Difference between revisions of "2006 iTest Problems/Problem 1"

(Added links to nearby problems)
 
Line 15: Line 15:
  
 
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.
 +
 +
==Obvious Solution==
 +
Since there is only one answer choice, the answer is <math>\boxed{\textbf{(A)}~8}.</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 00:08, 8 June 2021

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