Difference between revisions of "2006 iTest Problems/Problem 1"
Rockmanex3 (talk | contribs) (Solution to Problem 1) |
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | 2006 &= 2 \cdot 1003 | + | 2006 &= 2 \cdot 1003 \\ |
&= 2 \cdot 17 \cdot 59 | &= 2 \cdot 17 \cdot 59 | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 00:03, 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.
