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Question 79332: Hi, I've been trying to set up an appropriate system of equations for quite some time, and I've had no luck. The question is as follows;
The sum of the digits of a certain three digit number is nine. The sum of the hundreds and the tens digits is equal to the ones digit minus one. The number, divided by nine, equals three times the ones digit. What is the number? All I have so far is x+y+z=9. Could you show me how to set the problem up please?
Found 3 solutions by stanbon, scott8148, josmiceli:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The sum of the digits of a certain three digit number is nine. The sum of the hundreds and the tens digits is equal to the ones digit minus one. The number, divided by nine, equals three times the ones digit. What is the number? All I have so far is x+y+z=9. Could you show me how to set the problem up please?
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The number is x(10^2+y(10)+z
The digits are x,y,z
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EQUATIONS:
x+y+z=9
x+y=z-1
(100x+10y+z)/9 = 3z
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x+y+z=9
x+y-z=-1
100x+10y-26z=0
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Ans: x=1; y=3; z=5
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Cheers,
Stan H.
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! let h=hundreds digit, let t=tens digit, let u=ones digit ... h+t+u=9 ... h+t=u-1 ... (100h+10t+u)/9=3u
IF YOU ONLY WANT THE SETUP ... DO NOT READ BEYOND THIS LINE
from first two equations, u-1+u=9 ... 2u=10 ... u=5 ... t=4-h
using third equation, 100h+10(4-h)+5=3*5*9 ... 100h+40-10h+5=135 ... 90h=90 ... h=1 ... t=3
Answer by josmiceli(19441)  (Show Source):
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