__153 (une cent cinquante-trois)__Human wisdom is surpassed by the Holy Spirit.

How did the unlearned 2000 years ago know the mystery of 153? It needs knowledge about modular arithmetic.

Who actually wrote the Gospel of Christ Jesus? Or did Simon Peter know the mystery of 153 when he caught 153 fish following an instruction of Christ Jesus?

**SECTION I: 153 Makes Japan Holy**

Maybe, the number 153 is a holy number.

Joh 21:11 Simon Peter went up, and drew the net to land full of great fishes, an hundred and fifty and three: and for all there were so many, yet was not the net broken.

It is so, since 153 cab be disassembled as follows:

153 = 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17

It is also mysterious in the following way:

153 = (1 x 1 x 1) + (5 x 5 x 5) + (3 x 3 x 3) = 1 + 125 + 27 = 153

This type of equation can be only realized with four numbers 153, 370, 371, and 407, if between 100 and 999.

370 = (3 x 3 x 3) + (7 x 7 x 7) + (0 x 0 x 0) = 27 + 343 = 370

371 = (3 x 3 x 3) + (7 x 7 x 7) + (1 x 1 x 1) = 27 + 343 + 1 = 371

407 = (4 x 4 x 4) + (0 x 0 x 0) + (7 x 7 x 7) = 64 + 0 + 343 = 407

No other numbers satisfy this scheme of calculation. For example,

2592 =/= (2 x 2 x 2) + (5 x 5 x 5) + (9 x 9 x 9) + (2 x 2 x 2) = 8 + 125 + 729 + 8 = 870

Incidentally, situated on 153 degrees of east longitude is the eastmost territory of Japan: Minami Torishima Island of Ogasawara Village, Tokyo Prefecture,

So, with the above description of the Bible, Japan is prophesied and identified as situated at the easternmost point covered by Christ Jesus' glory.

**SECTION II: Miracle of 153**

Take any multiple of 3, say, 2592.

2592 =/= (2 x 2 x 2) + (5 x 5 x 5) + (9 x 9 x 9) + (2 x 2 x 2) = 8 + 125 + 729 + 8 =

870 =/= (8 x 8 x 8) + (7 x 7 x 7) + (0 x 0 x 0) = 512 + 343 + 0 =

855 =/= (8 x 8 x 8) + (5 x 5 x 5) + (5 x 5 x 5) = 512 + 125 + 125 =

762 =/= (7 x 7 x 7) + (6 x 6 x 6) + (2 x 2 x 2) = 343 + 216 + 8 =

567 =/= (5 x 5 x 5) + (6 x 6 x 6) + (7 x 7 x 7) = 125 + 216 + 343 =

684 =/= (6 x 6 x 6) + (8 x 8 x 8) + (4 x 4 x 4) = 216 + 512 + 64 =

792 =/= (7 x 7 x 7) + (9 x 9 x 9) + (2 x 2 x 2) = 343 + 729 + 8 =

1080 =/= (1 x 1 x 1) + (0 x 0 x 0) + (8 x 8 x 8) + (0 x 0 x 0) = 1 + 512 =

513 =/= (5 x 5 x 5) + (1 x 1 x 1) + (3 x 3 x 3) = 125 + 1 + 27 =

**153**

So, 153 is holy to any multiples of 3, since they all can be translated into 153.

Catch of the Day (153 Fishes)

A slightly more obscure property of 153 is that it equals the sum

of the cubes of its decimal digits. In fact, if we take ANY integer

multiple of 3, and add up the cubes of its decimal digits, then take

the result and sum the cubes of its digits, and so on, we invariably

end up with 153. For example, since the number 4713 is a multiple of

3, we can reach 153 by iteratively summing the cubes of the digits, as

follows:

starting number = 4713

4^3 + 7^3 + 1^3 + 3^3 = 435

4^3 + 3^3 + 5^3 = 216

2^3 + 1^3 + 6^3 = 225

2^3 + 2^3 + 5^3 = 141

1^3 + 4^3 + 1^3 = 66

6^3 + 6^3 = 432

4^3 + 3^3 + 2^3 = 99

9^3 + 9^3 = 1458

1^3 + 4^3 + 5^3 + 8^3 = 702

7^3 + 2^3 = 351

3^3 + 5^3 + 1^3 = 153 *------

The fact that this works for any multiple of 3 is easy to prove. First, recall that any integer n is congruent modulo 3 to the sum of its decimaldigits (because the base 10 is congruent to 1 modulo 3). Then, letting f(n) denote the sum of the cubes of the decimal digits of n, by Fermat's little theorem it follows that f(n) is congruent to n modulo 3. Also, we can easily see that f(n) is less than n for all n greater than 1999. Hence, beginning with any multiple of 3, and iterating the function f(n), we must arrive at a multiple of 3 that is less than 1999. We can then show by inspection that every one of these reduces to 153.

http://www.mathpages.com/home/kmath463.htm

Note: Fermat's little theorem

When p is a primary number and a has no common divisors with p,

a^(p-1) = 1 mod p.

If p = 3 and a =5, 5 to the 2nd power (= 25) leaves 1 when divided by 3 with a quotient 8.

This theory is applied to an encryption theory used on the Internet and PCs.

*** *** *** ***

Christ Jesus fasted for forty days in the wilderness, and was tempted by the devil (Matthew 4:1-2, Mark 1:12-13, Luke 4:1-2)

Christ Jesus was also in the tomb for about 40 hours.

And, after the Resurrection, Christ Jesus guided His followers for 40 days and then ascended to Heaven.

**Act 7:30 And when forty years were expired, there appeared to him in the wilderness of mount Sina an angel of the Lord in a flame of fire in a bush.**

Act 7:31 When Moses saw it, he wondered at the sight: and as he drew near to behold it, the voice of the LORD came unto him,

Act 7:32 Saying, I am the God of thy fathers, the God of Abraham, and the God of Isaac, and the God of Jacob. Then Moses trembled, and durst not behold.

Act 7:33 Then said the Lord to him, Put off thy shoes from thy feet: for the place where thou standest is holy ground.

Act 7:31 When Moses saw it, he wondered at the sight: and as he drew near to behold it, the voice of the LORD came unto him,

Act 7:32 Saying, I am the God of thy fathers, the God of Abraham, and the God of Isaac, and the God of Jacob. Then Moses trembled, and durst not behold.

Act 7:33 Then said the Lord to him, Put off thy shoes from thy feet: for the place where thou standest is holy ground.