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The tens digit of a three digit number exceeds its units digits by 3. If we were to swap the 100's digit and the unit's digit, the new number would be 297 more than the original number. If we were to remove the hundreds digit from the original number, the two digit number obtained would be 13 less than half the original number. What's the original number?

Respuesta :

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Answer: 174

Step-by-step explanation:

Let's define the notation:

U = units digit

T = tens digit

H = hundreds digit.

Where the number is HTU.

We know that:

T = U + 3

"UTH" - "HTU" = 297.

"TU" = "HTU"/2 - 13.

First, let's wrote this correctly, and make a system of equations:

T = U + 3.

U*100 + T*10 + H - H*100 - T*10 - U = (U - H)*99 = 297.

T*10 + U = (H*100 + T*10 + U)/2 - 13 = H*50 + T*5 + U*0.5 - 13

Then we have the system of equations:

T = U + 3.

(U - H)*99 = 297

T*10 + U = H*50 + T*5 + U*0.5 - 13.

First, we can replace the first equation into the third one, so now we have only two variables.

(U - H)*99 = 297

(U +3)*10 + U = H*50 + (U + 3)*5 + U*0.5 - 13.

Now we can isolate H in the first equation and get:

(U - H) = 297/99

H = -297/99 + U.

Now we can replace it into the other equation:

(U +3)*10 + U = (-297/99 + U)*50 + (U + 3)*5 + U*0.5 - 13.

Now let's solve this for U.

U*10 + 30 + U = U*50 -150 + 5*U + 15 + U*0.5 - 13

11*U + 30 = U*(50  + 5 + 0.5) - 148

30 +148 = U*(55.5 - 11) = U*44.5

U = 178/44.5 = 4.

Then using the previous equations:

T = U + 3 = 4 + 3 = 7

H = -3 + U = -3 + 4 = 1.

Then our number is:

174.

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