Question 253036: 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?
Found 2 solutions by richwmiller, MathTherapy:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! let a= the number
100x+10y+z=a
x=y+2
a=43*(x+y+z)
we also know that z=0 or z=5 since the number is divisible by 5
So we plug in 5 or 0 to test it. which ever gives us three integers for x, y and a is the right choice for z
hint z=5 works
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website! Let the units, tens, and hundreds digits be U, T, and H, respectively
Since the hundreds digits is 2 more than the tens digit, then the hundreds digit is T + 2
Therefore, the number is 100(T + 2) + 10(T) + U, or 100T + 200 + 10T + U, or 110T + U + 200
Also, the sum of the digits would be: T + 2 + T + U, or 2T + U + 2
Now, since the number (110T + U + 200) is 43 times the sum of the digits, or 2T + U + 2, then we have:
110T + U + 200 = 43(2T + U + 2)
110T + U + 200 = 86T + 43U + 86
24T – 42U = - 114
Now, since the number is divisible by 5, then its units digit, or U HAS to be either 0 or 5
Now, if we substitute 0 for U in 24T – 42U = 114, we will get T being equal to a negative number, which means that U CANNOT be 0. Therefore, U has to be 5
When 5 is substituted into the equation, 24T – 42U = -114, we get: 24T – 210 = -114, which means that 24T = 96, and T = 4
Since the tens digit, or T = 4, then the hundreds digit, or T + 2 = 6, and, as determined before, the units digit, or U = 5, which makes the 3-digit number, 
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