Difference between revisions of "2015 AIME II Problems/Problem 3"
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Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>. | Let <math>m</math> be the least positive integer divisible by <math>17</math> whose digits sum to <math>17</math>. Find <math>m</math>. | ||
− | ==Solution== | + | ==Solution 1== |
+ | The three-digit integers divisible by <math>17</math>, and their digit sum: <cmath> | ||
+ | \begin{array}{c|c} | ||
+ | m & s(m)\\ | ||
+ | 102 & 3 | ||
+ | 119 & 11 | ||
+ | 136 & 10 | ||
+ | 153 & 9 | ||
+ | 170 & 8 | ||
+ | 187 & 16 | ||
+ | 204 & 6 | ||
+ | 221 & 5 | ||
+ | 238 & 13 | ||
+ | 255 & 12 | ||
+ | 272 & 11 | ||
+ | 289 & 19 | ||
+ | 306 & 9 | ||
+ | 323 & 8 | ||
+ | 340 & 7 | ||
+ | 357 & 15 | ||
+ | 374 & 14 | ||
+ | 391 & 13 | ||
+ | 408 & 12 | ||
+ | 425 & 11 | ||
+ | 442 & 10 | ||
+ | 459 & 18 | ||
+ | 476 & 17 | ||
+ | \end{array} | ||
+ | </cmath> | ||
+ | |||
+ | Thus the answer is <math>\boxed{476}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2015|n=II|num-b=2|num-a=4}} {{MAA Notice}} | {{AIME box|year=2015|n=II|num-b=2|num-a=4}} {{MAA Notice}} |
Revision as of 13:00, 26 March 2015
Problem
Let be the least positive integer divisible by whose digits sum to . Find .
Solution 1
The three-digit integers divisible by , and their digit sum:
\[\begin{array}{c|c} m & s(m)\\ 102 & 3 119 & 11 136 & 10 153 & 9 170 & 8 187 & 16 204 & 6 221 & 5 238 & 13 255 & 12 272 & 11 289 & 19 306 & 9 323 & 8 340 & 7 357 & 15 374 & 14 391 & 13 408 & 12 425 & 11 442 & 10 459 & 18 476 & 17 \end{array}\] (Error compiling LaTeX. Unknown error_msg)
Thus the answer is .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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