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| (A) 1(B) 0(C) 49(D) 491(E) 50 | | (A) 1(B) 0(C) 49(D) 491(E) 50 |
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− | ==Solution 1== | + | ==Solution== |
| By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. 33⋅55⋅77⋅99⋅1111=1⋅1⋅1⋅1⋅1=(A)1 | | By the [[associative property]], we can rearrange the numbers in the numerator and the denominator. 33⋅55⋅77⋅99⋅1111=1⋅1⋅1⋅1⋅1=(A)1 |
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− | ==Solution 2 ==
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− | Multlipication gives:<cmath>\frac{15}{99}\cdot\frac{693}{105}=\frac{10395}{10395}=\boxed{\text{(A)}\ 1}.</cmath>
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− | ==Video Solution by BoundlessBrain!==
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− | https://youtu.be/eC_Vu3vogHM
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| ==See Also== | | ==See Also== |
Revision as of 19:22, 14 December 2023
Problem
9×113×5×3×5×77×9×11=
(A) 1(B) 0(C) 49(D) 491(E) 50
Solution
By the associative property, we can rearrange the numbers in the numerator and the denominator. 33⋅55⋅77⋅99⋅1111=1⋅1⋅1⋅1⋅1=(A)1
See Also
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.