If a positive odd integer N has p positive factors, how many positive factors will 2N have ? Choices:  A) p  B) 2p  C) P+1  D) 2p+1  E) Cannot be determined
no of factor of a=b^m * c^n * d^o * e^q.... = (m+1)*(n+1)*(o+1)*(q+1)...   Given that N is an odd integer. So N can be b^m * c^n * d^o * e^q, where b,c,d,e are odd numbers.   The number of factors of N = (m+1)*(n+1)*(o+1)*(q+1) = p   For a number 2N = 2^1 * b^m * c^n * d^o * e^p, number of factors will be (1+1)*(m+1)*(n+1)*(o+1)*(q+1) = 2*p   So, number of factors of 2N (1+1)*( +1)*( +1)*( +1)*( +1)= 
<gadget id="calculator">(1+1)*( +1)*( +1)*( +1)*( +1)</gadget>
<output>2</output>
2*p   So, number of factors of 2N= 2*p   Hence, answer will be 2p.
<result>B</result>
If a positive odd integer N has p positive factors, how many positive factors will 2N have ?