How many two-digit numbers, with a non-zero digit in the units place, are there which are more than thrice the number formed by interchanging the positions of its digits?
Started 10 months ago by Shashank in
Explanatory Answer
Let the two-digit number be xy, which can be expressed as 10x + y
Given that the two-digit number is more than thrice the number obtained by interchanging the
digits
So, 10x + y > 3 × (10y + x)
=> 10x + y > 30y + 3x
=> 7x > 29y
=> x > 29/7 × y
Approximately, x > 4y
Let us fix values for y and check for conditions,
For y = 1, x can take the values of 5 , 6 , 7 , 8 , 9 (As, 51 > (3 × 15), 61 > (3 ×16), 71> (3 × 17), 81
> (3 × 18), 91 > (3 × 19))
Similarly, for y = 2, the condition satisfies only for x = 9 (As, 92 > (3 × 29))
The remaining values of y does not satisfy the given conditions.
So, Total = 5 + 1 = 6 numbers
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