SOLUTION: Hi In a 3 digit number the units digit is 3 more than the tens digit and the sum of the digits is 11. If the units and hundreds digits are interchanged the number increased by 99

Question 1194539: Hi
In a 3 digit number the units digit is 3 more than the tens digit and the sum of the digits is 11. If the units and hundreds digits are interchanged the number increased by 99.
What is original number.
Thanks
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Answer: 425 is the original number

See below for the work shown and explanation.

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x = hundreds digit
y = tens digit
z = units or ones digit
These variables represent nonnegative single-digit whole numbers from the set {0,1,2,3,4,5,6,7,8,9}
Though I should be more careful to point out that x = 0 is not possible since something like 012 = 12 is a two digit number.
Luckily, zero is allowed for the other digits.

Any three-digit number is of the form 100x+10y+z
For example, if we had the following three values:
x = 1
y = 2
z = 3
then,
100x+10y+z = 100*1+10*2+3 = 123

We're told that "the units digit is 3 more than the tens digit", which means symbolically we would say z = y+3
Whatever y is, add on 3 to get z.

We also know that "the sum of the digits is 11", so,
x+y+z = 11
Let's replace z with y+3 and simplify
x+y+z = 11
x+y+y+3 = 11
x+2y+3 = 11
x+2y = 11-3
x+2y = 8
Then we can solve for x
x+2y = 8
x = 8-2y
which will be useful later in a substitution step.

Let A = 100x+10y+z be the original three-digit number, and B = 100z+10y+x be the reversed version of the original number (eg: A = 123 and B = 321)
As you can see, I swapped the hundreds digit and units digit to reverse the number.

We're told that "if the units and hundreds digits are interchanged the number increased by 99."
meaning,
new = old+99
B = A+99
100z+10y+x = 100x+10y+z+99

From here, we will plug in these items
x = 8-2y
z = y+3
so that everything is now in terms of y. This will allow us to solve for y.

100z+10y+x = 100x+10y+z+99
100(y+3)+10y+8-2y = 100(8-2y)+10y+y+3+99
100y+300+10y+8-2y = 800-200y+10y+y+3+99
108y+308 = 902-189y
108y+189y = 902-308
297y = 594
y = 594/297
y = 2

Then use this to find x and z
x = 8-2y = 8-2*2 = 4
z = y+3 = 2+3 = 5

The original number is:
A = 100x+10y+z
A = 100*4+10*2+5
A = 400+20+5
A = 425
The reversed number is
B = 100z+10y+x
B = 100*5+10*2+4
B = 500+20+4
B = 524

Notice how B-A = 524-425 = 99
which helps show that B = A+99 is correct.
Or you could verify like so:
B = A+99
524 = 425+99
524 = 524

Also,
x+y+z = 4+2+5 = 6+5 = 11
confirms the digits add up to 11

Lastly, the units digit (5) is exactly three more than the tens digit (2).

Therefore, we have fully confirmed the answer of 425

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

When the digits of a 3-digit number are reversed, the difference between the new number and the original number is 99 times the difference between the units digit and hundreds digit.

In this problem, the number is increased by 99 when the digits are reversed; that means in the original number the hundreds digit is 1 less than the units digit.

Then, given that the units digit is 3 more than the tens digit, we have

x = tens digit
x+3 = units digit (3 more than the tens digit)
x+2 = hundreds digit (1 less than the units digit)

The sum of the digits is 11:

x+x+3+x+2 = 11
3x+5=11
3x=6
x=2

The tens digit is x=2; the units digit is x+3=5; the hundreds digit is x+2=4.

ANSWER: 425

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