Which of the following numbers are divisible by 2, 3 and 5?
2345760
Understanding Divisibility Rules
To find a number that is divisible by 2, 3, and 5 simultaneously, we need to check each option against the divisibility rules for these numbers. A number is divisible by 2, 3, and 5 if and only if it satisfies the conditions for all three rules.
Divisibility Rule for 2
A number is divisible by 2 if its last digit is an even number (0, 2, 4, 6, or 8).
Divisibility Rule for 5
A number is divisible by 5 if its last digit is either 0 or 5.
Divisibility Rule for 3
A number is divisible by 3 if the sum of its digits is divisible by 3.
Checking the Options
Let's apply these rules to each given number:
Analyzing Option 1: 5467760
- Divisibility by 2: The last digit is 0, which is even. So, 5467760 is divisible by 2.
- Divisibility by 5: The last digit is 0. So, 5467760 is divisible by 5.
- Divisibility by 3: The sum of digits is $5 + 4 + 6 + 7 + 7 + 6 + 0 = 35$. Since 35 is not divisible by 3, 5467760 is not divisible by 3.
Option 1 is not divisible by 3, so it is not divisible by 2, 3, and 5 simultaneously.
Analyzing Option 2: 1345678
- Divisibility by 2: The last digit is 8, which is even. So, 1345678 is divisible by 2.
- Divisibility by 5: The last digit is 8. Since it is not 0 or 5, 1345678 is not divisible by 5.
Option 2 is not divisible by 5, so it is not divisible by 2, 3, and 5 simultaneously.
Analyzing Option 3: 2345760
- Divisibility by 2: The last digit is 0, which is even. So, 2345760 is divisible by 2.
- Divisibility by 5: The last digit is 0. So, 2345760 is divisible by 5.
- Divisibility by 3: The sum of digits is $2 + 3 + 4 + 5 + 7 + 6 + 0 = 27$. Since 27 is divisible by 3 ($27 \div 3 = 9$), 2345760 is divisible by 3.
Option 3 is divisible by 2, 5, and 3. Therefore, it satisfies the condition.
Analyzing Option 4: 2456732
- Divisibility by 2: The last digit is 2, which is even. So, 2456732 is divisible by 2.
- Divisibility by 5: The last digit is 2. Since it is not 0 or 5, 2456732 is not divisible by 5.
Option 4 is not divisible by 5, so it is not divisible by 2, 3, and 5 simultaneously.
Conclusion
Based on the analysis of each option using the divisibility rules for 2, 3, and 5, only the number 2345760 is divisible by all three numbers.
| Number | Divisible by 2 (Last digit) | Divisible by 5 (Last digit) | Divisible by 3 (Sum of digits) | Divisible by 2, 3 & 5? |
|---|---|---|---|---|
| 5467760 | Yes (0) | Yes (0) | No ($35 \div 3$) | No |
| 1345678 | Yes (8) | No (8) | N/A | No |
| 2345760 | Yes (0) | Yes (0) | Yes ($27 \div 3$) | Yes |
| 2456732 | Yes (2) | No (2) | N/A | No |
Revision Table: Divisibility Rules Recap
| Number | Rule | Condition |
|---|---|---|
| 2 | Last digit | Must be 0, 2, 4, 6, or 8. |
| 3 | Sum of digits | Must be divisible by 3. |
| 5 | Last digit | Must be 0 or 5. |
Additional Information: Combined Divisibility
If a number is divisible by two or more numbers that are coprime (meaning their greatest common divisor is 1), then the number is also divisible by the product of those numbers.
- The numbers 2, 3, and 5 are pairwise coprime:
- GCD(2, 3) = 1
- GCD(2, 5) = 1
- GCD(3, 5) = 1
- Since 2, 3, and 5 are pairwise coprime, a number divisible by 2, 3, and 5 must also be divisible by their product, which is $2 \times 3 \times 5 = 30$.
- A number is divisible by 30 if and only if it is divisible by both 10 and 3.
- Divisibility by 10: The last digit must be 0. This ensures divisibility by both 2 and 5.
- Divisibility by 3: The sum of the digits must be divisible by 3.
So, an alternative way to solve this problem is to check which number ends in 0 and has a sum of digits divisible by 3. Looking at the options, only 5467760 and 2345760 end in 0. We already calculated their sums of digits as 35 and 27, respectively. Only 27 is divisible by 3, confirming that 2345760 is the correct number.
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