Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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− | Let the next largest divisor be <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the gcd can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{( | + | Let the next largest divisor be <math>k</math>. Suppose <math>\gcd(k,323)=1</math>. Then, as <math>323|n, k|n</math>, therefore, <math>323\cdot k|n.</math> However, because <math>k>323</math>, <math>323k>323\cdot 324>9999</math>. Therefore, <math>\gcd(k,323)>1</math>. Note that <math>323=17\cdot 19</math>. Therefore, the smallest the gcd can be is <math>17</math> and our answer is <math>323+17=\boxed{\text{(C) }340}</math>. |
-tdeng | -tdeng |
Revision as of 17:47, 16 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.