A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A isimmediately after the digits of A, then prove that C is a multiple of 9.
twice the sum of the digits of B. If C is the number formed by writing the digits of B
So you're saying that C is a concatenation of BA whereby A = 7B and the sum of the digits of A is twice the sum of the digits of B.
You would have to express A and B in terms of series and you can use mathematical induction to prove. Have you gotten that
(I assume these numbers are written base ten.)
The sum of a number's digits is congruent to the number, mod 9.
. A similar result holds for B or any other positive integer.
Also, we have that , so, .
Adding these gives: .
Since 2 is relatively prime to 9, .
Previously, we saw that . Therefore, , so that
C is the number formed by writing the digits of B immediately after the digits of A, so .
Therefore, C is divisible by 9.
Another approach . . . a bit fuzzy, though.
Fact: If the digit-sum of a number is 0 or 9, the number is a multiple of 9.
Since , then: .
But we are told that: .
. . Hence: .
. . Then: .
That is, both and are multiples of 9.
Therefore, is a multiple of 9.