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04 Sep 2019, 01:03
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C
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If X is a positive integer and 405^4 is a multiple of 3^X, what is the largest possible value of X?
(A) 5.
(B) 12.
(C) 16.
(D) 20.
(E) 26
(A) 5.
(B) 12.
(C) 16.
(D) 20.
(E) 26
04 Sep 2019, 01:10
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IMO C
405 = 3^4 * 5
(405)^4 = (3^4 *5)^4
Largest value of X = 16
Posted from my mobile device
405 = 3^4 * 5
(405)^4 = (3^4 *5)^4
Largest value of X = 16
Posted from my mobile device
04 Sep 2019, 02:20
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Prime factorization of 405 is
3^4×5.
Therefore, (405)^4 = (3^4×5)^4
= 3^16×5^4. Since 3^16×5^4 is a multiple of 3^x, in other words, 3^x is a factor of 3^16×5^4. 3^x = 3^16, x = 16.
The largest possible value of x is 16.
Option C.
Posted from my mobile device
3^4×5.
Therefore, (405)^4 = (3^4×5)^4
= 3^16×5^4. Since 3^16×5^4 is a multiple of 3^x, in other words, 3^x is a factor of 3^16×5^4. 3^x = 3^16, x = 16.
The largest possible value of x is 16.
Option C.
Posted from my mobile device
