Sum of the digits of a number is nothing but the remainder when the number is divided by 9. Go on, try this out. Now, try to prove this.
Remainders - Sum of digits
Q.1: The sum of the digits of a number N is 23. The remainder when N is divided by 11 is 7. What is the remainder when N is divided by 33?
7
29
16
13
Correct Answer
Choice B. 29
Explanatory Answer
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Detailed Solution
Sum of digits = 23
Remainder when
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.
Answer choice (B)
Correct Answer: 29
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