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 6 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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