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phantom

I agree with rossim - why not 5? so 499

Remainder is 4 less than each divisor

Just expanding slightly on what phantom says, for a number to leave a remainder of 3 when divided by 7, it must be (7-3), ie 4, lower than an exact multiple of 7 (so 3, 10, 17...); likewise for a number to leave a remainder of 4 when divided by 8, it must be (8-4), ie 4, lower than an exact multiple of 8 (so 4,12,20...); similarly for 9, to leave 5 the number must be (9-5) lower than an exact multiple of 9 (5,14, 23...)

The lowest number which leaves the required remainders is therefore (7x8x9) - (7-3) [or (8-4), or (9-5)]

Spoffy

Thanks for the explanation. Very neat.

Thanks Phantom and Grunger.

... and to expand slightly on Spoffy's conclusion - "The lowest number which leaves the required remainders is therefore (7x8x9) - (7-3) [or (8-4), or (9-5)]", this is true only because the LCM of 7,8 and 9 is 7 x 8 x 9.

Replacing the " a remainder of 3 when it is divided by 7" with " a remainder of 2 when it is divided by 6" would require the lowest solution to be 4 less than the LCM of 6,8 and 9... which is 68

...yes indeed, where the divisors share a common factor other than unity those factors would need to be eliminated prior to multiplying up.

Spoffy, Tatters

Thanks for your postings which were very interesting to read.

Spoffy tatters

Thank you. I have been looking at this interesting problem. I think what you are saying is that the original problem could be expanded by "a remainder of 2 when it is divided by 6". So it is 500 for all divisors 6 to 9.

But why does it not work for 5 or 10?