In a 3-digit number, the hundreds digit is one more than the ones digit and the tens digit is twice the hundreds digit. If the sum of the digits is 11, find the number. Write and equation, where the hundredths digit is x.

The mentioned number in the exercise is:

362

Step-by-step explanation:

To obtain the mentioned number in the exercise, first you must write the equations you can obtain with it.

If:

x = hundredths digity = tens digitz = ones digit

We can write:

x = z + 1 (the hundreds digit is one more than the ones digit).y = 2x (the tens digit is twice the hundreds digit).x + y + z = 11 (the sum of the digits is 11).

Taking into account these data, we can use the third equation and replace it to obtain the number and the value of each digit:

x + y + z = 11(z + 1) + y + z = 11 (remember x = z + 1)z + 1 + y + z = 11z + z +y + 1 = 11 (we just ordered the equation)2z + y + 1 = 11 (z + z = 2z)2z + y = 11 - 1 (we passed the +1 to the other side of the equality to subtract)2z + y = 102z + (2x) = 10 (remember y = 2x)2z + 2x = 102z + 2(z + 1) = 10 (x = z + 1 again)2z + 2z + 2 = 104z + 2 = 104z = 10 - 24z = 8z = 8/4z = 2

Now, we know z (the ones digit) is 2, we can use the first equation to obtain the value of x:

x = z + 1x = 2 + 1x = 3

And we'll use the second equation to obtain the value of y (the tens digit):

y = 2xy = 2(3)y = 6

Organizing the digits, we obtain the number:

Number = xyzNumber = 362

As you can see, the obtained number is 362.

Iwould think that the correct answer is 8.1