Difference between revisions of "2006 iTest Problems/Problem 1"
Rockmanex3 (talk | contribs) m |
Rockmanex3 (talk | contribs) m |
||
| Line 14: | Line 14: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | 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 00:04, 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.
