Question
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?
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?
Answers
Damon
I am having the same trouble and do not see the trick.
Reiny
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 hundreds digit can be 2,5,7,or 9
You only considered 3 or 5
How about 1937 ?
It satisfies all the conditions
Damon
The tens and the ones add to 16 ?
Reiny
Thanks Damon, you got me.
Time for my third coffee.
Time for my third coffee.
MathMate
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 <b>11</b>, then among
3179, 5379, 7579, 9779,
1397, 3597, 5797, 7997
1397 satisfies all 5 requirements.
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 <b>11</b>, then among
3179, 5379, 7579, 9779,
1397, 3597, 5797, 7997
1397 satisfies all 5 requirements.