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Question 340893: The product of the three digits is 15.The sum of the first and third is twice the second.The sum of the second and third is one less than the first. Who Am I?
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! The product of the three digits is 15.The sum of the first and third is twice the second.The sum of the second and third is one less than the first. Who Am I?
3 digit number, product of the 3 digits is 15
1st + 3rd = 2 * 2nd
2nd + 3rd = 1st - 1
let 3 digit numbers product = abc = 15
a + c = 2b
b + c = a - 1
b + c = a - 1, solve for c
c = a - b - 1, substitute this into a + c = 2b
a + (a - b - 1) = 2b
2a - b - 1 = 2b
2a - 1 = 3b, solve for a
2a = 3b + 1
a = (3/2)b + 1/2, substitute into b + c = a - 1
b + c = (3/2)b + 1/2 - 1
b + c = (3/2)b - 1/2, solve for c
c = (1/2)b - 1/2
substitute a = (3/2)b + 1/2 & c = (1/2)b - 1/2 into abc = 15
((3/2)b + 1/2) * b * ((1/2)b - 1/2) = 15
((3/2)b + 1/2) * ((1/2)b^2 - (1/2)b) = 15
use FOIL (First Outer Inner Last)
3/2 * 1/2 * b^3 - 1/2 * 3/2 * b^2 + 1/2 * 1/2 * b^2 - 1/2 * 1/2 * b = 15
1/2 * (3/2 * b^3 - 3/2 * b^2 + 1/2 * b^2 - 1/2 * b) = 15
multiply both sides by 2
3/2 * b^3 - 3/2 * b^2 + 1/2 * b^2 - 1/2 * b = 30
3/2 * b^3 - 2/2 * b^2 - 1/2 * b = 30
3/2 * b^3 - b^2 - 1/2 * b = 30
multiply both sides by 2
3b^3 - 2b^2 - b = 60
b * (3b^2 - 2b - 1) = 60
b * (3b + 1)(b - 1) = 60
factor 60
60 = 2 * 30 = 2 * 2 * 15 = 2 * 2 * 3 * 5 = 3 * 2 * 10 = 6 * 10
b = 3, b - 1 = 2, 3b + 1 = 3 * 3 + 1 = 9 + 1 = 10, 3 * 10 * 2 = 30 * 2 = 60
substitute b = 3 into a = (3/2)b + 1/2
a = 3/2 * 3 + 1/2 = 9/2 + 1/2 = 10/2 = 5
substitute a = 5 & b = 3 into abc = 15
5 * 3 * c = 15
15 * c = 15
c = 1
the 3 digit number is 531
check:
3 digit number, product of the 3 digits is 15 --> 5 * 3 * 1 = 15, yes
1st + 3rd = 2 * 2nd --> 5 + 1 = 6 = 2 * 3 = 6, yes
2nd + 3rd = 1st - 1 --> 3 + 1 = 4 = 5 - 1 = 4, yes
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