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03 Oct 2019, 03:33
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:30) correct
24%
(01:47)
wrong
based on 314
sessions
History
Date
Time
Result
Not Attempted Yet
If x, y and z are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?
A. 7
B. 12
C. 13
D. 14
E. 16
M37-94
A. 7
B. 12
C. 13
D. 14
E. 16
M37-94
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06 Dec 2021, 07:11
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Official Solution:
If \(x\), \(y\) and \(z\) are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?
A. \(7\)
B. \(12\)
C. \(13\)
D. \(14\)
E. \(16\)
To answer the question we need to find out the largest integer powers of 2, 3, and 5 in \((7!)^2\).
\((7!)^2=(2*3*(2^2)*5*(2*3)*7)^2 =2^8*3^4*5^2*7^2\).
So, \(x_{max}=8\), \(y_{max}=4\), \(z_{max}=2\), and thus, \(x + y + z=8+4+2=14\).
Answer: D
If \(x\), \(y\) and \(z\) are largest positive integers for which \(2^x3^y5^z\) is a factor of \((7!)^2\), what is the value of \(x + y + z\) ?
A. \(7\)
B. \(12\)
C. \(13\)
D. \(14\)
E. \(16\)
To answer the question we need to find out the largest integer powers of 2, 3, and 5 in \((7!)^2\).
\((7!)^2=(2*3*(2^2)*5*(2*3)*7)^2 =2^8*3^4*5^2*7^2\).
So, \(x_{max}=8\), \(y_{max}=4\), \(z_{max}=2\), and thus, \(x + y + z=8+4+2=14\).
Answer: D
General Discussion
03 Oct 2019, 09:43
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Ans - D
Applied the divisibility formula which is applied in finding trailing zeroes.
Applied the divisibility formula which is applied in finding trailing zeroes.
