Sum of the digits of a …
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Posted by Haroon Ansari 5 years, 3 months ago
- 2 answers
Gaurav Seth 5 years, 3 months ago
Let the unit digit and tens digits of the number be x and y
Number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
⇒ x + y = 9 ... (i)
⇒ 9(10y + x) = 2(10x + y)
⇒ 88y - 11x = 0
⇒ -x + 8y =0 ... (ii)
Adding equation (i) and (ii), we get
⇒ 9y = 9
⇒ y = 1 ... (iii)
Putting the value in equation (i), we get
⇒ x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
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Gaurav Seth 5 years, 3 months ago
Let the unit digit and tens digits of the number be x and y
Number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
⇒ x + y = 9 ... (i)
⇒ 9(10y + x) = 2(10x + y)
⇒ 88y - 11x = 0
⇒ -x + 8y =0 ... (ii)
Adding equation (i) and (ii), we get
⇒ 9y = 9
⇒ y = 1 ... (iii)
Putting the value in equation (i), we get
⇒ x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
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