Question 508955: it is between 50 000 and 52 000.
there is a 2 in the tens place.
the sum of the digits is 23
Found 4 solutions by swincher4391, josmiceli, Edwin McCravy, MathTherapy:
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! So we know that the sum of the thousands, hundreds and ones is 16.
We also know that the thousands is between 0 - 2.
Thus we know that the sum between the hundreds and ones is 14-16.
50x2x
51x2x
52x2x
Things that add up to 14
5,9
6,8
7,7
Things that up to 15
6,9
7,8
Things that up to 16
7,9
8,8
It doesn't feel like we have enough information, but here are the numbers that fit the criteria.
-------
50729
50927
50828
51629
51926
51728
51827
52529
52925
52628
52826
52727
-------
Surely there is more info given in this problem?
Answer by josmiceli(19441)  (Show Source):
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website!
The first digit has to be 5, and the tens (fourth)
digit has to be 2, so it's like this
5 _ _ 2 _
Since the sum of the digits is 23, the remaining 3 digits
must have sum 23-7 or 16
The second digit can only be 0 or 1
If the second digit is 0, we have
5 0 _ 2 _
The 3rd and 5th digits must have sum 16
So the only possibilities are
5 0 7 2 9
5 0 8 2 8
5 0 9 2 7
If the second digit is 1, we have
5 1 _ 2 _
the 3rd and 5th digits must have sum 15
So the only possibilities are
5 1 6 2 9
5 1 7 2 8
5 1 8 2 7
5 1 9 2 6
So there are 7 possible different solutions
1. 50729
2. 50828
3. 50927
4. 51629
5. 51728
6. 51827
7. 51926
Edwin
Answer by MathTherapy(10556)  (Show Source):
You can put this solution on YOUR website!
_ _, _ _ _
5 T, H 2 U
5 0, H 2 U Thousands place can ONLY be 0 or 1.
Thousands place = 0
5 0, H 2 U. This means that H (hundreds) + U (units) = 16 [23 – (5 + 2)]. We therefore have 8 & 8, 9 & 7, or 7 & 9. this equates to 3 possibilities, which are:
50,828
50,927
50,729
Thousands place = 1
5 1, H 2 U. This means that H (hundreds) + U (units) = 15 [23 – (5 + 1 + 2)]. We therefore have 6 & 9, 9 & 6, 7 & 8, 8 & 7. This equates to 4 possibilities, which are:
51,629
51,926
51,728
51,827.
As seen, only 7 possibilities exist.
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