Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math> | + | Mary chose an even <math>4</math>-digit number <math>n</math>. She wrote down all the divisors of <math>n</math> in increasing order from left to right: <math>1,2,...,\dfrac{n}{2},n</math>. At some moment Mary wrote <math>323</math> as a divisor of <math>n</math>. What is the smallest possible value of the next divisor written to the right of <math>323</math>? |
<math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math> | <math>\textbf{(A) } 324 \qquad \textbf{(B) } 330 \qquad \textbf{(C) } 340 \qquad \textbf{(D) } 361 \qquad \textbf{(E) } 646</math> |
Revision as of 14:37, 18 February 2018
Contents
Problem
Mary chose an even -digit number . She wrote down all the divisors of in increasing order from left to right: . At some moment Mary wrote as a divisor of . What is the smallest possible value of the next divisor written to the right of ?
Solution 1
Prime factorizing gives you . Looking at the answer choices, is the smallest number divisible by or .
Solution 2
Let the next largest divisor be . Suppose . Then, as , therefore, However, because , . Therefore, . Note that . Therefore, the smallest the gcd can be is and our answer is .
-tdeng
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.