If ( 115C0 + 115C1 ) ( 115C1 + 115C2 ) ( 115C12 + 115C13 ) = 1314C0 14C1 14C12, then 30 is equal to .
Answer
32
Solution
General factor (r = 0, 1, , 12):
115Cr + 115Cr+1 = 15Cr+1 + 15Cr15Cr 15Cr+1 = 16Cr+115Cr 15Cr+1
16Cr+1 = 16r + 1 15Cr and 15Cr+1 = 15r + 1 14Cr
factor = 16(r + 1) 15Cr+1 = 1615 14Cr
product = ( 1615 )1314C0 14C1 14C12
= 1615
30 = 32
Answer: 32