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
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
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
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