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# If h (x) is the product of the integers from 1 to x, the

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Senior Manager
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10 Jan 2005, 10:05
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If h (x) is the product of the integers from 1 to x, the least prime factor of h (100)+1 is in which of the following ranges?

2 to 10
10 to20
20 to 30
30 to 40
above 40

How do we solve it? I have OA.
VP
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10 Jan 2005, 10:35
Wud go with 10 to 20, will explain if correct. What's the OA ?
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10 Jan 2005, 10:43
10-20 is incorrect
However you may want to explain your reasons..OA might be wrong and your explanation might give some lead.

OA is above 40
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10 Jan 2005, 10:43
Dan you are correct. Please explain.
Director
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10 Jan 2005, 10:45
Since h(100)+1 is 101, the range should be 'above 40'
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10 Jan 2005, 10:54
h(100)+1 cannot be 101
h(100) is product of first 100 ints.
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10 Jan 2005, 10:57
Take a smaller number as an example.

5!

is divisible by all numbers up to and including 5. This includes primes 2, 3 and 5. Now, if that's true, then 5! + 1 is not divisible by any of those numbers. 1 is not prime. and 5! is not divisible by 2 since 5! (or in the question 100!) is even and 5! + 1 is odd.

so for 100! + 1, the least prime has to be greater than not only 40 but also 100.
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10 Jan 2005, 14:43
Very interesting problem.

h (100)+1 = 100! +1

This will not be a factor of every integers from 2 to 100, so 0...40 makes sense.

Good one. Keep it coming!
CIO
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11 Jan 2005, 00:22
nocilis wrote:
Very interesting problem.

h (100)+1 = 100! +1

This will not be a factor of every integers from 2 to 100, so 0...40 makes sense.

Good one. Keep it coming!

yup, i agree. That number will be one more than any multiple of all the numbers between 2-100, so it will be divisible by none of them.
11 Jan 2005, 00:22
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