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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?


1. 19
2. 29
3. 9
4. None of the Above
5.

1 Answer

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Correct Answer - Option 2 : 29

Calculation:

⇒ 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}

 The remainder of N/3 = 2 {The 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. 

⇒ 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 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.

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