Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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MRENTHUSIASM (talk | contribs) (The quick solution does not really address the condition that n is a 4-digit number. So, I try to prioritize solutions based on rigor. Hope it is ok. :)) |
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+ | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #19]] and [[2018 AMC 10B Problems|2018 AMC 10B #21]]}} | ||
+ | |||
==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, | + | 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,\ldots,\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> | ||
− | ==Solution== | + | ==Solution 1 (Inequalities)== |
+ | |||
+ | Let <math>d</math> be the next divisor written to the right of <math>323.</math> | ||
+ | |||
+ | If <math>\gcd(323,d)=1,</math> then <cmath>n\geq323d>323^2>100^2=10000,</cmath> which contradicts the precondition that <math>n</math> is a <math>4</math>-digit number. | ||
+ | |||
+ | It follows that <math>\gcd(323,d)>1.</math> Since <math>323=17\cdot19,</math> the smallest possible value of <math>d</math> is <math>17\cdot20=\boxed{\textbf{(C) } 340},</math> from which <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath> | ||
+ | ~MRENTHUSIASM ~tdeng | ||
+ | |||
+ | ==Solution 2 (Inequalities)== | ||
+ | |||
+ | Let <math>d</math> be the next divisor written to the right of <math>323.</math> | ||
+ | |||
+ | Since <math>n</math> is even and <math>323=17\cdot19,</math> we have <math>n=2\cdot17\cdot19\cdot k=646k</math> for some positive integer <math>k.</math> Moreover, since <math>1000\leq n\leq9998,</math> we get <math>2\leq k\leq15.</math> As <math>d>323,</math> it is clear that <math>d</math> must be divisible by <math>17</math> or <math>19</math> or both. | ||
+ | |||
+ | Therefore, the smallest possible value of <math>d</math> is <math>17\cdot20=\boxed{\textbf{(C) } 340},</math> from which <cmath>n=\operatorname{lcm}(323,d)=17\cdot19\cdot20=6460.</cmath> | ||
+ | ~MRENTHUSIASM ~bjhhar | ||
+ | |||
+ | ==Solution 3 (Quick)== | ||
+ | The prime factorization of <math>323</math> is <math>17 \cdot 19</math>. Our answer must be a multiple of either <math>17</math> or <math>19</math> or both. Since <math>17 < 19</math>, the next smallest divisor that is divisble by <math>17</math> would be <math>323 + 17 = \boxed{\textbf{(C) } 340}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ||
+ | |||
+ | ==Solution 4 (Answer Choices)== | ||
+ | |||
+ | Since prime factorizing <math>323</math> gives you <math>17 \cdot 19</math>, the desired answer needs to be a multiple of <math>17</math> or <math>19</math>, this is because if it is not a multiple of <math>17</math> or <math>19</math>, <math>n</math> will be more than a <math>4</math> digit number. For example, if the answer were to instead be <math>324</math>, <math>n</math> would have to be a multiple of <math>2^2\cdot3^4\cdot17\cdot19</math> for both <math>323</math> and <math>324</math> to be a valid factor, meaning <math>n</math> would have to be at least <math>104652</math>, which is too big. Looking at the answer choices, <math>\textbf{(A)}</math> and <math>\textbf{(B)}</math> are both not a multiple of neither <math>17</math> nor <math>19</math>, <math>\textbf{(C)}</math> is divisible by <math>17</math>. <math>\textbf{(D)}</math> is divisible by <math>19</math>, and <math>\textbf{(E)}</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\boxed{\textbf{(C) } 340}</math> is the smallest number divisible by either <math>17</math> or <math>19</math> it is the answer. Checking, we can see that <math>n</math> would be <math>6460</math>, a <math>4</math>-digit number. Note that <math>n</math> is also divisible by <math>2</math>, one of the listed divisors of <math>n</math>. (If <math>n</math> was not divisible by <math>2</math>, we would need to look for a different divisor.) | ||
+ | |||
+ | -Edited by Mathandski | ||
+ | |||
+ | ==Solution 5 (Answer Choices)== | ||
+ | Note that <math>323</math> multiplied by any of the answer choices results in a <math>5</math> or <math>6</math>-digit <math>n</math>. So, we need a choice that shares a factor(s) with <math>323</math>, such that the factors we'll need to add to the prime factorization of <math>n</math> (in result to adding the chosen divisor) won't cause our number to multiply to more than <math>4</math> digits. | ||
+ | The prime factorization of <math>323</math> is <math>17\cdot19</math>, and since we know <math>n</math> is even, our answer needs to be | ||
+ | |||
+ | * even | ||
+ | |||
+ | * has a factor of <math>17</math> or <math>19</math> | ||
− | + | We see <math>340</math> achieves this and is the smallest to do so (<math>646</math> being the other). So, we get <math>\boxed{\textbf{(C) } 340}</math>. | |
+ | |||
+ | ~OGBooger (Solution) | ||
+ | |||
+ | ~Pearl2008 (Minor Edits) | ||
+ | |||
+ | ==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== | ==See Also== | ||
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{{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2018|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 11:04, 17 June 2023
- 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 (Inequalities)
Let be the next divisor written to the right of
If then
which contradicts the precondition that
is a
-digit number.
It follows that Since
the smallest possible value of
is
from which
~MRENTHUSIASM ~tdeng
Solution 2 (Inequalities)
Let be the next divisor written to the right of
Since is even and
we have
for some positive integer
Moreover, since
we get
As
it is clear that
must be divisible by
or
or both.
Therefore, the smallest possible value of is
from which
~MRENTHUSIASM ~bjhhar
Solution 3 (Quick)
The prime factorization of is
. Our answer must be a multiple of either
or
or both. Since
, the next smallest divisor that is divisble by
would be
Solution 4 (Answer Choices)
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
nor
,
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
-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 5 (Answer Choices)
Note that multiplied by any of the answer choices results in a
or
-digit
. So, we need a choice that shares a factor(s) with
, such that the factors we'll need to add to the prime factorization of
(in result to adding the chosen divisor) won't cause our number to multiply to more than
digits.
The prime factorization of
is
, and since we know
is even, our answer needs to be
- even
- has a factor of
or
We see achieves this and is the smallest to do so (
being the other). So, we get
.
~OGBooger (Solution)
~Pearl2008 (Minor Edits)
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.