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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Surprise, guys.
OA is 'C'.

OE:

take the remainder from each of 1044/33, 1047/33 and so on..

1044/33 gives remainder = 21
1047/33 gives remainder = 24
1050/33 gives remainder = 27
1053/33 gives remainder = 30

the net remainder is the product of above individual remainders. i.e = 21*24*27*30
break them into pairs 21*24/33 gives remainder 9
and 27*30/33 gives remainder 18

so 9*18/33 gives remainder 30.

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)

I don't understand why you can the following operation (adding the remainders):

15+18+19+20+21 = 93

Would you please explain?

Thanks!

(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

just ignore the numbers that are divisible by 33 to find the remainder !

don't think about it mathematically, apply logic.

also see gowani post.

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 !!!

I think it's easier: because all of the term divisible by 3 so the remainder of the multiply of them will be the remainder when after dividing by 3 then dividing by 11. This can only be 3 because all other choice is greater than 11

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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To find the remainder, we need to first find what are the remainders when 1044, 1047, 1050 and 1053 are divided by 33 individually.

1044/33=31.......21
1047/33=31.......24
1050/33=31.......27
1053/33=31.......30

Now we need to find the remainders when 21*24 and 27*30 are divided by 33.

21*24/33=15......9
27*30/33=24......18

Finally we need to find the remainder when 9*18 is divided by 33:

9*18/33=4........30

So the answer is (C)

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?

calculate the remainder when 1044 is divided by 33 = 21

so the eqn becomes 21*24*27*30 / 33

= 27*27*21/33

considering negative remainders we get -6*-6*21/33

= 36*21/33

= 3*21/33

= 63/33

= 30.

I will go with option C

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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Though I took a while to solve this but I was trying to recollect something which I learnt when I was in love with Remainder theorem.

(1044 x 1047 x 1050 x 1053)/33
(cancelling the common factor 3)
=>(1044 x 1047 x 1050 x 351)/11
Denoting remainder using []
[1044/11]= [10] is remainder. Similarly,
[1047/11]= [2]
[1050/11]= [5]
[351/11]= [10]
which gives: [(10 x 10 x 10)/11] = [10] as remainder.

Since we initially cancelled the common factor 3. The final remainder will be 10 x 3 =30 i.e. option C.
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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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Hi,

Math Experts....need help on this one....

We need to find the remainder in the above case. I started by finding whether any term is divisible by 33 and found the nearest multiple to be 1056 and changed the question to

(1056-12)*(1056-9)*(1056-6)*(1056-3)/33 which can be further reduced to

(-12)(-9)(-6)(-3)/ 33
On simplifying further we get -------> -12*-9*-6*-1/11-------> 648/11 ---Remainder 10.....its not even in the answers choices...

please suggest what's wrong with my approach
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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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Hmmm...got the point...So do we need to simplify the given expression only till
(-12)(-9)(-6)(-3)/ 33 and find out the answer. Basically not touching the denominator.

the answer will be 30.

Is that correct...and can it be generalized for such problems.
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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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