Difference between revisions of "1998 AHSME Problems/Problem 6"
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==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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:28, 5 July 2013
Problem
If is written as a product of two positive integers whose difference is as small as possible, then the difference is
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 as possible.
Since , the factors should be as close to or as possible.
Breaking down into its prime factors gives .
is relatively close to , and no numbers between and are factors of . Thus, the two factors are and , and the difference is , and the answer is
See Also
1998 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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 | ||
All AHSME Problems and Solutions |
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