A certain number between 10 and 100 is 8 times the sum of its digits and if the digits will be reversed . Find the number
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Answered by
1
72
Let the number between 10 and 100
= 10x + y
It is eight times the sum of its digits ⇒ 10x + y = 8(x + y) ⇒ 2x - 7y = 0 (1)
If 45 is subtracted from the number, the digits are reversed.
(10x + y) - 45 = (10y + x)
⇒ 9x - 9y = 45 ⇒ x - y = 5 (2)
By solving equations (1) and (2),
7y/2 - y = 5
5y/2 = 5
y = 2
So x = 7
we get x = 7 and y = 2.
Therefore, the original number is 72.
Answered by
0
Answer:
72
Let the number between 10 and 100
= 10x + y
It is eight times the sum of its digits ⇒ 10x + y = 8(x + y) ⇒ 2x - 7y = 0 (1)
If 45 is subtracted from the number, the digits are reversed.
(10x + y) - 45 = (10y + x)
⇒ 9x - 9y = 45 ⇒ x - y = 5 (2)
By solving equations (1) and (2),
7y/2 - y = 5
5y/2 = 5
y = 2
So x = 7
we get x = 7 and y = 2.
Therefore, the original number is 72.
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