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

(Solution to Problem 1)
 
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<cmath>\begin{align*}
 
<cmath>\begin{align*}
2006 &= 2 \cdot 1003
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2006 &= 2 \cdot 1003 \\
 
&= 2 \cdot 17 \cdot 59
 
&= 2 \cdot 17 \cdot 59
 
\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.
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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==
* [[2006 iTest Problems/Problem 2|Next Problem]]
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{{iTest box|year=2006|before=First Problem|num-a=2|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
* [[2006 iTest Problems]]
 
  
[[Category:Introductory Combinatorics Problems]]
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[[Category:Introductory Number Theory Problems]]

Latest revision as of 01:15, 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

2006 iTest (Problems)
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
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