AJIT AskiitiansExpert-IITD
Last Activity: 14 Years ago
Dear Akhil ,
consider the expression
(1 + a + a2 + a3 + a4 + a5 + a6 + a7) (1 + b + b2 + b3 + b4) (1 + c + c2 + c3) (1 + d) (1 + e) (1 + f)
If you observe the expression , you will find that each term in the expansion is a factor of N = a7b4c3def . Also the expansion starts from 1 and ends at the number itself , in between all the factors of numbers are located . so if we can find out the number of terms in the expansion -2 (for 1 and number itself) , we will get the answer, which can be easily seen as 8 X 5 X 4 X 2 X 2 X 2 = 1280 -2 = 1278
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