SOLUTION: When a certain 3-digit number is divided by the number with the digits reversed, the quotient is 2 and the remainder is 25. The tens digit is one less than twice the sum of the hun
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Question 1052913: When a certain 3-digit number is divided by the number with the digits reversed, the quotient is 2 and the remainder is 25. The tens digit is one less than twice the sum of the hundreds digit and units digit. If the units digit is subtracted from the tens digit the result is twice the hundreds digit. Find the number Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! let a = the 100's digit
let b = the 10's
let c = the units
:
Write an equiv equation for each statement, simplify as much as possible
:
"When a certain 3-digit number is divided by the number with the digits reversed, the quotient is 2 and the remainder is 25."
multiply by (100c+10b+a) and you have
100a + 10b + c - 25 = 2(100c + 10b + a)
100a + 10b + c - 25 = 200c + 20b + 2a
:
"The tens digit is one less than twice the sum of the hundreds digit and units digit."
b = 2(a+c) - 1
b = 2a + 2c - 1
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"If the units digit is subtracted from the tens digit the result is twice the hundreds digit. "
b - c = 2a
In the 2nd equation, replace 2a with(b-c)
b = (b-c) + 2c - 1
b - b = 2c - c - 1
0 = c - 1
therefore
c = 1
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Using the 1st equation simplified, replace c with 1
100a + 10b + 1 - 25 = 200(1) + 20b + 2a
100a - 2a + 10b - 20b - 24 = 200
98a - 10b = 200 + 24
98a - 10b = 224
Simplify, divide by 2
49a - 5b = 112
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In the 2nd equation, simplified, replace c with 1
b = 2a +2(1) - 1
-2a + b = 1
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Use elimination, multiply the above equation by 5 add
49a - 5b = 112
-10a+ 5b = 5
-------------------Adding eliminates b, find a
39a = 117
a = 117/39
a = 3
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Find b
we know b = 2a + 2c -1
replace a and c
b = 2(3) + 2(1) = 1
b = 6 + 2 - 1
b = 7
Therefore
our number = 371
:
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Check this in the first statement = = 2
You can check them in the other two statments
