How many two digit numbers that increased by 75% where their digits are reversed
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4 two digit numbers are possible such that, When increased by 75%, The digits are reversed.
Let a two digit number be, 10x + y.
75% of the number is,
= 75 % of (10 x + y)
= 75/100 × (10x+y)
= 3/4 × ( 10x + y)
According to the question, When increased by 75%, The digits must be reversed.
So,
(10 x + y) + 3/4 ( 10x + y) = 10y + x
4 (10x + y) + 3 ( 10x + y) = 4 ( 10y + x)
7(10x + y) = 4(10y + x)
70x + 7y = 40y + 4x
70x - 4x = 40 y - 7y
66x = 33y
2x = y
So,
x = 1, y = 2, 10x + y = 12
x = 2, y = 4, 10x + y = 24
x = 3, y = 6, 10x + y = 36
x = 4 , y = 8 , 10x + y = 48
x can't be 5, as y will be 10 ( y should be a single digit).
Therefore, 4 two digit numbers are possible.
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