To solve this problem, we need to apply the divisibility rules for 3 and 9.
We are looking for numbers that satisfy two conditions: they must be divisible by 3, AND they must NOT be divisible by 9.
Let's check each number from the list: 2142, 3423, 2430, 4521, 8433, 3258.
| Number | Sum of Digits | Divisible by 3? | Divisible by 9? | Satisfies Condition? (Divisible by 3 but not by 9) |
|---|---|---|---|---|
| 2142 | $2 + 1 + 4 + 2 = 9$ | Yes (9 is divisible by 3) | Yes (9 is divisible by 9) | No |
| 3423 | $3 + 4 + 2 + 3 = 12$ | Yes (12 is divisible by 3) | No (12 is not divisible by 9) | Yes |
| 2430 | $2 + 4 + 3 + 0 = 9$ | Yes (9 is divisible by 3) | Yes (9 is divisible by 9) | No |
| 4521 | $4 + 5 + 2 + 1 = 12$ | Yes (12 is divisible by 3) | No (12 is not divisible by 9) | Yes |
| 8433 | $8 + 4 + 3 + 3 = 18$ | Yes (18 is divisible by 3) | Yes (18 is divisible by 9) | No |
| 3258 | $3 + 2 + 5 + 8 = 18$ | Yes (18 is divisible by 3) | Yes (18 is divisible by 9) | No |
By examining the divisibility rules, we found that two numbers from the list (3423 and 4521) are divisible by 3 but not by 9. Therefore, the count is 2.
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Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero 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: