here is the official explanation

the bold part sums up what have said

hope it helps ...

Solution:

Any number that is divisible by â€˜6' will be a number that is divisible by both â€˜2' and â€˜3'. i.e. the number should an even number and should be divisible by three (sum of its digits should add to a multiple of â€˜3').

Any number that is divisible by â€˜9' is also divisible by â€˜3' but unless it is an even number it will not be divisibly by â€˜6'.

In the above case, â€˜9' is an odd number. Any power of â€˜9' which is an odd number will be an odd number which is divisible by â€˜9'.

Therefore, each of the terms 91, 92 etc are all divisibly by â€˜9' and hence by â€˜3' but are odd numbers.

Any multiple of â€˜3' which is odd when divided by â€˜6' will leave a reminder of â€˜3'.

For example 27 is a multiple of â€˜3' which is odd. 27/6 will leave a reminder of â€˜3'. Or take 45 which again is a multiple of â€˜3' which is odd. 45/6 will also leave a reminder of â€˜3'.

Each of the individual terms of the given expression 91 + 92 + 93 + ...... + 99 when divided by 6 will leave a reminder of â€˜3'. There are 9 such terms. The sum of all the reminders will therefore be equal to 9*3 = 27.

27/6 will leave a reminder of â€˜3'. _________________

when there is a will there is a way

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