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Question 104695: find a six-digit number in which the first is two less that the fifth, the second digit is one more that the fourth, and the fifth digit is four less than the last. The sum of the third and last digits equals the second and the sum of all the digits is 30.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! : find a six-digit number in which the first is two less that the fifth, the second digit is one more that the fourth, and the fifth digit is four less than the last. The sum of the third and last digits equals the second and the sum of all the digits is 30.
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Let the digits by a, b, c, d, e, f In that order.
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Write an equation for each statement and all it's combinations:
Designated * were used in the equation
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" the first is two less that the fifth,"
a = e-2
e = a+2
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"second digit is one more that the fourth,"
b = d+1
d = b-1
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"fifth digit is four less than the last."
e = f-4 *
f = e+4
From the 1st equation a = e-2
a = (f-4) -2
a = f-6 *
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"The sum of the third and last digits equals the second"
c + f = b
b = c+f *
f = b-c
c = b-f
From d = (b-1) and b = c+f, we can say
d = c+f-1 *
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"the sum of all the digits is 30."
a + b + c + d + e + f = 30
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Substitute all we can in terms of f from the above equations:
--a-------b-----c------d---- --e-----f
(f-6) + (c+f) + c + (c+f-1)+ (f-4) + f = 30
Combine like terms
f + f + f + f + f + c + c + c -6 - 1 -4 = 30
3c + 5f - 11 = 30
3c + 5f = 30 + 11
3c + 5f = 41
A manageable equation; we know that f has to be odd, it cannot be 9
Try f = 7 and solve for c
3c + 5(7) = 41
3c = 41 - 35
3c = 6
c = 2
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From c = 2, and f = 7, find the rest of the numbers:
b = c + f
b = 2 + 7
b = 9
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From d = b-1
d = 9 - 1
d = 8
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From e = f-4:
e = 7 - 4
e = 3
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From a = e-2
a = 3 - 2
a = 1
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Check and see if a b c d e f add up to 30
1 + 9 + 2 + 8 + 3 + 7 = 30
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Our number: 192837
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