Difference between revisions of "2018 AMC 10A Problems/Problem 22"
Williamgolly (talk | contribs) (→Solution 4 (Fastest)) |
Skyscraper (talk | contribs) (→Solution 4 (Fastest)) |
||
Line 47: | Line 47: | ||
==Solution 4 (Fastest)== | ==Solution 4 (Fastest)== | ||
− | Notice that <math>gcd (a,b,c,d)=gcd(gcd(a,b),gcd(b,c),gcd(c,d))=gcd(24,36,54)=6</math>, so <math>gcd(d,a)</math> must be a multiple of <math>6</math>. The only answer choice that gives a value between <math>70</math> and <math>100</math> when multiplied by 6 is <math>\boxed{\textbf{(D) } 13}</math>. - mathleticguyyy | + | Notice that <math>gcd (a,b,c,d)=gcd(gcd(a,b),gcd(b,c),gcd(c,d))=gcd(24,36,54)=6</math>, so <math>gcd(d,a)</math> must be a multiple of <math>6</math>. The only answer choice that gives a value between <math>70</math> and <math>100</math> when multiplied by 6 is <math>\boxed{\textbf{(D) } 13}</math>. - mathleticguyyy + einstein |
In the case where there can be 2 possible answers, we can do casework on gcd(d,a) | In the case where there can be 2 possible answers, we can do casework on gcd(d,a) |
Revision as of 13:17, 6 April 2020
Contents
Problem
Let and be positive integers such that , , , and . Which of the following must be a divisor of ?
Solution 1
The GCD information tells us that divides , both and divide , both and divide , and divides . Note that we have the prime factorizations:
Hence we have for some positive integers . Now if divdes , then would be at least which is too large, hence does not divide . Similarly, if divides , then would be at least which is too large, so does not divide . Therefore, where neither nor divide . In other words, is divisible only by primes that are at least . The only possible value of between and and which fits this criterion is , so the answer is .
Solution 2
We can say that and 'have' , that and have , and that and have . Combining and yields has (at a minimum) , and thus has (and no more powers of because otherwise would be different). In addition, has , and thus has (similar to , we see that cannot have any other powers of ). We now assume the simplest scenario, where and . According to this base case, we have . We want an extra factor between the two such that this number is between and , and this new factor cannot be divisible by or . Checking through, we see that is the only one that works. Therefore the answer is
Solution by JohnHankock
Solution 2.1
Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to because doing so would later the of and . This is why:
The is and the of is . However, the of (meaning both are divisible by 36). Therefore, is only divisible by (and no higher power of 3), while is divisible by only (and no higher power of 2).
Thus, the of can be expressed in the form for which is a number not divisible by or . The only answer choice that satisfies this (and the other condition) is .
Solution 3 (Better notation)
First off, note that , , and are all of the form . The prime factorizations are , and , respectively. Now, let and be the number of times and go into ,respectively. Define , , , and similiarly. Now, translate the s into the following: .
(Unfinished) ~Rowechen Zhong
Solution 4 (Fastest)
Notice that , so must be a multiple of . The only answer choice that gives a value between and when multiplied by 6 is . - mathleticguyyy + einstein
In the case where there can be 2 possible answers, we can do casework on gcd(d,a) ~Williamgolly
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.