SOLUTION: When 391758 and 394915 are divided by a certain three digit number, the three digit remainder is the same in each case. Find the divisor.

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Hi,
Let d be the divisor and r be the remainder. We know that for some integer and

and

Subtracting these gives

We also know that d is three digits. There are only two three digit factors of 3157, and they are 287 and 451. Substituting both of them into our above equations shows that 451 is the one we're after giving a remainder of 290.
Hope that helps,
Kev

You can put this solution on YOUR website!

When 391758 and 394915 are divided by a certain three digit number,
the three digit remainder is the same in each case. Find the divisor.

Let D = the required divisor
Let Q1 = quotient when 391758 is divided by D
Let Q2 = quotient when 394915 is divided by D
Let R - the remainder in each case

Then
391758 = Q1·D + R
394915 = Q2·D + R

Subtracting the 1st equation from the 2nd

3157 = Q1·D - Q2·D

3157 = D(Q2 - Q1)

D = 3157/(Q2 - Q1)

So Q2 - Q1 has to be a factor of 3157

3157 = 7×11×41

So the factors of 3157 are 1, 7, 11, 7×11, 7×41, 11×41, and 7×11×41

So Q2-Q1 is one of these: 1, 7, 11, 77, 287, 451, and 3157

So D = 3157, 451, 287, 41, 11, 7, or 1

Only two of these have three digits, 287 and 451.

When 391758 and 394915 are divided by 287, the remainder is 3 in both cases.

But the remainder must be a 3-digit number

When 391758 and 394915 are divided by 451, the remainder is 290 in both cases.

That's a 3-digit remainder. so 451 is the answer.

Edwin
AnlytcPhil@aol.com