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The number of integer value(s) of n when `3^512`-1 is divided by `2^n` is/are |
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Answer» `3^512 - 1 = (3^256)^2 - 1` `= (3^256 - 1)(3^256 + 1)` `= ((3^128)^2 - 1^2)(3^256 + 1)` `= (3^128 -1)(3^128 +1)(3^256 +1)` `3^512 - 1 = (3^64 - 1)(3^64 + 1)(3^128 + 1)(3^256+1)` `= (3^32 - 1)(3^32+1)` `= (3^16 - 1)(3^16+1)` `= (3^8 - 1)(3^8+1)` `= (3^4-1)(3^4+1)` `= (3^2-1)(3^2+1)` `3^512 - 1= (3-1)(3+1)3^2` `3^512 - 1= 2(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)(3^64+1)(3^128+1)(3^256+1)` `3^256 = (2+1)^256` `= .^256C_0(2)^256 (1)^0 + .^256C_1(2)^256(1)^1.......+1` `= 2(even) + 2(odd)` `= = 20^11` option c is correct answer |
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