# 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. |

==See Also== | ==See Also== |

## Revision as of 01:04, 26 November 2018

## 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.