a two digit no. is equal to 7 times the sum of its digits the no. formed by reversing its digit is less than the original no. by 18 find the original no.
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The original number = 10y + x
Number formed by interchanging the digits = 10x + y
Given (10y + x) = 7(x + y)
Þ 10y + x = 7x + 7y
Þ 7x + 7y – x – 10y = 0
Þ 6x – 3y = 0
Þ 2x – y = 0 → (1)
It is also given that on reversing the digits the number obtained is 18 less than the original number
Hence 10x + y = (10y + x) – 18
Þ 10x + y – 10y – x = – 18
Þ 9x – 9y = – 18
Þ x – y = – 2 → (2)
Subtract (2) from (1)
2x – y = 0
x – y = – 2
-------------
x = 2
Put x = 2 in 2x – y = 0
2(2) – y = 0
∴ y = 4
Therefore the original number is 42.
Answer:
42
Step-by-step explanation:
let original number be (10x+y)
given ,
(10x+y) = 7(x+y)
=> 10x +y = 7x + 7y
=> 10x-7x = 7y -y
=> 3x = 6y
=> x = 2y................1
also , (10x+y) = (10y+x) +18
=> 10x + y = 10y + x + 18
=> 10x-x = 10y-y +18
=> 9x = 9y + 18
=> x = y +2
.
putting value of x from 1
we get,
2y = y + 2
=> y = 2
x = 2y
=> x = 4
so ,the number is (10x+y) = (10×4+2)
number is 42