Que: If both 11 and \(7^2\) are factors of \(a\cdot4^5\cdot6^3\cdot7\), then what is the smallest possible value of a?
(A) 7
(B) 11
(C) 49
(D) 77
(E) 539
Que: If both 11 and \(7^2\) are factors of \(a\cdot4^5\cdot6^3\cdot7\), then what is the.....
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- Max@Math Revolution
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- Max@Math Revolution
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Solution: Express the given number \(a\cdot4^5\cdot6^3\cdot7\) in prime factors as \(a\cdot2^{10}\cdot2^3\cdot3^3\cdot7\)
For 11 to be the factor it should be contained in it. If we remove ‘a’ then we don’t have 11. Hence a = 11.
=> 7 is contained already and hence we need 1 more 7 so that \(7^2\) is the factor and hence 7 should also be in a.
Thus, smallest value of a = 7 * 11 = 77
Therefore, D is the correct answer.
Answer D
For 11 to be the factor it should be contained in it. If we remove ‘a’ then we don’t have 11. Hence a = 11.
=> 7 is contained already and hence we need 1 more 7 so that \(7^2\) is the factor and hence 7 should also be in a.
Thus, smallest value of a = 7 * 11 = 77
Therefore, D is the correct answer.
Answer D
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