What is the remainder when 1271 * 1275 * 1277 *1285 is divid

This is nice solution.... Does it hold true for all cases?

Yes this will hold true always.

I have a doubt. 1275 is divisible by 3, cannot we reduce question to 1271*405*1277*1285 divided by 4 ?

No, we cannot do that. For example, 6 divided by 15 yields the remainder of 6 but if you reduce by 3, then the remainder of 2 divided by 5 yields the remainder of 2. As you can see the remainders are not the same.

Did in the same way, with slight difference:

1271 = 1200 + 71 ; similar for other numbers as well

So just ignored 1200 as its divisible by 12 & worked upon only the last 2 digits

71
75
77
85

Result obtained is the same

When a number is expressed as a product of numbers, the remainder is the product of remainders.

Lets say : \(N = a * b * c\)

Then \(R (\frac{N}{D})\) = \(R (\frac{a}{D}) * R (\frac{b}{D}) * R (\frac{c}{D})\)

where : \(R (\frac{N}{D})\) represents remainder N leaves when divided by D and so on.

Eg : Lets say 500 is expressed as the product of 3 numbers i.e 500 = 5*10*10 , and we are looking for the remainder when divided by 3.

\(R (\frac{500}{3}) = 2\)
\(R (\frac{5}{3}) = 2\)
\(R (\frac{10}{3}) = 1\)
\(R (\frac{10}{3}) = 1\)

We can clearly see that : \(R = R_1 * R_2 * R_3\)

Also when a number is expressed as the sum of numbers, the remainder is the sum of individual remainders:

If \(N = a + b + c\), then

\(R (\frac{N}{D})\) = \(R (\frac{a}{D}) + R (\frac{b}{D}) + R (\frac{c}{D})\)

Eg : Let 100 be expressed as the sum of numbers i.e 100 = 25 + 25 + 50

\(R (\frac{100}{3}) = 2\)
\(R (\frac{25}{3}) = 1\)
\(R (\frac{25}{3}) = 1\)
\(R (\frac{50}{3}) = 2\)

In this case the sum of remainders is equal to 4. Since we are looking for remainder when divided by 3, remainder of 4 means remainder of 1.

Cheers

Hope this helps .

Not clear on this point Bunuel. I think the OP is saying that the question can be reduced to finding the Nr when 1271*405*1277*1285 / 4.


Ok I get it . So if we know that there is going to be a remainder , we can not reduce it like this. Would this be a correct understanding ?

What I meant is the remainder of 21 divided by 15, won't be the same as the remainder of 21/3=7 divided by 15/3=5.

Hi Bunuel,
Any specific rule when doing this. (1272 - 1)(1272 + 3)(1284 - 7)(1284 + 1)
Because I can also write it as
(1272 - 1)(1272 + 3)(1272 + 5)(1284 + 1) which leads to incorrect answer.

Or do we need to make sure that product is +.

It would be easier if the last term is positive, though this approach would also give you the correct answer: the remainder when dividing -15 by 12 is also 9: -15=-2*12+9.

Check similar question here: when-51-25-is-divided-by-13-the-remainder-obtained-is-130220.html

Hope it helps.

I tried doing this exercise the following way but it seems I do not get the same result.. So obviously I am doing something wrong. Could somebody possible tell me why this is not correct?

I just used the last 2 digits of each number as these would be the most important

So: 71*75*77*85.

The last 2 digits are 25, this means that 12 still goes twice into this number. With 1 remaining.
Is this exercise simply not solved by doing this or am I doing it completely wrong?

Fair question. Tedious, but fair.

You'll need to divide each number separately by 12 and then multiply the remainders together and divide by 12 again.

1271/12 = 105 R 11
1275/12 = 106 R 3
1277/12 = 106 R 5
1285/12 = 107 R 1

(11x15)/12 = 13 R 9

Thus D is the correct answer.

\(1271*1275*1277*1285\)

Because hundreds and thousands digits (12) bring no remainder upon division by 12 we need only to pay attention to

\(71*75*77*85\)

\(71 = -1(mod 12), 75 = 3 (mod 12), 77 = 5 mod (12), 85 = 1 (mod 12)\)

And we have:

\((-1)*3*5*1 = -15 = -3 (mod 12) = 9 (mod 12)\)

Remainder is 9.

my approach:
(1260 + 11) * (1272 +3) * (1272 + 5) * (1284 +1)
11*3*5*1 = 165
165/12 = quotient is 13, and remainder 9.
answer is D, 9.

Here's a sneaky approach.

The product isn't even. So, when you divide it by 12, it can't have an even remainder (because 12 is even, so something with an even remainder would have to have been even to begin with.)

The product is a multiple of 3, because 1275 is a multiple of 3. So, when you divide it by 12, the remainder will be a multiple of 3. (That's because 12 is a multiple of 3. In order for a number to be divisible by 3, and also be of the form 12x + remainder, the remainder also has to be divisible by 3. If it wasn't, then the whole thing wouldn't be divisible by 3, either.)

The only possible remainders, then, are 3 and 9.

Only one of those - 9 - is in the answer choices. The answer is (D).

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.