Difference between revisions of "1991 AHSME Problems/Problem 14"
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If <math>x</math> is the cube of a positive integer and <math>d</math> is the number of positive integers that are divisors of <math>x</math>, then <math>d</math> could be | If <math>x</math> is the cube of a positive integer and <math>d</math> is the number of positive integers that are divisors of <math>x</math>, then <math>d</math> could be | ||
| − | (A) | + | <math>\text{(A) } 200\quad\text{(B) } 201\quad\text{(C) } 202\quad\text{(D) } 203\quad\text{(E) } 204</math> |
| + | |||
== Solution 1: Number Sense== | == Solution 1: Number Sense== | ||
Solution by e_power_pi_times_i | Solution by e_power_pi_times_i | ||
| Line 14: | Line 15: | ||
| − | Since the divisors are from <math>x^3</math>, then the answer must be something in <math> | + | Since the divisors are from <math>x^3</math>, then the answer must be something in (mod <math>3</math>). Since <math>200</math> and <math>203</math> are the same (mod <math>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 == | ||
Latest revision as of 15:49, 14 March 2023
Problem
If 
is the cube of a positive integer and 
is the number of positive integers that are divisors of , then could be
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 (mod 
). Since 
and 
are the same (mod ), 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. 
