If you write out the sequence into prime factorization you get this:

1*2*3*4*5*6*7*8 = 1*2*3*(2*2)*5*(2*3)*7*(2*2*2)

If you need to know if the product is divisible by any number, say X, find the prime factors of that number and see if the same number of prime factors of x.

Is the product of all integers from 1 to 8 divisible by 350?

350

10 * 35

2 * 5 * 5 * 7

Now, in the sequence above for 1 to 8, start crossing out 2, 5, 5, and 7. If you have each of those, then yes, it will be divisible by that number.

I can't think of a very simple rule to state for this, but I hope my example shows the theory in use.

Lets try another one (My high school calc teacher always said this with a geeky-happy smile. Always made us laugh.)

Is the product of consecutive integers from 1 to 7 divisible by 648?

1 * 2 * 3 * 2 * 2 * 5 * 2 * 3 * 7

648

4 * 162

2 * 2 * 2 * 81

2 * 2 * 2 * 3 * 3 * 3 * 3

so

\(\frac{1 * 2 * 3 * 2 * 2 * 5 * 2 * 3 * 7}{2 * 2 * 2 * 3 * 3 * 3 * 3}\)

start canceling:

We need three 2's and four threes. The product of 1 to 7 is not divisible by 648 because we don't have enough 3's. We would have to have the product of the first 9 in order for it to be divisible by 648.

chan4312 wrote:

whenever we are multiplying consecutive numbers , then we can see whether any number 2^n

is a factor of the multiplication result without solving for the product.

let us say, we want to find if the product 2*3*4*5*6 contains 16(which is 2^4) as a factor.

here there are 5 numbers and 3 even numbers(which is 4-1) ..so divisible by 16

same product can not be divisible by 32 bcz.. number of even numbers should be 5-1 =4 even numbers

2*3*4*5*6*7*8 product is divisible by 32

can someone justify this pls.

I'm not sure about your 4-1 and 5-1. My way is certainly longer for very large examples. Say product of first 100 numbers and is this divisible by 123,456. But my example will certainly work.

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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