Question

PLEASE I NEED THIS FAST a three dight number has one more ten than it has hundreds, and it also has one more than twice as many units as tens the sum of the number and that number reversed is 31 less than 10 cubed find the reverse number

Answer 1

The reverse number of the three-digit number is 732

How to determine the reverse of the number?

Let the three-digit number be xyz.

So, the reverse is zyx

This means that

Number = 100x + 10y + z

Reverse = 100z + 10y + x

From the question, we have the following parameters:

y = x + 1

z = 1 + 2y

The sum is represented as:

100x + 10y + z + 100z + 10y + x = 10^3 - 31

100x + 10y + z + 100z + 10y + x = 969

Evaluate the like terms

101x + 101z + 20y = 969

Substitute y = x + 1

101x + 101z + 20(x + 1) = 969

101x + 101z + 20x + 20 = 969

Evaluate the like terms

101x + 101z + 20x = 949

121x + 101z = 949

Substitute y = x + 1 in z = 1 + 2y

z = 1 + 2(x + 1)

This gives

z = 2x + 3

So, we have:

121x + 101z = 949

121x + 101* (2x + 3) = 949

This gives

121x + 202x + 303 = 949

Evaluate the sum

323x = 646

Divide by 323

x = 2

Substitute x = 2 in z = 2x + 3 and y = x + 1

z = 2*2 + 3 = 7

y = 2 + 1 = 3

So, we have

x = 2

y = 3

z = 7

Recall that

Reverse = 100z + 10y + x

This gives

Reverse = 100*7 + 10*3 + 2

Evaluate

Reverse = 732

Hence, the reverse number of the three-digit number is 732

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