Question

find the greatest number of 6 digit which on dividing by 15 18 21 24 and 27 leaves 10,13,16, 19 and 22 as remainders​

Answer 1

Answer:

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.

Explanation:

hope it helps