Difference between revisions of "2004 AMC 10B Problems/Problem 6"
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<math> \mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101! </math> | <math> \mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101! </math> | ||
− | ==Solution== | + | ==Solution 1== |
Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write: | Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write: | ||
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We see that <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and because <math>11</math>, and <math>101</math> are primes, none of the other four choices are squares. | We see that <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math> is a square, and because <math>11</math>, and <math>101</math> are primes, none of the other four choices are squares. | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | Notice that <math>99! \cdot 99!</math> is a perfect square (obviously). Also, <math>100 = 10^2</math>. Thus, | ||
+ | <cmath>(99!)^2 \cdot 10^2 = 99! \cdot 100!</cmath> | ||
+ | is a perfect square, so the answer is <math>\boxed{\mathrm{(C) \ } 99! \cdot 100!}</math>. | ||
== See also == | == See also == |
Latest revision as of 17:48, 26 July 2020
Contents
Problem
Which of the following numbers is a perfect square?
Solution 1
Using the fact that , we can write:
We see that is a square, and because , and are primes, none of the other four choices are squares.
Solution 2
Notice that is a perfect square (obviously). Also, . Thus, is a perfect square, so the answer is .
See also
2004 AMC 10B (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 | ||
All AMC 10 Problems and Solutions |
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