Divising problem, people who like challenges!!?

Question

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.

Yeah, could someone explain to me how to work it out???


Like, I got as far as

3153=quotient(1st equation)*divisor - quotient(2nd equation)*divisor

if you don't understand that, take no notice!! But yeah. Thanks to all who help!!

Answer 1

Here goes...

Let D = The Divisor.
Let Q1 = Quotient when 391758 is divided by the divisor.
Let Q2 = Quotient when 394915 is divided by the divisor.
Let R = The Remainder.

The equations would be...

391758 = Q1 * D + R
394915 = Q1 * D + R

Now we'll subtract the first equation from the second...

3157 = Q1 * D - Q2 * D
3157 = D(Q2 - Q1) /* All I did here was factor out the D*/
D = 3157/(Q2 - Q1) /* Solve for D */

This tells us that 3157 MUST be a factor of Q2 - Q1.

3157 = 7 * 11 * 41

So its factors are 1, 7, 11, 7 * 11, 11 * 41, and 7 * 11 * 41.

This means that Q2 - Q1 MUST be one of the following: 1, 7, 11, 77, 287, 451, and 3157.

Only 2 of these numbers have 3 digits: 287 and 451. So try dividing both 391758 and 394915 by those numbers. If you do so, you find out that when you divide both 391758 and 394915 by 451 the remainder is 290 in both cases. 290 is a 3-digit number.

Therefore, your solution is 451.

Answer 2

Take the difference between 394915 - 391758 = 3157

We know that 3157 must be a multiple of the 3-digit divider because only this will yield the same 3-digit remainder for the 2 big numbers.

Factor out 3157 = 7 X 11 X 41

Since the divider must be 3 digit, it must be either
7 X 41 = 287 or
11 X 41 = 451

Divide 391758 by 287, the remainder is 3, not a 3-digit number
Divide 391758 by 451, the remainder is 290, a 3-digit number
Divide 394915 by 451, remainder is also 290

Therefore, the answer is 451

Answer 3

451.

For the numbers to have the same remainder when divided by a third number (which we'll call N), the difference between those numbers must be divisible by N. So N must be a factor of (394915-391758), which is 3157. The prime factorization of 3157 is (7, 11, 41), so any factors must be the result of multiplying some subset of those numbers together. The complete list of factors then, is (1, 7, 11, 41, 77, 287, 451, 3157). Since N has to be a three-digit number, we know that N must be either 287 or 451. However, dividing either of the original numbers by 287 yields a one-digit remainder, so N must be 451, and it is.

Edit: and I just got beaten by stormmedic85. Darn.

Answer 4

287 (remainder of 3)
451 (remainder of 290)

Used an excel spreadsheet to work it out by dividing the numbers by everything from 100 through to 999.

Answer 5

Divide both of them with 451, your remainder is 290.

Work it out using Excel or write a program to try all possible numbers.

Answer 6

not sure if you are looking for a pure analytical solution, but I found one with the help of Mathematica. First, let the circle be the unit circle centered at the origin. Place the first point at (1,0). Now moving counter clockwise around the circle label the first 3 angles as x,y, and z. Thus the 4th w angle is given by w = 360-x-y-z. Now the sample space is easily defined by the triple integral as the integral of 1 with x ranging from 0 to 360, y ranging from 0 to 360-x, and z ranging from 0 to 360-x-y. Now this works out to 7,776,000 now for part A) define the function F(x,y,z) as being 1 if Max(x,y,z,360-x-y-z)>=180 and 0 otherwise then we simply want to find the integral of F(x,y,z) with the same boundaries as above and this works out to be 3,888,000 and thus the probability is 3,888,000/7,776,000 = 1/2 thus the odds are 50% part B) now define F(x,y,z) as being 1 if Max(x,y,z,360-x-y-z) >= 3*Min(x,y,z,360-x-y-z) and again integrate this as before to get 6,998,400 and thus the probability is 6,998,400/7,776,000 = 9/10 thus the odds are 90% edit: I agree that intuition goes against my answers, yet I fail to find a flaw in my calculations. As for your remarks I make no assumptions about which of x,y,z,w is the smallest in either cases. I simply look at all possible values of x,y,z,w. My integral boundaries are based on the simple fact that x+y+z+w=360 thus we have 0

Answer 7

I know people have already beaten me to the answer, but I managed to do it without looking first, so I'm quite pleased with myself. I used the same method as the intelligent people (not the person who just tried every number in excel).

Answer 8

It may be possible to apply the Chinese remainder theorem to this, but I am not familiar enough with it to deal with the details.

Answer 9

I agree with all of the smart people shown above. (only doing A level maths first year and feels a little stupid)

Answer 10

451.