SOLUTION: For a certain 3-digit number, interchanging the first and last digits gives a 3-digit number 297 less than the original number. Doubling the first and last digits of the original n

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Question 402326: For a certain 3-digit number, interchanging the first and last digits gives a 3-digit number 297 less than the original number. Doubling the first and last digits of the original number gives a 3-digit number and increases the sum of the digits to 15. What is the number?
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
For a certain 3-digit number, interchanging the first and last digits gives a 3-digit number 297 less than the original number. Doubling the first and last digits of the original number gives a 3-digit number and increases the sum of the digits to 15. What is the number?

Let the number have the digits "htu" where 

h = its hundreds digit
t = its tens digit
u = its units (ones) digit

The original number = 100h + 10t + u

>>...interchanging the first and last digits...<<

That gives 100u + 10t + h 

>>...gives a 3-digit number 297 less than the original number...<<

That is 100h + 10t + u - 297, so we set them equal

100u + 10t + h = 100h + 10t + u - 297

277 = 99h - 99u 

99h - 99u = 277

That equation can be divided through by 99
  
  h - u = 3

>>...Doubling the first and last digits of the original number...<<

That gives a number with hundreds digit 2h, tens digit still t, and
units (ones) digit 2u

>>...gives a 3-digit number...<<

That tells us that 2h is NOT zero, but we knew that anyway.

>>...and increases the sum of the digits to 15...<<

So 2h + t + 2u = 15

So we have the equations:

      h - u = 3
2h + t + 2u = 15

Solve the first equation for h

h = 3 + u

Substitute in the second equation:

2(3 + u) + t + 2u = 15

6 + 2u + t + 2u = 15
         4u + t = 9
              t = 9 - 4u

   Since 

              0 < t < 9

            0 < 9 - 4u < 9

            -9 < -4u < 0

             9 < 4u < 0
           
            9/4 < u < 0 

           2.25 < u < 0 

And since the units (ones) digit must be 1 or larger, because

>>...interchanging the first and last digits gives a 3-digit number...<<

so it cannot be 0, for if it were 0, that interchange would make a 0 
for the first digit.

So u is either 1 or 2

If u = 1, then since h = 3 + u

h = 3 + 1 = 4

Substituting in

2h + t + 2u = 15

2(4) + t + 2(1) = 15

     8 + t + 2 = 15
        t + 10 = 15
             t = 5

So in this case the number "htu" is 451 

Checking:

interchanging the first and last digits gives 154

and that is a 3-digit number 297 less than the original number,

since 451 - 154 = 297 

>>...Doubling the first and last digits of the original number gives a 3-digit number and increases the sum of the digits to 15...<<

Doubling the first and last digit of 451 gives 852, and the sum
of its digits is 8+5+2=15

So 451 is a solution.

Now we try 

------

If u = 2, then since h = 3 + u

h = 3 + 2 = 5

Substituting in

2h + t + 2u = 15

2(5) + t + 2(2) = 15

    10 + t + 4 = 15
        t + 14 = 15
             t = 1

So in this case the number "htu" is 512 

Checking:

interchanging the first and last digits gives 215

and that is not a 3-digit number 297 less than the original number,

since 215 - 512 = -297 

So we must discard this possibility.  Also the last part

>>...Doubling the first and last digits of the original number gives a 3-digit number and increases the sum of the digits to 15...<<

You cannot double the first digit 5, for doubling 5 gives 10, 
not a single digit.

So there is just one solution, 451.

Edwin