2006 iTest Problems/Problem 1

Problem

Find the number of positive integral divisors of 2006.

$\mathrm{(A)}\, 8$

Solution

First, factor the number 2006.

$\begin{align*} 2006 &= 2 \cdot 1003 &= 2 \cdot 17 \cdot 59 \end{align*}$

A divisor could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are $2^3 = \boxed{\textbf{(A)} 8}$ positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start.

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2006 iTest Problems/Problem 1

Problem

Solution

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