Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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==Solution 4 (Using the answer choices)== | ==Solution 4 (Using the answer choices)== | ||
Note that 323 multiplied by any of the answer choices results in a 5 or 6 digit number. So, we need a choice that shares a factor with 323. The prime factorization of 323 is <math>17\cdot19</math>, and since we want the minimum choice, we are looking for an even answer with a factor of 17. We see 340 achieves this and is the smallest to do so (646 being the other). So, we get <math>\boxed{\text{(C) }340}</math> -- OGBooger | Note that 323 multiplied by any of the answer choices results in a 5 or 6 digit number. So, we need a choice that shares a factor with 323. The prime factorization of 323 is <math>17\cdot19</math>, and since we want the minimum choice, we are looking for an even answer with a factor of 17. We see 340 achieves this and is the smallest to do so (646 being the other). So, we get <math>\boxed{\text{(C) }340}</math> -- OGBooger | ||
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+ | ==Solution 5== | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 17:57, 9 August 2021
- The following problem is from both the 2018 AMC 12B #19 and 2018 AMC 10B #21, so both problems redirect to this page.
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
Since prime factorizing gives you
, the desired answer needs to be a multiple of
or
, this is because if it is not a multiple of
or
,
will be more than a
digit number. For example, if the answer were to instead be
,
would have to be a multiple of
for both
and
to be a valid factor, meaning
would have to be at least
, which is too big. Looking at the answer choices,
and
are both not a multiple of neither 17 nor 19,
is divisible by
.
is divisible by
, and
is divisible by both
and
. Since
is the smallest number divisible by either
or
it is the answer. Checking, we can see that
would be
, a four-digit number. Note that
is also divisible by
, one of the listed divisors of
. (If
was not divisible by
, we would need to look for a different divisor)
-Edited by Mathandski
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
.
Solution 3
Again, recognize . The 4-digit number is even, so its prime factorization must then be
. Also,
, so
. Since
, the prime factorization of the number after
needs to have either
or
. The next highest product after
is
or
.
You can also tell by inspection that , because
is closer to the side lengths of a square, which maximizes the product.
~bjhhar
Solution 4 (Using the answer choices)
Note that 323 multiplied by any of the answer choices results in a 5 or 6 digit number. So, we need a choice that shares a factor with 323. The prime factorization of 323 is , and since we want the minimum choice, we are looking for an even answer with a factor of 17. We see 340 achieves this and is the smallest to do so (646 being the other). So, we get
-- OGBooger
Solution 5
Video Solution 1
https://www.youtube.com/watch?v=qlHE_sAXiY8
Video Solution 2
https://www.youtube.com/watch?v=KHaLXNAkDWE
Video Solution 3
https://www.youtube.com/watch?v=vc1FHO9YYKQ
~bunny1
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.