A number N contains only the digit 5 and is exactly divisible by 373 . Then the remainder when n/373 is divided by 10000 is
Answers
Answer:
819 is the remainder..............
Given : a number N contains only the digit 5 and it is exactly divisible by 373
To Find : remainder when N/373 is divided by10000
A) 2535, B) 5235, C) 5325, D) 3525
Solution:
Number N = 55555.......
K = N/373
exactly divisible by 373
=> 373 * K = 55555........
unit Digit of K must be 5
=> 373 x 5 = 1865 Carry Over 186
Now Tens Digit = A
=> 373A + 186 Must end with 5
A = 3 is only possible
373 x 3 + 186 = 1305 Hence 130 is carry over
Now hundreds' digit = B
=> 373B + 130 Must end with 5
=> B = 5 is only possible
=> 373 * 5 + 130 = 1,995 Carry over 199
Thousands digit = C
=> 373C + 199 Must end with 5
Hence C = 2 is only possible
373 x 2 + 199 = 945 , carry over 94
Last four Digits of K are 2535
K = ......2535
=> K = 10000P + 2535
Hence When K Divided by 10000 Then remainder is 2535
and to Find number N repeat the above process until carry over is 55
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