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by sanju09 » Tue Feb 24, 2009 6:07 am
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18
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by willbeatthegmat » Tue Feb 24, 2009 6:31 am
What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Ans

there is a shortcut to deal with such questions..
when 1044/33..remainder is 21
1047/33...remainder is 24
1050/33...remainder is 27
1053/33....remainder is 30

now, take a pair of num..at a time
21*24/33 .... remainder is 9
27 * 30/33...remainder is 15

15* 9/33 ....remainder is 3

Therefore, a)3 is the ans.....
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by sureshbala » Tue Feb 24, 2009 6:43 am
Folks, we know that 33 = 11x3.

Clearly the given number is divisible by 3.

Also the remainder when 1044 is divided by 11 is -1 (take -ve remainder to make speed up ur calculation.)

So the remainder when 1044 * 1047 * 1050 *1053 is divided by 11 = -1 * 2 * 5 * 8 = -1 *10*8 (here again instead of 10 you can take the remainder as -1).

So we can conclude that the remainder is 8.

Now your option should satisfy both the conditions namely, it must be exactly divisible by 3 and it must leave a remainder 8 when divided by 11.

From the given choices it is 30
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Re: divided by 33

by billzhao » Tue Feb 24, 2009 7:10 am
sanju09 wrote:What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

A. 3
B. 27
C. 30
D. 21
E. 18
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)
Yiliang
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sorry 27 * 30/33...remainder is 18...ans is 30

by willbeatthegmat » Tue Feb 24, 2009 7:38 am
willbeatthegmat wrote:What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33?

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

Ans

there is a shortcut to deal with such questions..
when 1044/33..remainder is 21
1047/33...remainder is 24
1050/33...remainder is 27
1053/33....remainder is 30

now, take a pair of num..at a time
21*24/33 .... remainder is 9
27 * 30/33...remainder is 15

15* 9/33 ....remainder is 3

Therefore, a)3 is the ans.....
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by naaga » Tue Feb 24, 2009 9:25 am
clear explanation billzhao, thankyou.
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by GID09 » Tue Feb 24, 2009 10:37 am
The number divisible by 33 in the given range of numbers is 1056. Hence rewriting the equation as (1056-12)*(1056-9)*(1056-6) *(1056-3)

Now we know anything multiplied with 1056 will leave zero remainder.

Hence 12*9*6*3/33 = 1944/33 ...remainder is 30. Answer C.
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by x2suresh » Tue Feb 24, 2009 11:56 am
GID09 wrote:The number divisible by 33 in the given range of numbers is 1056. Hence rewriting the equation as (1056-12)*(1056-9)*(1056-6) *(1056-3)

Now we know anything multiplied with 1056 will leave zero remainder.

Hence 12*9*6*3/33 = 1944/33 ...remainder is 30. Answer C.
Nice work.
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