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.
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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.
  
 
==See Also==
 
==See Also==

Revision as of 01:04, 26 November 2018

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.

See Also

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