# Linear congruence

1. May 2, 2012

### Roodles01

I have a number with which I must use rules of congruence to find the remainder.
rules I must apply are;
split every 3 digits, starting from the right,
find the remainder of each 3 digit number on division by 7
form alternating sum of these remainders.

Thisnumber should be congruent to a modulo 7

the number is 2468135711201104

number 2 468 135 711 201 104
remain 2 6 2 4 5 6

Alternating sum 6-5+4-2+6-2 = 7
So a = 7 (mod 7)

I have gone through this several times with the same result
Would this be right?

Dividing through by 7 only to have a remainder of 7 seems a bit odd!!!

2. May 2, 2012

### tiny-tim

Hi Roodles01!
a is congruent to 7 mod 7 …

(btw, also works for remainders on division by 11 or 13 )

3. May 2, 2012

### HallsofIvy

Staff Emeritus
Not only is it "odd" to have a remainder of 7 when dividing by 7, it is impossible! What you are saying is that the remainder is 0. The calculation you have done is a test for divisibility by 7. Yes, 2468135711201104 is divisible by 7. The hard way to do that would have been to actually divide by 7: 2468135711201104 is 7 times 352590815885872!

4. May 2, 2012

### Curious3141

You've shown that a is divisible by 7, because a = 7 (mod 7) is equivalent to a = 0 (mod 7).

This is one of the tests for divisibility by 7. To see how it works, consider a = 1000x + y.

Now 1000 = 6(mod 7) = -1 (mod 7).

Hence 1000x = -x(mod 7).

So 1000x + y = (y - x) (mod 7).

Therefore splitting a large number into groups of threes ("thousands"), working out the residues modulo 7 and affixing alternate signs to them before summing allows a quick way to calculate the remainder upon division by 7.