The number of such 3-digit numbers is 20.

Quick Scoop: Core Idea

To be divisible by 5, a number must end in 0 or 5.
Here, the available digits are 2, 3, 5, 6, 7, and 9, so the only possible last digit is 5.

So we fix the units (last) place as 5.

  • Units place: 5 (only 1 option).
  • Remaining digits for hundreds and tens places: 2, 3, 6, 7, 9 (5 digits left).

Now we must form a 2-digit number (hundreds and tens) from these 5 digits, with no repetition.

  • Ways to choose the hundreds digit: 5 options.
  • After choosing the hundreds digit, ways to choose the tens digit: 4 options.

Total 3-digit numbers:

5×4×1=205\times 4\times 1=205×4×1=20

So, 20 different 3-digit numbers can be formed from 2, 3, 5, 6, 7, and 9 that are divisible by 5 with no repeated digits.

Mini Table: Position Choices

Position Available digits Number of choices
Hundreds 2, 3, 6, 7, 9 5
Tens Remaining after choosing hundreds 4
Units 5 (fixed, to be divisible by 5) 1
Total = 5 × 4 × 1 = 20.

Example View

A few sample numbers from the 20:

  • 235, 275, 295
  • 365, 695
  • 975, 765

Each uses digits from the set, ends in 5, and has no repeated digit.

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Learn how to find how many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? Step-by-step explanation with clear logic and examples. Information gathered from public forums or data available on the internet and portrayed here.