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Number properties
Source: Beat The GMAT — Problem Solving |
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simplyjat
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Break down 990 into prime factors.
990 = 11*3*3
Pick the biggest prime factor (11). So for the product of numbers from 1 to n, one product should be 11, if n is less than 11 then the product will not be a multiple of 990
990 = 11*3*3
Pick the biggest prime factor (11). So for the product of numbers from 1 to n, one product should be 11, if n is less than 11 then the product will not be a multiple of 990
simplyjat
Thanks Jay,
But why did you start looking at prime numbers? Is a strategy to use for this type of question?
I appreaciate your help and assistance.
But why did you start looking at prime numbers? Is a strategy to use for this type of question?
I appreaciate your help and assistance.
Probably you understand it better that way:
We are searching for numbers 1, 2, 3, ...., n such that
1*2*3*.....*n = m*990
Don't bother with the m on the right side, it's just for illustration that the product should be a multiple of 990.
We are actually looking for the numbers on the left side. We are asking ourselves which numbers must be on the left side in order to get a multiple of 990 as product?
Therefore we break 990 down to prime numbers (because from the prime factors you can make up all the other factors by multiplication). So let's take a look at our equation:
2*3*3*5*11=990
So if we have our numbers from 1 to n, we have to get at least to 11 in order to reach our 990:
1*2*3*4*5*6*7*8*9*10*11= m990
Because 11 is a prime factor, you cannot replace it with any combination of factors below. It is the biggest primefactor and therefore the numbers must go up at least to 11.
We are searching for numbers 1, 2, 3, ...., n such that
1*2*3*.....*n = m*990
Don't bother with the m on the right side, it's just for illustration that the product should be a multiple of 990.
We are actually looking for the numbers on the left side. We are asking ourselves which numbers must be on the left side in order to get a multiple of 990 as product?
Therefore we break 990 down to prime numbers (because from the prime factors you can make up all the other factors by multiplication). So let's take a look at our equation:
2*3*3*5*11=990
So if we have our numbers from 1 to n, we have to get at least to 11 in order to reach our 990:
1*2*3*4*5*6*7*8*9*10*11= m990
Because 11 is a prime factor, you cannot replace it with any combination of factors below. It is the biggest primefactor and therefore the numbers must go up at least to 11.
