

InterviewSolution
Saved Bookmarks
1. |
Find (i) the last digit, (ii) the last two digits, and (iii) the lastthree digits of `17^(256)dot` |
Answer» We have `17^(256) = (17^(2))^(128) = (289)^(128) = (290-1)^(128)` `:. 17^(256) = .^(128)C_(0)(290)^(128)-.^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `- .^(128)C_(125)(290)^(3)+.^(128)C_(126)(290)^(2)-.^(128)C_(127)(290)+1` `=[.^(128)C_(0)(290)^(128) - .^(128)C_(1)(290)^(127)+.^(128)C_(2)(290)^(126)-"....."` `-.^(128)C_(125)(290)^(3)]+.^(128)C_(126)(290)^(2) -.^(128)C_(127)(290)+1` `=1000m+((128)(127))/(2)(290)^(2)-128xx290+1` `= 1000 m + (128)(127)(290)(145)-128xx290-1` `= 1000m+(128)(290)(127xx145-1)+1` `=1000m+(128)(290)(18414)+1` `=1000m+683527680+1` `= 1000m+683527000+680+1` `= 1000(m+683527)+681` Hence, the last three digits of `17^(256)` are `6,8,1`. As a result, the last two digits of `17^(256)` are `8, 1` and the last digit of `17^(256)` is 1. |
|