Quandaries
and Queries 

Hi My name is Richard and I am a high school student interested in understand this problem: The sum of the digits was calculated for the number 2^{100}, then the sum of the digits was calculated for the resulting number and so on, until a single digit is left. Which digit is this?. My teacher said that the result was 7 and we should use remainders modulo 9. Although I asked for it to classmates I do not understand it yet. Pease can you help me. 

Hi Richard, I expect that you know that both an integer and the sum of its digits, have the same remainder when divided by 9. There is an argument to show this is true in the answer to a previous question. The development there is for the case where the remainder is zero, but argument is the same regardless of the remainder. For example
These numbers, 2^{20}, 31 and 4 all have the same remainder when divided by 9. The question then is how can you determine that the remainder when dividing 2^{20} by 9 is 4, without expanding 2^{20}? I want to illustrate with a different example. I want to use 7^{107} rather than 2^{100}, so to restate the question
Equivalently
The technique is to expand 7^{107} but to do so modulo 9.
Thus if I continue with higher powers of 7 and do the arithmetic modulo 9, then every time I have three 7's the product is 1 (modulo 9). 107 = 3 35 + 2 and hence if I continue expanding the powers of seven 107 times I get one 35 times and then two more sevens. That is, doing the arithmetic modulo 7
In the second line there are 35 groups of 3 sevens and the two more sevens. Thus if 7^{107} is divided by 9 the remainder is 4. Penny 

