Question

How many numbers are there from 0 to 1000 which on division by 2 4 6 8 leaves a remainder 1 3 5 7 respectively?

Answer 1

Hi there !
Here's the answer:

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Here,
Clearly required No.s are odd as when divided by 2 leaving a remainder of 1

If a No. required as per the condition is denoted by n
Then we see that n+1 is divisible by 2, 4, 6, 8

Hence (n+1) must be divisible by LCM(2, 4, 6,8).

LCM(2, 4, 6, 8) = 24

=> n+1 = 24k for some integer k
=> n = 24k-1 for some integer k

For n>0 ; for k = 0 to 24k - 1

where 24k-1 = 1000
24k = 1001
=> k = 1001/24 = 41.7…

So k = 1 to 41 yields Positive Integer solutions between 0 & 1000

=> There are 41 such No.s

The No.s which when divided by 2,4,6 and 8 leaves a remainder of 1,3,5 and 7 respectively are :

23 47 71 95 119
143 167 191 215 239
263 287 311 335 359
383 407 431 455 479
503 527 551 575 599
623 647 671 695 719
743 767 791 815 839
863 887 911 935 959
983

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Hope it helps