Math, asked by rupalipderle, 6 months ago

A three digit number is equal to 17 times the sum of the digits. if the digits are reversed the new number is 198 more than the original number. the sum of the extreme digits is 1 less than middle digit. find the original number. ​

Answers

Answered by sanatarajan55
1

Step-by-step explanation:

digit number be xy2

400x+10y+z=17(x+y+z)83x=7y+16z−0

⇒100z+10y+x=198+100x+10y+399

(3−x)=1983−x=23=x+2−(2)

⇒x+z=y−1→(3) 2 in 1,3

83x=7y+16x+32

67x=7y+32−6

14x=7y−21−(6) solve 4,6

number is 153

Answered by pm62341
12

Answer:

153

Step-by-step explanation:

let the hundred place digit be x, tens digit be y and unit digit be z

So number = 100x+10y+z

We are given that A three digit number is equal to 17 times the sum of its digit

100x + 10y + z = 17 (x + y + z)

100x + 10y + z = 17x + 17y + 17z

100x - 17x + 10y - 17y + z - 17z = 17x + 17y + 17z

83x - 7y - 16z = 0 ------------------ (1)

if the digits reserved, the new number is 198 more than the old number

100x + 10y + z + 198 = 100z + 10y + x

100x -- x + 10y - 10y + z - 100z - 198 = 0

99x - 99z + 198 = 0

z - x = 2

z - x = 2 -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- (2)

the sum of extreme digit is less than the middle digit by unity

And,

z + x = y- 1

By putting (2)

z + z - 2 = y - 1

y = 2z - 1

Putting (3) and (2) in (1) we get

83(z - 2) - 7 (2z - 1) -- 16z = 0

83(z - 2) - 7 (2z - 1) -- 16z = 0

z = 3

Put z = 3 in (3)

y = 2(3) -1

y = 5

Put z = 3 in (2)

x = 3-2= 1

So. number = 100(1) + 10(5) + 3 = 153

So, number = 153

Hence the number is 153

hope it helps you

Please mark me as brainlist

Similar questions