+121 If an integer ends in the digit $0$ and the sum of its digits is divisible by $3$, then how many of the numbers $2, 3, 4, 5, 6, 8, 9$ necessarily divide it?
+641 We can analyze this problem by considering the divisibility rules for 2, 3, 4, 5, 6, 8, and 9:
Divisibility Rules:
A number is divisible by 2 if the last digit is even (0, 2, 4, 6, or 8).
A number is divisible by 3 if the sum of its digits is divisible by 3. (Given in the problem statement)
A number is divisible by 4 if the last two digits are divisible by 4. (Not applicable here since the number ends in 0)
A number is divisible by 5 if the last digit is a 0 or a 5.
A number is divisible by 6 if it's divisible by both 2 and 3.
A number is divisible by 8 if the last three digits are divisible by 8. (Not applicable here since the number ends in 0)
A number is divisible by 9 if the sum of its digits is divisible by 9. (This is similar to the divisibility rule for 3, but a stronger condition)
Analysis:
Ending in 0: Since the number ends in 0, it is automatically divisible by 2 and 5 (based on the divisibility rules).
Sum of Digits Divisible by 3: The problem states that the sum of the digits is divisible by 3. This means the number itself must also be divisible by 3 (based on the divisibility rule for 3).
Divisibility by Other Numbers:
Looking at the remaining numbers (4, 6, 8, and 9), we can see that:
4: Not necessarily divisible. Just because the last digit is 0 (divisible by 2), it doesn't guarantee divisibility by 4 (requires the last two digits to be divisible by 4).
6: Since the number is divisible by both 2 (ending in 0) and 3 (given in the problem statement), it is automatically divisible by 6.
8: Not necessarily divisible. Similar to 4, just because the number ends in 0 (divisible by 2), it doesn't guarantee divisibility by 8 (requires the last three digits to be divisible by 8).
9: Not necessarily divisible. The sum of digits being divisible by 3 is a weaker condition than being divisible by 9. So, a number divisible by 3 might not necessarily be divisible by 9.
Conclusion:
Therefore, a number ending in 0 with a sum of digits divisible by 3 is necessarily divisible by 2, 3, and 6 (out of the given numbers). There are 3 such divisors.
