Question 1142439: am five-digit number greater than 40,000 but less than 70,000. My ones digit is more than my ten thousands digit. All my other digits are the same. The sum of my digits is 19. What number am I?
Found 2 solutions by math_helper, josmiceli:
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website!
Let the number be ABBBC
A can be any one of 4,5, or 6
C can only be 5, 6, 7, 8, or 9, subject to C > A
Finally, A+3B+C = 19
We want 19-A-C to be divisible by 3, subject to C > A, there are three possible cases to consider:
19-10 = 9
19-13 = 6
19-16 = 3
So we need to check cases where A+C = 10, A+C = 13, and A+C = 16
A+C = 10:
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A=4, C=6: 19-4-6 = 9 ---> B = 9/3 = 3, so 43336 is one solution
A+C = 13:
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A=4, C=9: 19-4-9 = 6 ---> B = 2, so 42229 is another solution
A=5, C=8: 19-5-8 = 6 ---> B = 2, so 52228 is a solution
A=6, C=7: 19-6-7 = 6 ---> B = 2, so 62227 is a solution
A+C = 16:
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There are no cases for A+C = 16, for even if we choose the maximum value of A, which is A=6, we find C=10 and that is not a single digit.
I think these are all of the solutions: 43336, 42229, 52228, and 62227
Answer by josmiceli(19441)  (Show Source):
You can put this solution on YOUR website! The number is
N x x x M
M > N
M is either 4, 5, or 6
N + M + 3x = 19
x cannot be 9
x cannot be 8
x cannot be 7
x cannot be 6
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Suppose x is 5
N + M + 3*5 = 19
N + M = 4
M must be 4 and N must be 0
The number is
05554 -cannot be
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Suppose x is 4
N + M + 3*4 = 19
N + M = 7
If M is 6, then N is 1
number is:
14446 -out of range
If M is 4, the N is 3 -also out of range
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Suppose x is 3
N + M + 3*3 = 19
N + M = 10
If M is 6, then N is 4
number is:
43336
the number is in range
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If x is 2, then
M + N + 6 = 19
M + N = 13
52228 works
62227 works
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If x is 1
M + N + 3*1 = 19
M + N = 16
this makes number out of range
( N is 10 or greater )
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The possible numbers are:
43336
52228
62227
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get a 2nd opinion if needed
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