2006 iTest Problems/Problem 1

Revision as of 23:08, 7 June 2021 by Aops-g5-gethsemanea2 (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.

Obvious Solution

Since there is only one answer choice, the answer is $\boxed{\textbf{(A)}~8}.$

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
First Problem
Followed by:
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 U1 U2 U3 U4 U5 U6 U7 U8 U9 U10