What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18

This one really needs some calculations and I don't know how it can be done more quickly:

Let's find the closest multiple of 33 to these numbers: it's 1056.

(1056-12)(1056-9)(1056-6)(1056-3) every term after simplification will have 1056 as its multiple except the last one which will be 12*9*6*3.

So the remainder will be the same when 12*9*6*3 is divided by 33.

The same way here 12*6*3=216 remainder when divided by 33 =18,
9*18=162 remainder=30

Done

Answer: C.

I believe there is a easier solution...
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one trick is to break the number into workable parts.

you can tell that all 1044,1047,1050,1053 are all dividable by 3

since 1+0+4+4 = 9
and 1+0+4+7 = 12

so

348*3*349*3*350*3*1053*3

81*(33*10+18)*(33*10+19)*(33*10+20)*(33*10+21)

(33*2+15)*(33*10+18)*(33*10+19)*(33*10+20)*(33*10+21)

15+18+19+20+21 = 93

33*2+27 = 93

the remainder is 27 (B)

(33*2+15)*(33*10+18)*(33*10+19)*(33*10+20)*(33*10+21)

(33*2) can be cancelled out since it is multiplied for all the sets

hence you only have 15, 18, 19, 20, and 21 left

well - I can only find comfort in the words of Sir Winston Churchill:

"Success is the ability to go from one failure to another with no loss of enthusiasm"

good question! and good official explanation!

actually, after reading umbdude post I had my doubts. But now I know why !!!

That's not right:

Consider this 6*6*6 divided by 33. What is remainder?

According to you logic: as all of them are divisible by 3 remainder must be less than 11, but in this case remainder is 18>11.

For the original question the answer IS 30. But my point was that the solution I provided is not easy, so I wonder if there is some easier way to do the same.
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Key to this remainder problem is the following properties,
you can multiply the remainders as long as you correct the excess remainders.

1044/33 yields R21
1047/33 yields R24 (Don't need to calculate as 1047 is 3 integers away from 1044)
1050/33 yields R27
1053/33 yields R30

Now, multiply remainders and correct the excess remainders.

R21* R24 * R27 * R30
Solve the problem till you get a remainder less than 33.

since 1044 and 33 are big numbers, it is possible to make arithmetic mistake here. Is there away to solve this problem
efficeintly?

We can play with the questions the way we like to.... That's the beauty of remainder questions.

When 1044, 1047, 1050, and 1053 divided by 33 individually will give the remainders as 21, 24, 27, and 30 respectively.

Now as per rule, remainders are always non-negative, but still we can consider the negative remainders for the calculation as long as we convert them is positive remainder at the end.

So 21, 24, 27, and 30 when divided by 33 will give the negative remainders as -12, -9, -6, -3

(-12 * -9)(-6 * -3) --------> 108 * 18 ------->Positive Remainder -----> 9 * 18 --------> 162/33 ------Negative Remainder -----> -3 -------> Positive Remainder -----> -3 + 33 = 30 (Always add the divisor in to the negative remainder to obtain positive(correct) remainder)
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Yeah, This is also a fantastic method and, if I remember correctly, has been discussed in Mr. Arun Sharma's CAT QA book
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I will give you a hint, because you really have got it. Just one mistake ;

Remainder of \(\frac{14}{4} = \frac{7}{2} = 1\) Is this correct?
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Actually you just have to multiply the remainder you got by the common factor which you cancelled.

For example, \(\frac{14}{4}\), here you factored out 2. Keep that seperate. Now, remainder of \(\frac{7}{2} = 1\). Thus, the final remainder is 1*2 = 2.

Apply this to your method above and get the same, i.e. 3*10 = 30.
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