Difference between revisions of "2005 PMWC Problems/Problem I15"
(sol) |
m (→Solution: typo) |
||
| Line 6: | Line 6: | ||
<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
| − | 6 && A | + | 6 && A\\ |
+ B && 3 | + B && 3 | ||
\end{eqnarray*}</cmath> | \end{eqnarray*}</cmath> | ||
| Line 12: | Line 12: | ||
[[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.
. 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 
.
. 
, 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 | ||
