If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are
positive integers such that x = 2i3k5m7p, then i + k + m + p =
Please note that i, k , m and p are exponents of 2, 3 , 5 and 7 respectively
ANS-
A. 4
B. 7
C. 8
D. 11
E. 12
Correct Answer D
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positive integers
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MAAJ
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x = the product of the positive integers from 1 to 8
x = 1 * 2 * 3 * (2*2) * 5 * (2*3) * 7 * (2*2*2)
x = (2^7)(3^2)(5)(7); Also we know that
x = (2^i) (3^k) (5^m) (7^p); Thus
i=7; k=2; m=1; p=1
i + k + m + p = ?
7 + 2 + 1 + 1 = 11
Correct Answer [spoiler](D)[/spoiler]
x = 1 * 2 * 3 * (2*2) * 5 * (2*3) * 7 * (2*2*2)
x = (2^7)(3^2)(5)(7); Also we know that
x = (2^i) (3^k) (5^m) (7^p); Thus
i=7; k=2; m=1; p=1
i + k + m + p = ?
7 + 2 + 1 + 1 = 11
Correct Answer [spoiler](D)[/spoiler]
phelps wrote:If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are
positive integers such that x = 2i3k5m7p, then i + k + m + p =
Please note that i, k , m and p are exponents of 2, 3 , 5 and 7 respectively
ANS-
A. 4
B. 7
C. 8
D. 11
E. 12
Correct Answer D
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cans
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x=8! = 2^7 * 3^2 * 5^1 * 7^1
thus sum = 11
IMO D
thus sum = 11
IMO D
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