Difference between revisions of "2005 PMWC Problems/Problem I15"

(sol)
 
m (Solution: typo)
 
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<cmath>\begin{eqnarray*}
 
<cmath>\begin{eqnarray*}
6 && A
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6 && A\\
 
+ B && 3  
 
+ B && 3  
 
\end{eqnarray*}</cmath>
 
\end{eqnarray*}</cmath>
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[[Casework]]:
 
[[Casework]]:
  
*<math>s = 3</math>. It quickly becomes apparent that <math>c = 111</math>, which gives us <math>A = 8, B = 4</math>.
+
*<math>s = 3</math>. It quickly becomes apparent that <math>c = 111</math>, which gives us <math>A = 8, B = 4 (8,4)</math>.
 
*<math>s = 12</math>. Suppose <math>A \ge 7</math>. Then <math>A + 3</math> gives either <math>0, 1, 2</math>, and we carry over the one to <math>6 + B</math>. So the sum of the digits of <math>7 + B</math> must add up to <math>> 10</math>, which quickly shows us that this isn't possible.
 
*<math>s = 12</math>. Suppose <math>A \ge 7</math>. Then <math>A + 3</math> gives either <math>0, 1, 2</math>, and we carry over the one to <math>6 + B</math>. So the sum of the digits of <math>7 + B</math> must add up to <math>> 10</math>, which quickly shows us that this isn't possible.
 
:Hence, <math>A < 7</math>. [[Greedy algorithm]]: if <math>A = 6</math>, then <math>A + 3 = 9</math>, so the sum of the digits of <math>B + 6</math> must be 12. So <math>B = 6 \longrightarrow (6,6)</math>. The other possible pairs are <math>(5,7)(4,8)(3,9)(2,1)(1,2)</math>.
 
:Hence, <math>A < 7</math>. [[Greedy algorithm]]: if <math>A = 6</math>, then <math>A + 3 = 9</math>, so the sum of the digits of <math>B + 6</math> must be 12. So <math>B = 6 \longrightarrow (6,6)</math>. The other possible pairs are <math>(5,7)(4,8)(3,9)(2,1)(1,2)</math>.
  
Quickly taking the product of these, we find that <math>(6,6) \Longrightarrow 36</math> gives us the largest product of <math>AB</math>.  
+
Quickly taking the product of these, we find that <math>(6,6) \Longrightarrow 36</math> gives us the largest product of <math>AB</math>.
  
 
== See also ==
 
== See also ==
 
{{PMWC box|year=2005|num-b=I14|num-a=T1}}
 
{{PMWC box|year=2005|num-b=I14|num-a=T1}}

Latest revision as of 17:23, 2 October 2007

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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