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
[hide]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 |