SOLUTION: 1. A three-digit number divisible by 5 has a hundreds digit that is 2 more than the tens digit. If the number is 43 times the sum of the digit , what is the number? Thank y

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Question 252616: 1. A three-digit number divisible by 5 has a hundreds digit that is 2 more than the tens digit. If the number is 43 times the sum of the digit , what is the number?
Thank you very much for answering my first question . It helps A lot. Thanks again
Found 2 solutions by palanisamy, JimboP1977:
Answer by palanisamy(496) About Me  (Show Source):
You can put this solution on YOUR website!
The given number is divisible by 5.
So the unit place is either 0 or 5
First let us take the unit place to be 0.
Let the tenth place be x.
Given, hundreds digit that is 2 more than the tens digit.
So the hundreth place = x+2.
The sum of the digits = x+2+x+0 = 2x+2
The valve of the number = (x+2)*100+x*10+0
= 100x+200+10x
= 110x+200
Given, the number is 43 times the sum of the digit
110x+200 = 43*(2x+2)
110x-86x = 86-200
24x = -114
x = -114/24
This is not possible because x is an inteder.
Next,
let us take the unit place to be 5.
Let the tenth place be x.
Given, hundreds digit that is 2 more than the tens digit.
So the hundreth place = x+2.
The sum of the digits = x+2+x+5 = 2x+7
The valve of the number = (x+2)*100+x*10+5
= 100x+200+10x+5
= 110x+205
Given, the number is 43 times the sum of the digit
110x+205 = 43*(2x+7)
110x-86x = 301-205
24x= 96
x= 96/24
x=4
So the tenth place = 4.
So the hundreth place = x+2 = 4+2 = 6
Therefore the given number is 645

Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
This is how I would tackle the problem:
The three digit number can be represent as 100x%2B10y%2Bz where 100x is the hundreds, 10y is the tens and z is the units.
We know that 100x%2B10y%2Bz+=+43%28x%2By%2Bz%29. We know that the three digit number is divisible by 5 which means that z must be either 0 or 5.
We know that x-2 = y.
Lets assume that z = 0. Collect terms in the equation 100x%2B10y%2Bz+=+43%28x%2By%2Bz%29
100x%2B10y+=+43x%2B43y
57x-33y+=+0
Substitute x-2 in for y to give 57x-33%28x-2%29+=+0
57x-33x%2B66+=+0
24x%2B66+=+0
x=+-66%2F24
x must be an integer, so we know z must be 5.
Collect terms in the equation 100x%2B10y%2Bz+=+43%28x%2By%2Bz%29
100x%2B10y%2B5+=+43x%2B43y%2B215
57x-33y-210=+0
Substitute x-2 in for y to give 57x-33%28x-2%29-210+=+0
57x-%2833x-66%29-210+=+0
24x%2B66-210+=+0
24x+=+144
x+=+6
S0 we know that the 3 digit number is 645.