Big number
If N=1223334444.......and is a 100-digit number, find the remainder when N is divided by 16.
Note: The number appears times, except possibly for the last digit.
The answer is 9.
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1 solution
The number as we can see follows the pattern that 1 appears 1 time, 2 appear 2 times, n appears n times... Now digit 9 appearing 9 times will be ending after n(n+1)/2= 9×(9+1)/2=45 digits...now the next numbers are 2 digit numbers. So the next number 10 appearing 10 times will take 20 digits (2×10) and 11 will take 22 digits (2×11). So at the end of 11 there will have been 45+20+22= 87digits. Now the number 12 will appear, so the 88th digit will be 1 and 89th digit will be 2 ....thus leaving us with last 4 digits of the 100 digit number as 2121.
Lets consider all digits before 2121as x. So N =10000x+2121. Now dividing it by 16 we will get (10000x+2121)÷16. Now we know that 10000 is divisible by 16, therefore 10000x is also divisible by 16. So the remainder when N is divided by 16 will be the remainder when 2121 is divided by 16 which is 9. Therefore the remainder when N is divided by 16 is 9. Hence the answer is 9.