Difference between revisions of "2005 PMWC Problems/Problem I15"
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<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
− | 6 && A | + | 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, and , is divisible by . What is the largest possible product of and ?
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, , of the digits of , satisfies . The only reasonable values for .
- . It quickly becomes apparent that , which gives us .
- . Suppose . Then gives either , and we carry over the one to . So the sum of the digits of must add up to , which quickly shows us that this isn't possible.
- Hence, . Greedy algorithm: if , then , so the sum of the digits of must be 12. So . The other possible pairs are .
Quickly taking the product of these, we find that gives us the largest product of .
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 |