Question

a number leaves a remainder of 1 in each case when divided by 2, 3, 4, 5, 6. Find the number if it is divisible by 7.

1. thongdat2
301
Step-by-step explanation:
The desired number will be a multiple of 7 that is 1 more than some multiple of the least common multiple of 2, 3, 4, 5, and 6. The value can be found by trial and error, or by using inverses modulo 7.

### LCM

The least common multiple of 2, 3, 4, 5, 6 will be the product of their prime factors, each to the highest power:
2 = 2; 3 = 3; 4 = 2²; 5 = 5; 6 = 2·3
Then the LCM is 2²·3·5 = 60.

### Divisible by 7

For some multiple n of 60, we want …
(60n +1) mod 7 = 0
This is equivalent to …
4n mod 7 = 6
The solution to this can be found by trial and error, or by making use of the mod-7 inverse of 4 (which is also found by trial and error).
Here are a few trials:
4·1 mod 7 = 4
4·2 mod 7 = 1 . . . . . . this tells us the mod 7 inverse of 4 is 2
4·3 mod 7 = 5
4·4 mod 7 = 2
4·5 mod 7 = 6 . . . . . . the value of n we’re looking for is 5
That is …
(60·5 +1) mod 7 = 301 mod 7 = 0
The (smallest positive) number with the division properties we’re looking for is 301.
__