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Question 808961: the tens digit of a three-digit number is 3 less than 5 times the unit digit. three times the sum of the digits is 2 more than 4 times the hundreds digit. if the digits are reversed, the new number is 549 less than than the original number. what is the original number?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! a = the 100's digit
b = the 10's
c = the units
then
100a+10b+c = the number
:
Write an equation for each statement:
:
" the tens digit of a three-digit number is 3 less than 5 times the unit digit."
b = (5c-3)
:
" three times the sum of the digits is 2 more than 4 times the hundreds digit."
3(a+b+c) = 4a + 2
3a + 3b + 3c = 4a + 2
3a - 4a + 3b + 3c = 2
-a + 3b + 3c = 2
:
"if the digits are reversed, the new number is 594 less than than the original number."
100c + 10b + a = 100a + 10b + c - 594
100c - c + 10c - 10c + a - 100a = -594
99c - 99a = -594
simplify, divide by 99
c - a = -6
mult by -1
a - c = 6
a = (c+6)
:
Now we have three equations
b = (5c-3); -a + 3b + 3c = 2; a = (c+6)
In the middle equation, replace a and b
-(c+6) + 3(5c-3) + 3c = 2
-c - 6 + 15c - 9 + 3c = 2
Combine like terms
17c - 15 = 2
17c = 2 + 15
17c = 17
c = 1
then
b = 5(1) - 3
b = 2
and
a = 1 + 6
a = 7
:
Our number then is 721
:
;
Check this in the 3rd statement:
127 = 721 - 594
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