present the following: 3 x 40 = 120. 120 x 120 x 10 = 144,000. (23) If we add any two of these circuits, we get 80 cubits (40+40=80). 80 x 18 x 10 x 10= 144,000. There being three of such circuit sums possible, we number our next (26). A perfect cube, the most holy, represents the Christ, who is represented in Rev. 21: 16 as a perfect cube of 12,000 cubits height, length and width respectively. In each perfect cube there are 12 lines, or edges. Taking the 12 lines of the most holy each one to represent 12,000, as this is the case in the symbolic cube, New Jerusalem of. Rev. 21: 16; compare with 7: 4-8, we get 12 x 12,000 = 144,000.

(47) Now for some of the most holy's symbolizations that the Christ will bless the world during the Millennium: (1) Each of its three circuits is 40 cubits, which is 1,000 inches (40 x 25=1,000). This appearing three times, we indicate our next point as (4). (4) Each of the six perimeters of the six sides of the most holy is 40 cubits (10+10+10+10 = 40), or (40 x 25) 1,000 inches. This occurring six times, we will number our next point (10). (10) The area of each of its six sides is 100 square cubits (10 x 10). 100 x 10 =1,000. This occurring six times, we will number our next point (16). (16) The cubic contents of the most holy (10 x 10 x 10) are 1,000 cubic cubits. (17) The square of each one of its six sides' diagonals is 200. 200 x 5 =1,000. This occurring 12 times, we number our

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next point (29). (29) Any two of the main dimensions (length, height and width) added (10 + 10) yield 20 cubits. 20 x 10 x 5 = 1,000. This occurs three times, making 31 points in all. God has embedded His name Jehovah in the most holy, as the Architect of this structure, which we can recognize by various uses of its length, height and width: (1) The length is 10 cubits (Yod). The product of its length and height (10 x 10 = 100) divided by the sum of its length and height (10 + 10 = 20) = 5(He). The sum of its height, length and width (30) multiplied by the sum of its height and width (10 + 10 = 20) and divided by the product of its height and width (10 x 10 =100) gives us (30 x 20 ÷ 100) 6(Vav). And the product of the length and width (10 x 10 =100) divided by the sum of the height and width (10+10 =20) gives us (100 ÷ 20) 5(He). To simplify matters let us abbreviate the word length by the letter L, the word height by the letter H and the word width by the letter W. We can make four substitutions for the second process of point (1): Lx H ÷ (L+H). This will give us, combined with the other three processes as they stand, by which we got (1), four other ways of finding the name Jehovah. The first of these is L x W ÷ (L+H) = 5 (He). The second of these is H x W ÷ (L+H) = 5 (He). The third of these is H x W ÷ (H+W) = 5 (He). The fourth of these is LxW ÷ (L+W) = 5 (He). This will make us number our next point (6). It will be noted that the fourth, or last, process of (1) was LxW ÷ (H+W) = 5. For this we can also make a substitution, as follows: (6) L x H (10 x 10 = 100) ÷ (H+W) (10+10 = 20) = 5. Then (7) L x H ÷(L+W) = 5. So also (8) HxW ÷ (L+W) = 5. These seven substitutions do not repeat the same combination. In the most holy there are, accordingly, 26 symbolizations of the Christ as having 144,000 members, 31 of the Millennium as the 1,000 years of the Christ blessing the race, 8 of God's name, 65 in all.

(48) We will now proceed to the study of Ex. 37. It will be noted that only our Lord (Bezaleel) and

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the star-members (Aholiab) worked on the antitypical Tabernacle's and Court's furniture and their appurtenances (Ex. 37: 1-38: 23); but Jesus, the star-members, their special helpers and other assistants worked on other features of the Tabernacle and Court, especially the antitypical curtains and coverings. Or to put it in other ways, all the more important things have been done by Jesus and the star-members alone, while in the less important things the others helped them, while Jesus and the star-members did the main parts of these less important things (36: 8-38). From the fact that the mercy seat and its cherubim represent God in His attributes of justice, power and love and from the fact that the chest of the ark represents the Christ beyond the veil, and since God and His attributes have always been, and since God has been resurrecting the Christ in the Divine nature, we construe that building the ark by Bezaleel and Aholiab does not type Jesus and the star-members creating God in His justice, power and love and raising the Head and Body to the Divine nature, but types their giving the pertinent teachings of God in such attributes and of the Christ beyond the veil, while their making the other antitypical furniture and its appurtenances means their developing these antitypes and setting forth the pertinent teachings. It might be further remarked that the wood used in constructing various pieces of furniture is not typical, the type being in the golden or copper coverings, the wood being used doubtless to lighten the weight of the various pieces of furniture (v. 2). The rings and staves having already been given in antitype, it will be unnecessary to repeat them whenever hereafter mentioned. Let us now note the way that the Lord in the ark symbolized the Christ as being 144,000 members, the Millennium 1,000 years in which the Christ blesses the race and the name of God.

(49) The length of the ark was 2½ cubits (v. 1), or (2½x 25) 62½ inches long: (1) 2.5x18x10x10x10x10x10x10x10 ÷ (5x5x5x5x5) = 144,000. (2) 62.5x18x

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10x10x10x10x10x10x10 ÷ (5x5x5x5x5x5x5) = 144,000. (3) The long circuit was 8 cubits (2½+1½+2½+1½ =8), or (8x25) 200 inches. 8x18x10x10x10 = 144,000. (4) 200x18x10x10x10 ÷ (5x5) = 144,000. This is based upon measuring across top and bottom and the two heights; but measuring along the four sides gives us the same figures: 8 cubits, or 200 inches. Hence we get thereby two more points: (5) and (6). (7) The perimeters of the two long sides (2½+1½+2½+1½=8; of both=16) are 16 cubits. The perimeter of each end (4x1½) is 6 cubits; hence for both ends is 12 cubits. The perimeter of the bottom (2½+ 1½+2½+1½) is 8 cubits. The total of the perimeters (16+ 12+8) =36 cubits. This is viewed from the standpoint of an open box, hence the perimeter of the top is not included in these figures. 36x10x10x10x10x10 ÷ (5x5) =144,000. (8) Considered as a bottomless box, the perimeter involved will also total 36 cubits, and the figures just given will give us another symbolism. (9) The square of the end diagonal is (the square of each of the two short sides of the involved right-angled triangle is 1.5x1.5, or 2.25, and the sum of the squares of both its short sides would be 2.25+ 2.25=4.5) 4.5 square cubits, which is the square of each end's diagonal, since the diagonal is the hypotenuse of a right-angled triangle. There are four diagonals in the ark's two ends, hence 4x4.5=18, the main key number. 18x10x10x10x10x10x10 ÷ (5x5x5) = 144,000. (10) The perimeter of each long side of the ark is 8 cubits (2½+ l½+ 2½+l½ = 8), or 200 inches. 200x18x10x10x10 ÷ (5x5)=144,000. This is also true of the perimeters of the other three sides, hence we number our next point (14). (14) The half of the perimeter of any one of the four long sides of the ark is (2½+1½) 4 cubits. 4 x 18 x 10 x 10 x 10 x 10 ÷ 5=144,000. This occurs two times in each of the four long sides; hence the next point is numbered (22). (22) The short perimeter is 4x1½, or 6 cubits, whose square (6x6) is 36 square cubits. 36x10x10x10x10x10 ÷ (5x5) = 144,000.

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This short perimeter appears in the two ends, yielding points (23) and (24).

(50) Now will be pointed out how the ark symbolizes that the Christ will bless the world in the Millennium: (1) The long circuit, as pointed out above, was 8 cubits, or 200 inches. 200x5=1,000. This occurs twice, as shown in the preceding paragraph. (3) As seen above, a half of each long perimeter is 4 cubits, or 100 inches. 100x10=1,000. This appears eight times in the four long perimeters, hence there are at least ten features in the ark symbolizing that the Christ will bless the world during the Millennium. The dimensions of the ark in its relation to the dimensions of the most holy enable us to find the name of God, Jehovah, in combinations of these sets of dimensions: (a) By multiplying the length of the ark by the length of the most holy (2½x10) we get 25 square cubits. (b) By multiplying the width of the ark by the width of the most holy (1½x10) we get 15 square cubits. (c) By multiplying the height of the ark by the height of the most holy (1½x10) we get 15 square cubits. Using the letters a, b and c to stand respectively for these three sets of cubit operations as indicated in connection with them, and in certain cases using the sum of the key numbers 10 and 5, we get the following: (1) a-b(25–15) = 10(Yod). b+c-a (15+15–25) = 5 (He). b+c÷5([15+15]÷5) = 6(Vav). a÷5(25÷5) = 5(He). It will be noted that the results found in the operations used in a, b, and c are three sides of the four sides of each of the four perimeters of the ark multiplied by the length, height and width of the most holy respectively. But there are three other long sides of the ark of the same dimensions, which will enable us also to find the name Jehovah by the same calculation, once in each of these three sides. Hence the name Jehovah occurs 4 times in the ark. Thus the ark has 38 symbolizations (24+10+4).

(51) A number of the ark's (vs. 6-9) symbolisms having been given in (49) and (50), they will not be

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treated here. The rest of the antitypes of the mercy seat and its cherubim having already been explained, they will not be repeated here. But some may ask, How de we know that the cherubim represent power and love? We give the following reply to this question: In Heb. 9: 5 they are called the cherubim[s] of glory. The word glory applied to God usually refers to His character (2 Cor. 3: 18) of wisdom, power, justice and love. Since Christ, who is our Mercy Seat (Rom. 3: 25), as the righteousness (justice) of God (Rom. 3: 22, 25, 26), i.e., the righteousness that God provided for us, the mercy seat represents God's justice, which is also manifest from the Blood of atonement being sprinkled thereon. The Shekinah represents God Himself and the light shining out of it represents His wisdom, through which He, the antitypical Shekinah, is the Light of the universe. Thus this process of elimination proves that the cherubim represent His power and love. We will now study the table of shewbread (vs. 10-16), insofar as not hitherto studied. The dimensions of the table as given in v. 10 were: length, 2 cubits; height, 1 ½ cubits; width, 1 cubit. The table's length (2x25) was 50 inches. (1) 50x18x10x10x10x10x10÷(5x5x5x5)=144,000. This length appears four times, hence we will number the next point (5). Its width in inches (1x25) was 25 inches. (5) 25x18x10x 10x10x10x10x10 ÷ (5x5x5x5x5) = 144,000. This width appears four times: twice on the top and twice on the bottom; hence the next point will be numbered (9). The top diagonal, like all the table's other diagonals, was the hypotenuse of a right-angled triangle, the sum of the squares of whose short sides equals the square of the hypotenuse, and measured 2.2361 cubits (2x2 = 4; 1x1 = 1; and 4+1=5, whose square root is 2.2361). (9) 2.2361x2.2361(5)x18x10x10x10x10x10x10 ÷ (5x5x5x5) =144,000. It had four such diagonals, all of the same length. Hence the next point will be numbered (13). The square of the side diagonal [1½x1½ (2.25) + 2x2(4)] is 6.25 square cubits, whose square root is