The sum of the digits of a positive integer N is 23. The remainder whe

Question

The sum of the digits of a positive integer N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?

A. 7
B. 13
C. 17
D. 16
E. 29

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GMATinsight · Answer

N = 11a+7

i.e. Possible values of N may be 18, 29, 30, 41, 52, 63, 74, 85, 96, 107, 118, 129, 140, 151, 162, 173, 184, 195 and so on

11a may be a 3 digit number such as 121, 242, 363,  979

N = 979+7 = 986

9+8+6 = 23  Satisfied!

986 when divided by 33 leaves remainder = 29

Answer: Option E

nick1816 · Answer

The sum of the digits of a positive integer N is 23.

N or sum of digits of N leaves same remainder when divided by 9. N leaves 5 remainder when divided by 9

N = 9k+5 = (9k+3)+2

N leaves 2 remainder when divided by 3 and 7 when divided by 11.

N = 33x + 7 or 33x + 18 or 33x+29

Only 29 leaves 2 remainder when divided by 3.

preetamsaha · Answer

23 can be expressed as a sum of one or two digits 
so, taking as 3 digits   =7 
so, 23 = 7+8+8 
The remainder when N is divided by 11 is 7. 
N= 11*80+7 = 887 
the remainder when N is divided by 33 = 29

correct option E

preetamsaha · Answer

Sum of digits = 23
Remainder when N is divided by 9
N/9 = 5 {Remainder on division by 9 for any number is equal to the remainder of dividing the sum of the digits of the number by 9} 
=> Remainder of N/3 = 2 {A number of the form 9k + 5 divided by 3 leaves a remainder 2} 
N = 11k + 7 
N = 3m + 2 
11k + 7 => Possible numbers are 7, 18, 29, 40, 51
3m + 2 => Possible numbers are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29

The number that is of the form 11k + 7 and 3m + 2 should be of the form 33b + 29. How did we arrive at this result?

The first natural number that satisfies both properties is 29.
Now, starting with 29, every 11th number is of the form 11k + 7, and every 3rd number is of the form 3m + 2. 
So, starting from 29, every 33rd number should be on both lists (33 is the LCM of 11 and 3). 
Or, any number of the form 33b + 29 will be both of the form 11K + 7 and 3m + 2, where b, k, m are natural numbers.
The remainder when the said number is divided by 33 is 29.

jacob46 · Answer

Hi,

Great Answer!
Could you please help me with the logic, I understood until "N leaves 2 remainder when divided by 3 and 7 when divided by 11."
But I could not understand logic of how below it started with 7, 18 and 29 as remainders when N is divided by 33
N = 33x + 7 or 33x + 18 or 33x+29

Thank you

siddhantvarma · Answer

The sum of the digits of a positive integer N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?

Here's a "lucky" way to solve this question because of the answer choices. 
We know,
N = 11a + 7
Also, N = 33b + r
Therefore we can say 11a + 7 = 33b + r => r = 11(a-3b) + 7
We know 11(a-3b) will be a multiple of 11 => 11(a-3b) = 11, 22, 33, 44, etc
Let's put 11(a-3b) = 11 => r = 11+ 7 = 18 (Not present in answer choices but this also eliminates A, B, C and D)
Putting 11(a-3b) = 22 => r = 22 + 7 = 29 (Option E)

I say "lucky" because the options really helped out here so I didn't need to use the information that the sum of the digits is 23. Ofc, this isn't a legitimate solution but for this specific problem and given the answer choices, it works.

The sum of the digits of a positive integer N is 23. The remainder whe

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The sum of the digits of a positive integer N is 23. The remainder whe

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