what is the least possible value for n

mcheytan

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If n=positive integer, the product of all integers 1 to n including is a multiple of 990, what is the least possible value of n.

I did this by multiplying the numbers from 1 to 10 and see if it works, then to 11 and found out that n = 11, but is there an easier way and shorter to do this?
 
Re: what is the leaset possible value for n

note that 990 = 9*10*11 = 3[sup:2dxo82xb]2[/sup:2dxo82xb]*2*5*11

since 11 is a prime factor of 990, it is also a factor for any multiple of 990. since you need the 11 and all the other prime factors are included in the lesser integers, then n = 11.

had the number been 90 = 3[sup:2dxo82xb]2[/sup:2dxo82xb]*2*5, then the smallest value for n would be 6 ... see why?
 
Re: what is the leaset possible value for n

Sorry, I could not understand for the example you gave me....I thought it would have been n=5 for the smallest possible number of n, because .... as in the case above, 990=3^2*2*5*11, and here is 90=3^2*2*5???
 
Re: what is the leaset possible value for n

mcheytan said:
Sorry, I could not understand for the example you gave me....I thought it would have been n=5 for the smallest possible number of n, because .... as in the case above, 990=3^2*2*5*11, and here is 90=3^2*2*5???

90 = 3[sup:39izrhmf]2[/sup:39izrhmf]*2*5

for 1*2*3*4*5 ... where is that second factor of 3?
 
Re: what is the leaset possible value for n

I really don't get it now....
 
Re: what is the leaset possible value for n

mcheytan said:
I really don't get it now....
1*2*3*4*5 =120 >>>>not divisible by 90.....................(1)

1*2*3*4*5*6 = 720 >>>>> divisible by 90 ....................(2)

In (1) you have only 1 number divisible by 3 (that was 3) - it is not divisible by 9.

In (2) you have 2 numbers divisible by 3 (that was 3 and 6 ) - it is divisible by 9.
 
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