The number N, represented by the decimal numeral ABC, is divided by the decimal numeral AC. The quotient is 9, and the remainder is 0. What is the maximum possible value for N?
+121 The maximum possible value for N is 405, in which case AC would have to be 45.
The way I reasoned went like this:
When ABC is divided by AC the quotient is 9 and the remainder 0, so
ABC = 9(AC).
This makes ABC divisible evenly by both 9 and AC. Also, expanding the numbers we get
C+10B+100A =9(C+10A)=9C+90A.
Simplifying, we have,
10(A+B)= 8C, or 5(A+B)= 4C.
Now this last equation implies that A+B is a multiple of 4.
Moreover, A+B cannot be a bigger multiple than 4, since A+B=8 would result in C=10, which it cannot be.
Now since we are looking for the largest N, I set A=4 and B=0, forcing C=5, and low and behold it worked. QED.
