Difference between revisions of "2018 AMC 10B Problems/Problem 21"
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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 1== | + | ==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 (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 | + | ==Solution 4 (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 06:22, 19 October 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 (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 (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 4 (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.