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When the digits of a two-digit natural number are interchanged, the original number is greater than three times the new number so obtained. How many such natural numbers are there? (Ignore the numbers which have 0 in the units place.)

1. 5
2. 6
3. 7
4. 8

Correct answer: 6

Solution

Let the number be 10a + b and the reversed number be 10b + a. The condition gives 10a + b > 3(10b + a), which simplifies to 7a > 29b. Checking valid digit pairs with b ≠ 0 gives 6 numbers satisfying the inequality.

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