If 60 s is a factor of (2^2)(3^3)(5) , and s is a positive integer, how many possible values are there for s ?
A. None
B. One
C. Two
D. Three
E. Four
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kvcpk
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product given is 540.
we need to find S so that 60s divides 540.
Given 60s is factor. So 60s*x =540
we know that 540 = 60*9 = 60*1*9=60*3*3=60*9*1
therefore there are 3 possible values for S
hope this helps!!
we need to find S so that 60s divides 540.
Given 60s is factor. So 60s*x =540
we know that 540 = 60*9 = 60*1*9=60*3*3=60*9*1
therefore there are 3 possible values for S
hope this helps!!
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papgust
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Nice to see you again, Komal! Let me try this question.
60*s = 2^2 * 3 * 5 * s
60s is a factor of (2^2)(3^3)(5)
s could take a max value of 9 as 3^2 is left while cancelling out 60. So, s could be 1,3, or 9 (s is a positive integer) if 60s factors out (2^2)(3^3)(5) completely.
Thus, 3 possible values of s.
60*s = 2^2 * 3 * 5 * s
60s is a factor of (2^2)(3^3)(5)
s could take a max value of 9 as 3^2 is left while cancelling out 60. So, s could be 1,3, or 9 (s is a positive integer) if 60s factors out (2^2)(3^3)(5) completely.
Thus, 3 possible values of s.
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amising6
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60 s=2*2*3*5 *s
=2^2*3*5*s
now since 60 *s is a factor of (2^2)(3^3)(5)
(2^2)(3^3)(5) should be completely divisible by 60 *s
so ((2^2)(3^3)(5) ) /2^2*3*5*s
we can say s can either take value 1 or take value 3 and 3^2
so altogether s can take 3 value 1,3,3^2
so answer is D
so s can take value of 1and 3
=2^2*3*5*s
now since 60 *s is a factor of (2^2)(3^3)(5)
(2^2)(3^3)(5) should be completely divisible by 60 *s
so ((2^2)(3^3)(5) ) /2^2*3*5*s
we can say s can either take value 1 or take value 3 and 3^2
so altogether s can take 3 value 1,3,3^2
so answer is D
so s can take value of 1and 3
