Question 515831: The tenth digit of a two digits is three greater than the units digit. When the digit are reversed,the number is reduced by 27.what is the number.
Found 2 solutions by josmiceli, drcole:
Answer by josmiceli(19441)   (Show Source):
Answer by drcole(72)  (Show Source):
You can put this solution on YOUR website! Let x be the ten digit of the two-digit number, and let y be the units digit. Then the two-digit number is equal to
10x + y
(For example, 38 = 10(3) + 8, and 67 = 10(6) + 7).
If the digits are reversed, then the new two-digit number is:
10y + x
(For example, if we reverse the digits of 38, we get 83 = 10(8) + 3).
We know that reversing the digits reduces the number by 27. We can express this fact as an algebraic equation in the following way:
(New Number with Reversed Digits) = (Original Number) - 27
10y + x = (10x + y) - 27
10y - 9x = y - 27 (subtract 10x from both sides)
9y - 9x = -27 (subtract y from both sides)
We also know that, for the original number, the tens digit is three greater than the units digit. We can express this fact as an algebraic equation as well:
(Tens Digit) = (Units Digit) + 3
x = y + 3
So now we have two equations in two unknowns. The second equation is already solved for x, so we can substitute y + 3 for x in the first equation, yielding an equation involving y only:
9y - 9x = -27
9y - 9(y + 3) = -27 (substituting y + 3 for x)
Now we solve for y:
9y - 9y - 27 = -27 (distributing -9)
-27 = -27 (combining like terms)
So we get an interesting situation: our equation reduced down to -27 = -27, which is always true. In other words, any y solves this equation. Now, let's remember the original problem: y represents the units digit of a two digit number. So y can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. In other words, y must be an integer between 0 and 9 (including 0 and 9). Each of these y values gives us a corresponding x value using the equation x = y + 3. So we have:
y = 0, x = 0 + 3 = 3
y = 1, x = 1 + 3 = 4
y = 2, x = 2 + 3 = 5
y = 3, x = 3 + 3 = 6
y = 4, x = 4 + 3 = 7
y = 5, x = 5 + 3 = 8
y = 6, x = 6 + 3 = 9
y = 7, x = 7 + 3 = 10
y = 8, x = 8 + 3 = 11
y = 9, x = 9 + 3 = 12
Now remember that x represents the tens digit of a two digit number. So x = 10, x = 11, and x = 12 cannot work, since 10, 11, and 12 are not digits. So we have to reject y = 7, y = 8, and y = 9 as solutions. The rest of the y-values give us solutions:
y = 0, x = 3 ---> number is 30
y = 1, x = 4 ---> number is 41
y = 2, x = 5 ---> number is 52
y = 3, x = 6 ---> number is 63
y = 4, x = 7 ---> number is 74
y = 5, x = 8 ---> number is 85
y = 6, x = 9 ---> number is 96
So the possible numbers are 30, 41, 52, 63, 74, 85, and 96. Let's check each of these to confirm that, when the digits are reversed, the number is reduced by 27:
30 - 03 = 27
41 - 14 = 27
52 - 25 = 27
63 - 36 = 27
74 - 47 = 27
85 - 58 = 27
96 - 69 = 27
So all seven of these numbers are correct solutions.
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