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A three digit number "ABC" is divided by the two digit number "AC". The quotient is 16 with no remainder.
What is the largest possible number "ABC"?
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From the problem, we have
100*A + 10*B + C = 16*(10A + C), (1)
1 <= A <= 9, 0 <= B, C <= 9. (2)
Equation (1) is equivalent to
100*A + 10*B + C = 160*A + 16C,
60*A - 10B + 15C = 0,
12*A - 2B + 3C = 0. (3)
From the last equation (3), we conclude that
(a) C is an even number 0, 2, 4, 6, or 8;
(b) B is a multiple of 3, so B is either 0, or, 3, or 6, or 9;
(c) 2B - 3C >= 12.
From (c), if C= 0, then admittable values for B are 6 or 9;
if C= 2, then admittable value for B is 9;
C can not be 4, or 6, or 8 ( from inequalities (c) )
So, the only possible blocks of two digits BC can be 60, 90, 92.
If BC is 60, then from (3) 12*A - 2*6 + 3*0 = 0, hence, A = 1. So, AC = 10, ABC = 160.
If BC is 90, then from (3) 12*A - 2*9 + 3*0 = 0, hence, A = 18/12 = 1.5, which is not an integer number, so this case does not work.
If BC is 92, then from (3) 12*A - 2*9 + 3*2 = 0, hence, A = 12/12 = 1. So, AC = 12, ABC = 192.
Thus, the maximum possible number ABC is 192.
ANSWER. The only maximum possible number ABC under given condition is 192.
Solved.
