Remainders

rlevochkin · **Posted:** Fri Oct 23, 2009 2:04 pm


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

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

Bunuel · **Posted:** Fri Oct 23, 2009 11:22 pm

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...

hariharakarthi · **Posted:** Mon Oct 26, 2009 6:52 pm

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?

eresh · **Posted:** Fri Oct 23, 2009 10:35 pm

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

ngoctraiden1905 · **Posted:** Sat Oct 24, 2009 6:35 am

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

Bunuel · **Posted:** Sat Oct 24, 2009 8:22 am

ngoctraiden1905 wrote:

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.

Economist · **Posted:** Sat Oct 24, 2009 9:51 am

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)

eresh · **Posted:** Sat Oct 24, 2009 8:02 pm

Economist wrote:

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 :lol:

Still a bit long solution......

rlevochkin · **Posted:** Sat Oct 24, 2009 9:46 pm

OA is C.

eresh · **Posted:** Sat Oct 24, 2009 11:41 pm

rlevochkin wrote:

OA is C.

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

deepakraam · **Posted:** Mon Oct 26, 2009 11:49 pm

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

eresh · **Posted:** Mon Oct 26, 2009 11:57 pm

deepakraam wrote:

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.

vibhaj · **Posted:** Tue Oct 27, 2009 8:54 pm

useful insight..
thanks for the good question.

gmatcracker1 · **Posted:** Sun Nov 08, 2009 7:13 pm

smtripathi wrote:

I guess, we can prove this by taking two numbers in general format :

N1=dQ1+R1 ......1
N2=dQ2+R2 ......2

so, if we multiply them to find the remainder, it will become :
N1*N2=d^2*Q1Q2+dQ2*R1+dQ1*R2+R1*R2
or, =d[d*Q1*Q2+Q2*R1+Q1*R2] + R1*R2
so, we have remainder R1*R2 , but this may be bigger than d, so we need to rationalize R1*R2 to get the actual remainder.

I tend to remember things if I prove them .. so thought of adding my two cents.

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

thanks

amit2k9 · **Posted:** Thu May 12, 2011 6:37 am

good method by Bunuel.

Remainders

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