Question:medium

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7? [This Question was asked as TITA]

Updated On: Jan 15, 2026
  • 22
  • 12
  • 21
  • 11
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The Correct Option is C

Solution and Explanation

To determine the count of 3-digit numbers where the product of their digits falls between 2 and 7 (exclusive), we analyze combinations yielding products of 3, 4, 5, and 6. A 3-digit number is defined by digits a, b, and c, with the constraint that a ≠ 0.
Product = 3: The digit combination is {1,1,3}, forming 3 unique numbers (113, 131, 311).
Product = 4: The digit combinations are {1,1,4} and {1,2,2}. These combinations form a total of 6 unique numbers (114, 141, 411, 122, 212, 221).
Product = 5: The digit combination is {1,1,5}, forming 3 unique numbers (115, 151, 511).
Product = 6: The digit combinations are {1,1,6} and {1,2,3}. These combinations form a total of 9 unique numbers (116, 161, 611, 123, 132, 213, 231, 312, 321).
Grand total: The sum of numbers for each product category is 3 + 6 + 3 + 9 = 21.
Consequently, there are 21 3-digit numbers whose digit product is greater than 2 and less than 7.
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