SOLUTION: A number consist of three digits of whose sum is 17 the middle digit exceeds the sum of two digits by one if the digits of the number are reversed the number dimini shes by 396 fin
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Question 619003: A number consist of three digits of whose sum is 17 the middle digit exceeds the sum of two digits by one if the digits of the number are reversed the number dimini shes by 396 find the number?
Found 2 solutions by ankor@dixie-net.com, MathTherapy:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Write an equation for each statement
:
A number consist of three digits
a, b, c
100a + 10b + c = "the number"
:
of whose sum is 17
a + b + c = 17
:
the middle digit exceeds the sum of two digits by one
b = a + c + 1
It's easy to find b now, rearrange the above and add to the first equation
a + b + c = 17
-a +b - c = 1
---------------Adding eliminates a and c, find b
2b = 18
b = 9
:
if the digits of the number are reversed the number diminishes by 396
100a + 10b + c - (100c + 10b + a) = 396
100a + 10b + c - 100c - 10b - a = 396
Combine like terms
100a - a + 10b - 10b + c - 100c = 396
99a - 99c = 396
Simplify, divide by 99
a - c = 4
a = (c+4)
:
Find the number
:
Using the 1st equation: a + b + c = 17, replace a and b
(c+4) + 9 + c = 17
2c + 13 = 17
2c = 17-13
2c = 4
c = 2
then
a = c + 4
a = 2 + 4
a = 6
:
692 is the number
:
:
Check this 692 - 296 = 396
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
A number consist of three digits of whose sum is 17 the middle digit exceeds the sum of two digits by one if the digits of the number are reversed the number dimini shes by 396 find the number?
Let the hundreds, tens, and units digits be H, T, & U, respectively
Since the 3 digits sum to 17, then: H + T + U = 17 ------ eq (i)
Since the middle digit (T) exceeds the sum of the other 2 digits by 1, then:
T = H + U + 1
H T + U = - 1 ---- eq (ii)
Since when number is reversed the number diminishes by 396, then:
100U + 10T + H = 100H + 10T + U - 396
H - 100H + 10T 10T + 100U U = - 396
99H + 99U = - 396
99(- H + U) = 99(- 4) ----- Factoring out GCF, 99
- H + U = - 4 ----- eq (iii)
H + T + U = 17 ---- eq (i)
H T + U = - 1 ---- eq (ii)
2H + 2U = 16 ----- Adding eqs (i) & (ii)
2(H + U) = 2(8) ------ Factoring out GCF, 2
H + U = 8 ------- eq (iv)
- H + U = - 4 ----- eq (iii)
2U = 4 ----- Adding eqs (iv) & (iii)
U, or units digit = , or
H + 2 = 8 ----- Substituting 2 for U in eq (iv)
H, or hundreds digit = 8 2, or
6 + T + 2 = 17 ------ Substituting 2 for U, and 6 for H in eq (i)
T = 17 8
T, or tens digit =
The original number is
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