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Multplication of n consecutive numbers.

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by Anurag@Gurome » Sun Mar 18, 2012 10:15 pm
puneetluthra198508 wrote:what's the best way to calculate the solution for multiplication of first n numbers..
ex ... p(100) -> 1*2*3*4___99*100..
Product of first n numbers is 1 * 2 * 3 * ... * n = n! as n! = n * (n - 1) * ... 3 * 2 * 1

In this case, 1 * 2 * 3 * 4 ... 99 * 100 = 100!
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by Whitney Garner » Fri Mar 23, 2012 1:43 pm
puneetluthra198508 wrote:what's the best way to calculate the solution for multiplication of first n numbers..
ex ... p(100) -> 1*2*3*4___99*100..
Hi puneetluthra198508!

Anurag@Gurome is exactly right in that you're actually looking at 100!. But to make sure it is clear for your question, there is NO way to calculate this actual value without a heavy duty calculator. You will never be asked to do so on the test as the number has 157 significant digits (it equals approximately 9.3 x 10^157.

The test might ask you how many trailing 0s it has (the number of zeros between the decimal and the first non-zero digit - ie, the number 97000 has 3 trailing 0s).

I'll let you puzzle over how to figure that one out and you can yell back for help if you get stuck!

Hope this helps!
:)
Whit
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