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30 Aug 2009, 13:55
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50% (01:55) correct
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Please explain, thank you.
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30 Aug 2009, 15:40
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H(100) = 2*4*6...*100
H(100) + 1 = 2*4*6...*100 + 1
= 2^50(1*2*3*4...*50) + 1
Now, if we divide this value by any number, lessthan equal to 50 , then,
2^50(1*2*3*4...*50) will give remainder 0 , but some fractional part will come from +1/any number less than equal to 50..
So, there may be a possibility that 2^50(1*2*3*4...*50) + 1 will be divisible by any prime number greater than 50, as in that case, 2^50(1*2*3*4...*50) won't be divisible by that number.
Hence, the smallest number will be grater than 50 ..Answer E..
I know, it's difficult to understand, but just try some numbers and you will find out what i want to explain..
Thanks
H(100) + 1 = 2*4*6...*100 + 1
= 2^50(1*2*3*4...*50) + 1
Now, if we divide this value by any number, lessthan equal to 50 , then,
2^50(1*2*3*4...*50) will give remainder 0 , but some fractional part will come from +1/any number less than equal to 50..
So, there may be a possibility that 2^50(1*2*3*4...*50) + 1 will be divisible by any prime number greater than 50, as in that case, 2^50(1*2*3*4...*50) won't be divisible by that number.
Hence, the smallest number will be grater than 50 ..Answer E..
I know, it's difficult to understand, but just try some numbers and you will find out what i want to explain..
Thanks
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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