Difference between revisions of "2018 AMC 10A Problems/Problem 22"
m |
|||
Line 21: | Line 21: | ||
for some positive integers <math>w,x,y,z</math>. Now if 3 divdes <math>w</math>, then <math>\gcd(a,b)</math> would be at least <math>2^3\cdot 3^2</math> which is too large, hence 3 does not divide <math>w</math>. Similarly, if 2 divides <math>z</math>, then <math>\gcd(c,d)</math> would be at least <math>2^2\cdot 3^3</math> which is too large, so 2 does not divide <math>z</math>. Therefore, | for some positive integers <math>w,x,y,z</math>. Now if 3 divdes <math>w</math>, then <math>\gcd(a,b)</math> would be at least <math>2^3\cdot 3^2</math> which is too large, hence 3 does not divide <math>w</math>. Similarly, if 2 divides <math>z</math>, then <math>\gcd(c,d)</math> would be at least <math>2^2\cdot 3^3</math> which is too large, so 2 does not divide <math>z</math>. Therefore, | ||
<cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath> | <cmath>\gcd(a,d)=2\cdot 3\cdot \gcd(w,z)</cmath> | ||
− | where neither 2 nor 3 divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> | + | where neither 2 nor 3 divide <math>\gcd(w,z)</math>. In other words, <math>\gcd(w,z)</math> is divisible only by primes that are at least 5. The only number of this form between 70 and 100 is <math>78=2\cdot3\cdot13</math>, so the answer is <math>\boxed{\textbf{(D) }13}</math>. |
== Solution 2 == | == Solution 2 == |
Revision as of 00:36, 11 December 2018
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 24 divides , both 24 and 36 divide , both 36 and 54 divide , and 54 divides . Note that we have the prime factorizations:
Hence we can write for some positive integers . Now if 3 divdes , then would be at least which is too large, hence 3 does not divide . Similarly, if 2 divides , then would be at least which is too large, so 2 does not divide . Therefore, where neither 2 nor 3 divide . In other words, is divisible only by primes that are at least 5. The only number of this form between 70 and 100 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
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.