SOLUTION: I can't solve the following problem about a three-digit number:The sum of the three digits is 12 . The tens digit exceeds the hundred digit by the same amount that the units digit
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Question 722524: I can't solve the following problem about a three-digit number:The sum of the three digits is 12 . The tens digit exceeds the hundred digit by the same amount that the units digit exceeds the tens digit. If the digits are reversed, the new number exceeds the original number by 198. Find the original number. For the first equation I got h+t+u=12,but I just can't find the second.Please help!!! Found 2 solutions by mananth, lwsshak3:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! h+t+u=12...............(1)
t-h=u-t
2t-h-u=0..................(2)
100u+10t+h-198 = 100h+10t+u
99u-99h=198
/99
u-h=2...................(3)
solve for h t u
345 is the number
You can put this solution on YOUR website! I can't solve the following problem about a three-digit number:The sum of the three digits is 12 . The tens digit exceeds the hundred digit by the same amount that the units digit exceeds the tens digit. If the digits are reversed, the new number exceeds the original number by 198. Find the original number. For the first equation I got h+t+u=12,but I just can't find the second.
***
let u=units digit
let t=tens digit
let h=hundreth digit
..
h+t+u=12
t-h=u-t
..
h+t+u=12
-h+2t-u=0
add
3t=12
t=4
..
100h+10t+u=100u+10t+h+198
99h-99u=198
divide by 99
h-u=2
h=2+u
..
h+t+u=12
2+u+4+u=12
2u=6
u=3
..
h+t+u=12
h+4+3=12
h+7=12
h=5
..
original number: 100h+10t+u=500+40+3=543
reversed number:100u+10t+h=300+40+5=345
543-345=198
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