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]]}} | ||
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==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,...,\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> | + | 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,...,\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> | ||
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==Solution 1== | ==Solution 1== | ||
− | + | 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 * 3^4 * 17 * 19</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>\text{(A) }324</math> and <math>\text{(B) }330</math> are both not a multiple of neither 17 nor 19, <math>\text{(C) }340</math> is divisible by <math>17</math>. <math>\text{(D) }361</math> is divisible by <math>19</math>, and <math>\text{(E) }646</math> is divisible by both <math>17</math> and <math>19</math>. Since <math>\boxed{\text{(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 four-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) | |
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+ | -Edited by Mathandski | ||
==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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>. |
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+ | ==Solution 3== | ||
+ | Again, recognize <math>323=17 \cdot 19</math>. The 4-digit number is even, so its prime factorization must then be <math>17 \cdot 19 \cdot 2 \cdot n</math>. Also, <math>1000\leq 646n \leq 9998</math>, so <math>2 \leq n \leq 15</math>. Since <math>15 \cdot 2=30</math>, the prime factorization of the number after <math>323</math> needs to have either <math>17</math> or <math>19</math>. The next highest product after <math>17 \cdot 19</math> is <math>17 \cdot 2 \cdot 10 =340</math> or <math>19 \cdot 2 \cdot 9 =342</math> <math>\implies \boxed{\text{(C) }340}</math>. | ||
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+ | You can also tell by inspection that <math>19\cdot18 > 20\cdot17</math>, because <math>19\cdot18</math> is closer to the side lengths of a square, which maximizes the product. | ||
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+ | ~bjhhar | ||
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+ | ==Video Solution 1== | ||
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+ | https://www.youtube.com/watch?v=qlHE_sAXiY8 | ||
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+ | ==Video Solution 2== | ||
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+ | https://www.youtube.com/watch?v=KHaLXNAkDWE | ||
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+ | ==Video Solution 3== | ||
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+ | https://www.youtube.com/watch?v=vc1FHO9YYKQ | ||
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+ | ~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}} | ||
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+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 21:59, 31 December 2020
- 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
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.