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Question 214070: Find the six-digit number in which the first digit is one less than the second digit and half the third; the fourth digit is one more than the third and is also the sum of the first and second digits; and the fifth and sixth digits make a number that is the sum of the first four digits. The sum of all the digits is 19.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Write an equation for each statement
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Find the six-digit number
u, v, w, x, y, z
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in which the first digit is one less than the second digit
u = v-1
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and half the third;
u = .5w
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the fourth digit is one more than the third
x = w+1
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and is also the sum of the first and second digits;
x = u + v
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the fifth and sixth digits make a number that is the sum of the first four digits.
10y + z = u + v + w + x
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The sum of all the digits is 19.
u + v + w + x + y + z = 19
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We can replace the 1st 4 digits with 10y + z
10y + z + y + z = 19
11y + 2z = 19
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From this we can write an equation that has only one positive integer solution
2z = 19-11y
z = 
y = 1
z = 
z = 
z = 4
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u + v + w + x + y + z = 19
Replace y and z
u + v + w + x + 1 + 4 = 19
u + v + w + x = 19 - 5
u + v + w + x = 14
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Replace u with .5w, replace x with w+1
.5w + v + w + (w+1) = 14
2.5w + v = 14 - 1
2.5w + v = 13
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We know that u = v-1 and we know that u = .5w, therefore
.5w = v - 1
or
v = .5w + 1
Replace v with (.5w+1)
2.5w + .5w + 1 = 13
3w = 13 - 1
3w = 12
w = 4
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Find v
v = .5(4) + 1
v = 3
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So far we have:
v=3, w=4, y=1, z=4
find x
x = w+1
x = 4+1
x = 5
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find u
u = .5w
u = .5(4)
u = 2
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The number: 234514
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Check this:
2 + 3 + 4 + 5 + 1 + 4 = 19
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