Question

The middle digit of a three digit number is half the sum of the other two digit the number is 7 more than 35 times the sum of its extreme digits the new number obtained by reversing the digits is 396 more than the original number find the original number​

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kannu2408

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this is your answer

let a = 100's digit

let b = the 10's

let c = the units

then

100a+10b+c = the original number

:

Write an equation for each statement, simplify as much as possible

:

the middle digit of a three digit number is half the sum of the other two digits.

b = .5(a+c)

b = .5a + .5c

:

The number is 20.5 times the sum of its digits.

100a+10b+c = 20.5(a+b+c)

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The new number obtained by interchanging the digits in the unit's and hundred's places is more than the original number by 594.

100a + 10b + c + 594 = 100c + 10b + a

100a - a + 10b - 10b = 100c - c - 594

99a = 99c - 594

simplify, divide by 99

a = c - 6

We know that c has to be 7, 8, or 9

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in the first equation, b = .5a + .5c, replace a with (c+6)

b = .5(c-6) + .5c

b = .5c - 3 + .5c

b = c - 3

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If c = 9, then the number is 369

if c = 8, then the number is 258

if c = 7, then the number is 147

All of these numbers, when reversed, have a difference is 594?

Which of these satisfy the 2nd statement

369/18 = 20.5

258/15 = 17.2

147/12 = 12.25  

Find the original number 369

Step-by-step explanation: