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1 year ago

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PLEASE HELP a three dight number has one more ten than it has hundreds, and it also has one more than twice as many units as ten

s the sum of the number and that number reversed is 31 less than 10 cubed find the reverse number

1 answer:

bija089 [108]1 year ago

8 0

The reverse number of the three-digit number is 732

<h3>How to determine the reverse of the number?</h3>

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

Read more about digits at:

brainly.com/question/731916

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