1. ## Problem Solving

A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
twice the sum of the digits of B. If C is the number formed by writing the digits of B
immediately after the digits of A, then prove that C is a multiple of 9.

2. Originally Posted by matgrl
A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
twice the sum of the digits of B. If C is the number formed by writing the digits of B
immediately after the digits of A, then prove that C is a multiple of 9.
"If C is the number formed by writing the digits of B immediately after the digits of A,..." Please explain this part further.

3. A= 6 B= 3 C = BA or 36 which is a multiple of 9 we need to prove this somehow

4. ## My thoughts

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
far?

5. No i havent....but that is a start

6. ## Sum of digits.

Originally Posted by matgrl
A and B are positive integers such that (i) A=7B and (ii) the sum of the digits of A is
twice the sum of the digits of B. If C is the number formed by writing the digits of B
immediately after the digits of A, then prove that C is a multiple of 9.

(I assume these numbers are written base ten.)

The sum of a number's digits is congruent to the number, mod 9.

$\text{Sum of A's digits }\equiv A (\bmod 9)$. A similar result holds for B or any other positive integer.

$\text{Sum of A's digits }\equiv 2\times (\text{Sum of B's digits })(\bmod 9)$

So, $A \equiv 2B\ (\bmod 9)$.

Also, we have that $A=7B$, so, $A \equiv 7B\ (\bmod 9)$.

Adding these gives: $A \equiv 2B\ (\bmod 9)$.

Since 2 is relatively prime to 9, $A \equiv 0 \ (\bmod 9)$.

Previously, we saw that $A \equiv 2B\ (\bmod 9)$. Therefore, $0 \equiv 2B\ (\bmod 9)$, so that $B \equiv 0 \ (\bmod 9)$

C is the number formed by writing the digits of B immediately after the digits of A, so $\text{Sum of C's digits }= (\text{Sum of A's digits })+(\text{Sum of B's digits })$.

$C \equiv A+B\ (\bmod 9) \ \ \implies\ \ C \equiv 0+0\ (\bmod 9)\ \ \implies\ \ C \equiv 0\ (\bmod 9)$.

Therefore, C is divisible by 9.

7. Hello, matgrl!

Another approach . . . a bit fuzzy, though.

$A\text{ and }B\text{ are positive integers such that (i) }A=7B$
$\text{ and (ii) the digit-sum of }A\text{ is twice the digit-sum of }B.$

$\text{If }C\text{ is the number formed by writing the digits of }B\text{ immediately}$
$\text{after the digits of }A\text{, then prove that }C\text{ is a multiple of 9.}$

Fact: If the digit-sum of a number is 0 or 9, the number is a multiple of 9.

Let: . $\begin{Bmatrix}d(A) &=& \text{digit-sum of }A \\
d(B) &=& \text{digit-sum of }B \end{Bmatrix}$

Since $A \,=\,7B$, then: . $d(A) \:=\:7\!\cdot\!d(B)$

But we are told that: . $d(A) \:=\:2\!\cdot\! d(B)$

. . Hence: . $7\!\cdot d(B) \:=\:2\!\cdot d(B) \quad\Rightarrow\quad d(B) \:=\:0$

. . Then: . $d(A) \:=\:0$

That is, both $\,A$ and $\,B$ are multiples of 9.

$\text{Since }C \:=\:\{AB\},\,\text{ then: }\,d(C) \:=\:d(A) + D(B) \:=\:0$

Therefore, $\,C$ is a multiple of 9.