hiplease anyone can solve this find the remainder when 123412341234.............................................................................upto 400 digits divided by 909ans is 685plz anyone let me know the method to solve this
Take 1234 out...
1234*10^396 + 1234*10^392 + .....
1234*[10^396 + 10^392 + ....10^0]
now, rem[1234 * [10^396 + 10^392 + ....10^0] ] whn divided by 101
1234 mod 101 * [10^396 + 10^392 + ....10^0] mod 101=> 22*[100^198 + 100^196 + ....10^0]mod 101=> 22*[1+1+.....100 times]=> 2200 mod 101 = 79
Little STar
n/a
@ Little Star Hi but ans for this is 685 here you divide 909 into 101*9 but you dint specify in solution wt u did with the 9 so plesae clarify it
thanks
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Take 1234 out...
1234*10^396 + 1234*10^392 + .....
1234*[10^396 + 10^392 + ....10^0]
now, rem[1234 * [10^396 + 10^392 + ....10^0] ] whn divided by 101
1234 mod 101 * [10^396 + 10^392 + ....10^0] mod 101
=> 22*[100^198 + 100^196 + ....10^0]mod 101
=> 22*[1+1+.....100 times]
=> 2200 mod 101 = 79
Little STar
n/a
@ Little Star Hi but ans for this is 685 here you divide 909 into 101*9 but you dint specify in solution wt u did with the 9 so plesae clarify it
thanks
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