NYC5648 wrote:
When the product of 3,070,956 and n is divided by 720 there will be no remainder. If n > 1, what is the smallest value of n?
A. 10
B. 20
C. 30
D. 40
E. 60
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.
P.S. Please read and follow: 11-rules-for-posting-133935.html (points 3 and 8).Your explanations blow my mind, Bunuel. In other words, they are outstanding. All I knew was that this is a prime factorization problem, but wasn't sure how to break down the problem. Would you say this is about a 650+ difficulty?