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whenever we are multiplying consecutive numbers , then we

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chan4312
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whenever we are multiplying consecutive numbers , then we [#permalink] New post   18 Aug 2008, 13:51
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.



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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   18 Aug 2008, 13:57
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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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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   19 Aug 2008, 05:47
Ok.
tell me whether 1*2*3*4*5*6*....58*59*60 is divisible by 2147483648 (which is \(2^{31}\))

we can not just sit and expect our brain to calculate such a complex calculation.!!

I dont expect such kind of questions in gmat,...But the method is very good and it is important.

since there are 31-1=30(the power of 2 minus 1) even numbers in the product sequence , i would say it is divisible.
if 60 was not there in the product sequence, it would not have been divisible by 2147483648.

your explanation is also helpful for me.
I rest my case.!!
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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   19 Aug 2008, 06:07
I just realized that I think we're talking about the same principle/rule but in different ways.

Your number is \(2^{31}\) and the product is of the numbers 1 through 60. If you take all of the even numbers, you can divide 2 into each one of them. So count how many even numbers you have and that will give you a start on how many 2's there are in the product of all numbers 1 - 60. Divide 60 by 2 = 30. There are 30 even numbers, each of them is divisible by 2, so we know we have at least thirty 2's in the product of all numbers 1 - 60. We know there is 31 because 4 is there, and that's divisble twice. So when you do the 31-1=30 even numbers and say it's divisible by 2^31, i really think we're saying something very, very similar.

Tell me this...how does your method help you solve this question:

I think we both agree that 1*2*3...*59*60 is divisible by \(2^{31}\). What is the smallest set of consecutive, positive integers, the product of which is divisible by \(2^{31}\)?
Example: 1*2*3*4*...n is divisible by \(2^{31}\). What is the smallest value of n that is divisible by \(2^{31}\)?


chan4312 wrote:
Ok.
tell me whether 1*2*3*4*5*6*....58*59*60 is divisible by 2147483648 (which is \(2^{31}\))

we can not just sit and expect our brain to calculate such a complex calculation.!!

I dont expect such kind of questions in gmat,...But the method is very good and it is important.

since there are 31-1=30(the power of 2 minus 1) even numbers in the product sequence , i would say it is divisible.
if 60 was not there in the product sequence, it would not have been divisible by 2147483648.

your explanation is also helpful for me.
I rest my case.!!
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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   19 Aug 2008, 06:27
jallenmorris wrote:
I just realized that I think we're talking about the same principle/rule but in different ways.

Your number is \(2^{31}\) and the product is of the numbers 1 through 60. If you take all of the even numbers, you can divide 2 into each one of them. So count how many even numbers you have and that will give you a start on how many 2's there are in the product of all numbers 1 - 60. Divide 60 by 2 = 30. There are 30 even numbers, each of them is divisible by 2, so we know we have at least thirty 2's in the product of all numbers 1 - 60. We know there is 31 because 4 is there, and that's divisble twice. So when you do the 31-1=30 even numbers and say it's divisible by 2^31, i really think we're saying something very, very similar.

Tell me this...how does your method help you solve this question:

I think we both agree that 1*2*3...*59*60 is divisible by \(2^{31}\). What is the smallest set of consecutive, positive integers, the product of which is divisible by \(2^{31}\)?
Example: 1*2*3*4*...n is divisible by \(2^{31}\). What is the smallest value of n that is divisible by \(2^{31}\)?


chan4312 wrote:
Ok.
tell me whether 1*2*3*4*5*6*....58*59*60 is divisible by 2147483648 (which is \(2^{31}\))

we can not just sit and expect our brain to calculate such a complex calculation.!!

I dont expect such kind of questions in gmat,...But the method is very good and it is important.

since there are 31-1=30(the power of 2 minus 1) even numbers in the product sequence , i would say it is divisible.
if 60 was not there in the product sequence, it would not have been divisible by 2147483648.

your explanation is also helpful for me.
I rest my case.!!


for \(2^31\) to be factor of product series there should be at least 31 minus 1 = 30 even numbers in the series.
So n should be 2*30 =60 numbers ...59 wont work..bcz it only has 29 evens.

yes..u r right ..we are talking about same thing here..
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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   19 Aug 2008, 06:31
For ease of reference, I'll refer to 1*2*3*...60 as 60! because it's the same thing. And n! for the variable we're trying to find.
\(2^{31}\) will divide into n! where n! is much less than 60.
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Re: whenever we are multiplying consecutive numbers , then we [#permalink] New post   19 Aug 2008, 06:46
jallenmorris wrote:
For ease of reference, I'll refer to 1*2*3*...60 as 60! because it's the same thing. And n! for the variable we're trying to find.
\(2^{31}\) will divide into n! where n! is much less than 60.



I guess ..I am wrong ... :(

number of evens ---- dividing factor

2 -------- 8
3 -------- 16
4 ----------- 32
4 ---------- 64
4 --------- 128
5 ------ 256
6 ------- 512
6 -------- 1024
7 -------- 4096
do you see any pattern here.



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Re: whenever we are multiplying consecutive numbers , then we [#permalink]
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