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

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

Opps! Really a hard one. I cannot figure out how to quickly solve it!! Please help someone.

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: 30

I believe there is a easier solution...

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.

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)

Thanks, this is indeed a easier way. It also seems to work for any number

Still a bit long solution......

OA is C.

Any OE? Coz answer C is apparent but how to solve is the key here.

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

Wow!! Everyday I am learning new thing is this forum. Thanks.

useful insight..
thanks for the good question.

good one . this can be extended to the SUM as well.

thanks