Iam an even multiple of 2,if you reverse my digits ,I become a multiple of 9.who am I?
By an even multiple of 2, do you mean a multiple of 4? As in 6 is an odd (3) multiple of 2, but 24 is an even (12) multiple of 2?
In any case, just pick any multiple of 2 or 4, as required, whose digits sum to 9.
Such as 72, 234, 69732, etc.
Read about "casting out nines".
18
To solve this riddle, let's break it down step by step:
1. We know that the number is an even multiple of 2. This means that the last digit must be an even number (0, 2, 4, 6, or 8).
2. If we reverse the digits of the number and it becomes a multiple of 9, this means that the sum of the digits must be a multiple of 9.
3. Let's try different even numbers for the last digit and reverse their digits while checking if their sum is a multiple of 9:
- If the last digit is 0, then reversing it would still be 0. But 0 is not a multiple of 9, so it doesn't satisfy the condition.
- If the last digit is 2, then reversing it would be 2. The sum of 2 and 2 is 4, which is not a multiple of 9, so it doesn't satisfy the condition.
- If the last digit is 4, then reversing it would be 4. The sum of 4 and 4 is 8, which is not a multiple of 9, so it doesn't satisfy the condition.
- If the last digit is 6, then reversing it would be 6. The sum of 6 and 6 is 12, which is not a multiple of 9, so it doesn't satisfy the condition.
- If the last digit is 8, then reversing it would be 8. The sum of 8 and 8 is 16, which is not a multiple of 9, so it doesn't satisfy the condition.
After trying all the even numbers for the last digit, none of them satisfy the condition. Therefore, there is no number that is an even multiple of 2 and becomes a multiple of 9 when its digits are reversed.