SOLUTION: The tens digit of a two digit number is 3 times greater than the units digit. Eight times the sum of the digits is 1 less than the number. Find the number.
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Question 262168: The tens digit of a two digit number is 3 times greater than the units digit. Eight times the sum of the digits is 1 less than the number. Find the number.
I've been provided the answer and have ran through numerous equations, but can't seem to get the correct equation that solves for the answer.
I appreciate any help you can provide... Found 2 solutions by mananth, richwmiller:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! The tens digit of a two digit number is 3 times greater than the units digit. Eight times the sum of the digits is 1 less than the number. Find the number.
I've been provided the answer and have ran through numerous equations, but can't seem to get the correct equation that solves for the answer.
I appreciate any help you can provide...
Let the units digit be x .
The tens digit will be 3x
Eight times the sum = 8*(x+3x)
1 less than the number = 30x+x -1
8*(x+3x) = 30x +x -1
32x = 31x-1
x=-1
tens digit =-3 The number is -31
You can put this solution on YOUR website! let t=ten's digit and u equal units digit.
t=3u
8(t+u)=10t+u-1
8t+8u=10t+u-1
7u=2t-1
7u=2*(3u)-1
7u=6u-1
u=-1
t=-3
the number is -31
check
8(-3-1)=-31=1
8*-4=-32
-32=-32
So why are you hiding the answer you were given?
It turns out that the original problem says three greater and not three times greater
so picking up the problem
7u=2t-1
t=3+u
7u=2(3+u)-1
7u=6 +2u-1
5u=5
u=1
t=3+u
t=3+1
t=4
check
8*(4+1)=41-1
40=40
ok
