Difference between revisions of "1998 AHSME Problems/Problem 6"

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==Solution==
 
==Solution==
{{solution}}
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If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to <math>\sqrt{1998}</math> as possible.
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Since <math>45^2 = 2025</math>, the factors should be as close to <math>44</math> or <math>45</math> as possible.
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Breaking down <math>1998</math> into its prime factors gives <math>1998 = 2\cdot 3^3 \cdot 37</math>. 
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<math>37</math> is relatively close to <math>44</math>, and no numbers between <math>38</math> and <math>44</math> are factors of <math>1998</math>.  Thus, the two factors are <math>37</math> and <math>2\cdot 3^3 = 54</math>, and the difference is <math>54 - 37 = 17</math>, and the answer is <math>\boxed{C}</math>
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==See Also==
 
==See Also==
  
 
{{AHSME box|year=1998|num-b=5|num-a=7}}
 
{{AHSME box|year=1998|num-b=5|num-a=7}}

Revision as of 21:29, 7 August 2011

Problem

If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is

$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$

Solution

If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.

Since $45^2 = 2025$, the factors should be as close to $44$ or $45$ as possible.

Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$.

$37$ is relatively close to $44$, and no numbers between $38$ and $44$ are factors of $1998$. Thus, the two factors are $37$ and $2\cdot 3^3 = 54$, and the difference is $54 - 37 = 17$, and the answer is $\boxed{C}$

See Also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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