Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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==Solution 1== | ==Solution 1== | ||
− | Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. Looking at the answer choices, <math>\fbox{( | + | Prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>. Looking at the answer choices, <math>\fbox{(C) 340}</math> is the smallest number divisible by <math>17</math> or <math>19</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 16:47, 16 February 2018
Contents
[hide]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 |
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