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

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Director
Joined: 14 Jan 2007
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What is the remainder when 1044*1047*1050*1053 is divided by  [#permalink]

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20 May 2007, 17:19
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45% (02:09) correct 55% (02:08) wrong based on 260 sessions

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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
Math Expert
Joined: 02 Sep 2009
Posts: 48037

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24 Oct 2009, 00:22
6
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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

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

I believe there is a easier solution...
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22 May 2007, 09:29
6
4
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.
##### General Discussion
VP
Joined: 08 Jun 2005
Posts: 1139

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20 May 2007, 21:24
3
2
vshaunak@gmail.com wrote:
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Can somebody suggest a trick to solve this question.

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)

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Joined: 05 Jun 2003
Posts: 47

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21 May 2007, 13:26
KillerSquirrel wrote:

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

Thanks!
Manager
Joined: 17 May 2007
Posts: 69

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21 May 2007, 13:51
1
umbdude wrote:
KillerSquirrel wrote:

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

15+18+19+20+21 = 93

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
VP
Joined: 08 Jun 2005
Posts: 1139

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21 May 2007, 14:38
1
umbdude wrote:

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

15+18+19+20+21 = 93

Thanks!

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.

VP
Joined: 08 Jun 2005
Posts: 1139

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22 May 2007, 10:41
vshaunak@gmail.com wrote:
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.

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 !!!

Manager
Joined: 11 Aug 2008
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24 Oct 2009, 07:35
1
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
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Posts: 48037

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24 Oct 2009, 09:22
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.
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24 Oct 2009, 10:51
1
1
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

Intern
Joined: 13 Jul 2009
Posts: 19

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26 Oct 2009, 19:52
2
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?
Manager
Joined: 15 Sep 2009
Posts: 118

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27 Oct 2009, 00:49
1
1
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
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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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23 Aug 2013, 03:46
rlevochkin wrote:
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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23 Aug 2013, 09:31
1
1
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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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23 Aug 2013, 10:19
vabhs192003 wrote:
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.

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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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15 Dec 2013, 23:58
vshaunak@gmail.com wrote:
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Can somebody suggest a trick to solve this question.

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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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16 Dec 2013, 00:05
WoundedTiger wrote:
vshaunak@gmail.com wrote:
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Can somebody suggest a trick to solve this question.

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

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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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16 Dec 2013, 01:49
mau5 wrote:
WoundedTiger wrote:
vshaunak@gmail.com wrote:
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Can somebody suggest a trick to solve this question.

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

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?

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.

Is that correct...and can it be generalized for such problems.
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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is  [#permalink]

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16 Dec 2013, 02:59
1
WoundedTiger wrote:

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.

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

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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Re: What is the remainder when 1044 * 1047 * 1050 * 1053 is &nbs [#permalink] 16 Dec 2013, 02:59

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