Help, Please! You'll get points =D?

Question

When 823519 is divided by a nukber bigger than 1, the remainder is 3times the remainder by dividing 274658 by the same number. Please help me to work this out. =D

Please someone explain what "mod" means, and dont give an answer with "mod" in it unless you can tell me what "mod" is.

Answer 1

Here is the analytical solution:
First we assign variables: 'x' being the number we divide by...
'y' being the remainder of the division, and we use constants 'a' and 'b'. No we plot our equations:

ax+y=274,658
bx+3y=823,519

Then we 'add' them so to eliminate the 'y' variable, so:
3(ax+y)=3(274,658)
-1(bx+3y=823519(-1)

3ax+3y=823,974
-bx-3y=-823,519
------------------------------
3ax-bx=455
So:
(3a-b)x=455
x=455/(3a-b)

Since (3a-b) is an integer (otherwise the division wouldn't be exact) we then factorize 455. 455=13*7*5 and the solution can only be the product of two of those factors because the third factor is equal to (3a-b) So:
13*7=91
13*5=65
7*5=35
by substitution, you arrive to the answer '91'

Answer 2

I wrote a program to solve this:

Iterate i from 1 to 100,000.
For each value of i, calculate the two remainders:
a = Mod[823519, i]
b = Mod[274658, i]
If a = 3*b, print out i.

There is one answer, and it's 91.
823519 divided by 91 leaves a remainder of 60.
274658 divided by 91 leaves a remainder of 20.

Answer 3

The no. is 91.

Proof:
1st number:
823519 / 91 = 9049 remainder 60

2nd number:
274658 / 91 = 3018 remainder 20

The remainder in the 1st number is 3 times the remainder in the 2nd number.

Answer 4

91. I did this computationally. I would love to see an analytic solution.