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Consider all 3−digit numbers (without repetition of digits) obtained using three nonzero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
  1. S is always divisible by 74.
  2. S is always divisible by 9.
select the correct answer using the code given below:
  • a)
    1 only
  • b)
    2 only
  • c)
    Both 1 and 2
  • d)
    Neither 1 nor 2
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
Consider all3−digit numbers (without repetition of digits) obtai...
Given,
Digits are non-zero and multiple of 3 i.e., 3, 6, 9.
Number doesn't have repetition of digits.
The numbers thus formed are
369, 396, 639, 693, 936, 963
Sum of these numbers =3996 and it is divisible by both 74 as well as 9.
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Community Answer
Consider all3−digit numbers (without repetition of digits) obtai...

Explanation:

Sum of 3-digit numbers with three nonzero digits
- The three nonzero digits which are multiples of 3 are 3, 6, and 9.
- We can form 6 numbers with these digits: 369, 396, 639, 693, 936, 963.

Calculating the sum (S)
- Sum of all these numbers is S = 369 + 396 + 639 + 693 + 936 + 963 = 3996.
- S = 3996, which is divisible by 74 and 9.

Verification of Divisibility
- To check if S is divisible by 74, calculate S % 74. If the remainder is 0, then it is divisible by 74.
- 3996 % 74 = 0, so S is divisible by 74.
- To check if S is divisible by 9, calculate the sum of the digits of S. If the sum is divisible by 9, then S is also divisible by 9.
- Sum of digits of 3996 = 3 + 9 + 9 + 6 = 27, which is divisible by 9.

Conclusion
- Both statements are correct.
- Therefore, the correct answer is option C - Both 1 and 2.
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