2005 PMWC Problems/Problem I15

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Problem

The sum of the two three-digit integers, $\text{6A2}$ and $\text{B34}$, is divisible by $18$. What is the largest possible product of $\text{A}$ and $\text{B}$?

Solution

A number is divisible by 18 iff it is divisible by 2 and 9. Divisibility by 2 is already satisfied, so we need the number to be divisible by 9; the divisibility rule for 9 states that we only need the sum of the digits to be divisible by 9. The units digit is 6; so the sum, $s$, of the digits of $c = (6+b) \cdot 10 + (A+3)$, satisfies $s \equiv 3 \pmod{9}$. The only reasonable values for $s = 3, 12$.

\begin{eqnarray*} 6 && A\\ + B && 3  \end{eqnarray*}

Casework:

  • $s = 3$. It quickly becomes apparent that $c = 111$, which gives us $A = 8, B = 4 (8,4)$.
  • $s = 12$. Suppose $A \ge 7$. Then $A + 3$ gives either $0, 1, 2$, and we carry over the one to $6 + B$. So the sum of the digits of $7 + B$ must add up to $> 10$, which quickly shows us that this isn't possible.
Hence, $A < 7$. Greedy algorithm: if $A = 6$, then $A + 3 = 9$, so the sum of the digits of $B + 6$ must be 12. So $B = 6 \longrightarrow (6,6)$. The other possible pairs are $(5,7)(4,8)(3,9)(2,1)(1,2)$.

Quickly taking the product of these, we find that $(6,6) \Longrightarrow 36$ gives us the largest product of $AB$.

See also

2005 PMWC (Problems)
Preceded by
Problem I14
Followed by
Problem T1
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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