When the product of 3,070,956 and n is divided by 720 there

Given: \(\frac{3,070,956*n}{720}=integer\).

Factorize the divisor: \(720=2^4*3^2*5\).

Now, since 56 (the last two digits of 3,070,956) is divisible by 4, then 3,070,956 is divisible by 4=2^2, and since 956 (the last three digits of 3,070,956) is NOT divisible by 8, then 3,070,956 is NOT divisible by 8=2^3. That means that n must have 2^2=4 as its factor (3,070,956 is divisible only by 2^2 so in order 3,070,956*n to be divisible by 2^4 n must have 2^2 as its factor);

Similarly since the sum of the digits of 3,070,956 is 3+0+7+0+9+5+6=30 then 3,070,956 divisible by 3 but not by 3^2, so n must have 3 as its factor;

And finally since the units digit of 3,070,956 is 6 then 3,070,956 is not divisible by 5, so n must have 5 as its factor.

Therefore the least value of \(n\) is \(2^2*3*5=60\).

Answer: E.

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