Difference between revisions of "2006 iTest Problems/Problem 1"
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A [[divisor]] could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are <math>2^3 = \boxed{\textbf{(A) } 8}</math> positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start. | A [[divisor]] could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are <math>2^3 = \boxed{\textbf{(A) } 8}</math> positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start. | ||
| + | |||
| + | ==Obvious Solution== | ||
| + | Since there is only one answer choice, the answer is <math>\boxed{\textbf{(A)}~8}.</math> | ||
==See Also== | ==See Also== | ||
Latest revision as of 23:08, 7 June 2021
Contents
Problem
Find the number of positive integral divisors of 2006.
Solution
First, factor the number 2006.
A divisor could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are 
positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start.
Obvious Solution
Since there is only one answer choice, the answer is 
See Also
| 2006 iTest (Problems, Answer Key) | ||
| Preceded by: First Problem |
Followed by: Problem 2 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10 | ||
