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}} | ||
Revision as of 20:29, 7 August 2011
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 | ||
