7oops. yes ketan, it will be 15 0s r of the order(2n -1) but i m nt sure of ur ans.
202 x 20002 x 200000002 x 20000...(13 zeroes)...2x 2... (31 zeroes)...2. Find the sum of the digits of the product thus formed.
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3 Answers
1057first of all...i think the 4th term will have 15 zeroes and not 13...
so the product is (2x102+2)(2x104+2)(2x108+2)(2x1016+2)(2x1032+2)
note that if u try and multiply the first two....
it turns out to be like this...
4x106+4x104+4x102+4
which is nothing but 4040404
this pattern will remain with 8 and the product will be 808080808080808
and the next product will be 16161616161616161616161616161616 and the next consisting of all 32s....
so answer is 32(3+2)=160....
71057http://www.wolframalpha.com/input/?i=202+x+20002+x+200000002+x+20000000000000002x+200000000000000000000000000000002
isko dekh....