Number properties

[vinviper1]

Thanks!

[Luca80]

11

[simplyjat]

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

[vinviper1]

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.

[KICKGMATASS123]

CAN YOU EXPLAIN FURTHER?

[2ndShot]

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.