Question
When 315! is divided by 1215x, remainder = 0. What is the maximum possible value for x?
Answer: Option A
:
A
The question is based on highest power of a number in a factorial.
Here since 1215 is composite number, prime factorize 1215 i.e 1215=35×5.
Required answer will be the highest power of 35 in 315!
(No need to find to find the highest power of 5 in 315! as that will always be more than that of 35)
To find out highest power of 35, we will first find the highest power of 3 and then divide it by 5.
Highest power of 3 in 315! = 155 (105 + 35 + 11 + 3 + 1)
Highest power of 35 in 315! = 31 (highest power of 5 we will get is 77)
Required answer is 31.
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:
A
The question is based on highest power of a number in a factorial.
Here since 1215 is composite number, prime factorize 1215 i.e 1215=35×5.
Required answer will be the highest power of 35 in 315!
(No need to find to find the highest power of 5 in 315! as that will always be more than that of 35)
To find out highest power of 35, we will first find the highest power of 3 and then divide it by 5.
Highest power of 3 in 315! = 155 (105 + 35 + 11 + 3 + 1)
Highest power of 35 in 315! = 31 (highest power of 5 we will get is 77)
Required answer is 31.
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