A couple of challenging number theory problems!! Any ideas?

1) No. To see this, let = denote congruence.

Then each step can be generalized as

10x + y -> x + 5y.

Now suppose 10x + y = 0 (mod 7).

y = -10x = 4x (mod 7)

It follows that x + 5y = x + 5(4x) = 21x = 0 (mod 7).

That is, if we start with a number which is divisible by 7, then we'll always have a number divisible by 7. Obviously 7^2007 is divisible by 7, but

2007^7 = (-2)^7 = -128 = 5 (mod 7)

so it isn't divisible by 7.

*** Something interesting: there is at least one number that is fixed by the above process: 49 -> 4 + 5*9 = 49. It is easily seen that there are only finitely many such numbers, but are there any more?

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2) Consider it mod 100 = 4 * 25.

In mod 4, we have immediately:

2^20000009 = 0 mod 4

6^20000009 = 0 mod 4

7^1 = 3 mod 4, 7^2 = 1 mod 4, so that

7^20000009 = (7^2)^10000004 * 7 = 3 mod 4.

so the sum = 0 + 0 + 3 = 3 mod 4.

In mod 25, we have phi(25) = (5-1)*5^(2-1) = 20, so that by Euler's theorem:

2^20000009 = (2^20)^1000000 * 2^9 = 512 = 12 (mod 25)

6^20000009 = 6^9 = 21 (mod 25)

7^20000009 = 7^9 = 7 (mod 25)

so the sum = 12 + 21 + 7 = 15 (mod 25)

The sum is 3 mod 4 and 15 mod 25, which gives us 15 mod 100.

Last two digits = 15

*** It's not hard to find the last three digits in a similar fashion. Try it ... note that phi(125) = 100.

The answer should be 015.

1) Yes

2) 521