The sum of digits of the telephone number aaabbbb equals the two-digit number ab. What is the sum $a+b?$
$(A) \space8$
$(B)\space 9 $
$(C) \space10 $
$(D)\space 11$
$(E) \space 12$
So far I have tried using the formula $3a + 4b = k$, where $k$ is made up of the digits $a, b.$ From my understanding, to get the maximum value of $a+b$. We would have to maximize the value of $k$. So by plugging in $9$ to be $a$ and $b$ we get $k = 63.$ Therefore $a + b = 9.$
I understand why this is wrong, as for some number $< 63$, there exists an $a + b $ where $a + b$ is greater than $9$. My doubt now is, how do I find this number.
Consider: $$ 3a + 3b + b = 10a + b $$ implies that $$ 10 | 3a + 3b $$ So you’re only in interested in those solutions for $a+b$ so that $3(a+b)$ is divisable by $10$. In your case this is only possible with option C. This does then imply that $$ 3\cdot 10 = 30 = a0$$ so $$ a = 3 $$ and thus $$ b = 10 - 3 = 7 $$
If we did not know these possible solutions, consider: If $3(a+b)=X$, then $10\leq X< 3\cdot 20 = 60$ and $10|X$ and $3|X$. Thus $30|X$, so the only possible value for $X$ is $30$.