math puzzle: i am less than 3000. all four digits are odd. all four digits are different. the sum of my tens digit and ones digit is 16. i am divisible b 13. who am i?
Ans: the thousand digit must be 1, the hundreds digit is 3 or 5. The tens and ones digits are 9,7 or 7, 9. So I guess the number is 1397, 1379, 1597 or 1579, but none is divisible by 13. Where did I go wrong?
I am having the same trouble and do not see the trick.
The thousand digit can only be 1
the hundreds digit can be 2,5,7,or 9
You only considered 3 or 5
How about 1937 ?
It satisfies all the conditions
The tens and the ones add to 16 ?
Thanks Damon, you got me.
Time for my third coffee.
Agree with Damon.
The only 4 digit number for which
1. all digits are odd,
2. sum of last two digits equal 16
3. divisible by 13
are
7397, 9997, 7579
each of which does not satisfy the requirements
4. digits being distinct,
5. number is less than 3000.
However, if the divisibility requirement (3) had been divisible by 11, then among
3179, 5379, 7579, 9779,
1397, 3597, 5797, 7997
1397 satisfies all 5 requirements.
To solve this math puzzle, you are on the right track. However, there seems to be a small mistake in your approach. Let's break down the puzzle again and identify where you might have gone wrong.
The given conditions are:
- The number is less than 3000.
- All four digits are odd.
- All four digits are different.
- The sum of the tens digit and ones digit is 16.
- The number is divisible by 13.
Based on these conditions, let's go through the steps again:
1. The thousand digit must be 1, as the number is less than 3000.
2. The hundreds digit can be either 3 or 5 because all four digits must be different.
3. The sum of the tens digit and ones digit is 16. Since all four digits are odd, there are only two possible combinations for the tens and ones digit: (9, 7) or (7, 9).
4. Now, we have the following possible numbers:
- 1397
- 1379
- 1597
- 1579
5. To check if any of these numbers are divisible by 13, you can add the digits in odd positions and subtract the sum of digits in even positions. If the result is divisible by 13, then the number is divisible by 13.
Let's check one of the numbers, 1397:
- Adding the digits in odd positions: 1 + 9 = 10
- Subtracting the sum of digits in even positions: (3 + 7) - 9 = 1
- Difference: 10 - 1 = 9
Since 9 is not divisible by 13, 1397 is not divisible by 13.
6. Repeat step 5 for the other numbers on the list to see if any of them are divisible by 13.
By properly applying these steps, you should be able to find the correct number that satisfies all the given conditions.