SOLUTION: Find the greatest number of 6 digit dividing by 15,18,21,24,27 leaves the remainder 10,13,16,19,and 22.

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Question 768603: Find the greatest number of 6 digit dividing by 15,18,21,24,27 leaves the remainder 10,13,16,19,and 22.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The remainders are always 5 short of the divisor.
For example, if the number, N is divided by 15, the remainder is 10, so that for some integer p,
N=15p%2B10-->N%2B5=15p%2B15=15%28p%2B1%29,
which makes N%2B5 a multiple of 15.
The same happens with 18, 21, 24, and 27.
That number, N, is such that N%2B5 is a multiple of 15, 18, 21, 24, and 27.
That means N%2B5 is a multiple of the least common multiple of
15=3%2A5,
18=3%2A2%2A3=2%2A3%5E2,
21=3%2A7,
24=3%2A2%5E3, and
27=3%5E3,
and that least common multiple is
2%5E3%2A3%5E3%2A5%2A7=7560
Dividing 1,000,000 by 7560 we find a quotient of 132 plus a remainder of 2080.
That means that 132%2A7560%2B2080=1000000-->132%2A7560=1000000-2080=997920
and 997,920 is a multiple of 7560.
The next multiple of 7560 is 133%2A7560=1005480
That means that 997920 is the largest N%2B5 multiple of 15, 18, 21, 24, and 27 that will yield a number N with 6 digits,
So N%2B5=997920 --> highlight%28N=997915%29.