JustusJPS wrote:
Hi, I used the following approach: \(4800=2^{6}*3*5^{2}; 12=2^{2}*3\)
\(\frac{4800}{12}=\frac{2^{6}*3*5^{2}}{2^{2}*3}\)
\(=2^{4}*5^{2}\)
Number of Factors:\((4+1)*(2+1)= 15\)
I do not know if this approach works for similar problems. Can someone help me?
HiJustusJPS .
I used the same approach as above ,it works good.
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