Difference between revisions of "1991 AHSME Problems/Problem 14"
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(A) <math>200</math> (B) <math>201</math> (C) <math>202</math> (D) <math>203</math> (E) <math>204</math> | (A) <math>200</math> (B) <math>201</math> (C) <math>202</math> (D) <math>203</math> (E) <math>204</math> | ||
| − | == Solution == | + | == Solution 1: Number Sense== |
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
Notice that if <math>x</math> is expressed in the form <math>a^b</math>, then the number of positive divisors of <math>x^3</math> is <math>3b+1</math>. Checking through all the answer choices, the only one that is in the form <math>3b+1</math> is <math>\boxed{\textbf{(C) } 202}</math>. | Notice that if <math>x</math> is expressed in the form <math>a^b</math>, then the number of positive divisors of <math>x^3</math> is <math>3b+1</math>. Checking through all the answer choices, the only one that is in the form <math>3b+1</math> is <math>\boxed{\textbf{(C) } 202}</math>. | ||
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| + | ==Solution 2: Answer Choices== | ||
| + | Solution by e_power_pi_times_i | ||
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| + | |||
| + | Since the divisors are from <math>x^3</math>, then the answer must be something in <math>(mod 3)</math>. Since <math>200</math> and <math>203</math> are the same <math>(mod 3)</math>, as well as <math>201</math> and <math>204</math>, <math>\boxed{\textbf{(C) } 202}</math> is the only answer left. | ||
== See also == | == See also == | ||
Revision as of 11:56, 13 December 2016
Problem
If 
is the cube of a positive integer and 
is the number of positive integers that are divisors of , then could be
(A) 
(B) 
(C) 
(D) 
(E) 
Solution 1: Number Sense
Solution by e_power_pi_times_i
Notice that if is expressed in the form 
, then the number of positive divisors of 
is 
. Checking through all the answer choices, the only one that is in the form is 
.
Solution 2: Answer Choices
Solution by e_power_pi_times_i
Since the divisors are from , then the answer must be something in 
. Since and are the same , as well as and , is the only answer left.
See also
| 1991 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
