least possible value

[lenagmat]

If the product of the integers from 1 to n is divisible by 490, what is the least possible value of n?
(A) 7
(B) 14
(C) 21
(d) 28
(E) 35

Here is the answer, that I found. But I completly do not understand why it is not 7 (7<14) ?

490 = 7 x 7 x 5 x 2 = 14 x 7 x 5

Let the number be 'N'

N/490 = integer

Therefore, if the denominator is to be completely cancelled out from the numerator, N must contain 14, 7 and 5.

=> the smallest value of n must be 14.

[neelgandham]

Here is the answer, that I found. But I completly do not understand why it is not 7 (7<14) ?

1)Let us say the answer = 7, so 1*2*3*4*5*6*7 is divisible by 490. Let us check if it really is.

490=7*7*5*2, so 1*2*3*4*5*6*7/(7*7*5*2) = (1*3*4*6/7), (dividing numerator and denominator by (7*5*2))
1*3*4*6 is not divisible by 7 => 1*2*3*4*5*6*7 is not divisible by 490(7*7*5*2).

2)Let us say the answer = 14, so 1*2*3*4*5*6*7*8*9*10*11*12*13*14 is divisible by 490. Let us check if it really is.

490=7*7*5*2, so 1*2*3*4*5*6*7*8*9*10*11*12*13*14/(7*7*5*2) = (1*3*4*6*8*9*10*11*12*13*2), (dividing numerator and denominator by (7*5*2*7)). => 1*2*3*4*5*6*7*8*9*10*11*12*13*14 is divisible by 490

[kmittal82]

The key here is:
"If the product of the integers from 1 to n", which is basically n!

If we use 7!, we can cancel out 7,5,2 from the denominator, which still leaves us with an extra 7 in the denominator, hence thats not the answer

Using 14, we cancel out all the terms of the denominator, hence (B)