How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?
12
Calculation:
Number of 3-digit numbers without repeatition of digits in which each digit is odd and the number is divisible by 5.
We have odd numbers as 1, 3, 5, 7, 9
In 3-digit numbers we have three positions of digits i.e. units, tens and hundreds in which units digit will alwasy be 5 because number is divisible by 5.
For remaining two positions we have 4 numbers i.e. 1, 3, 7, 9. So, for arranging these numbers we use permutation as:
4 P2 = 4! / (4 - 2)! = (4 × 3 × 2 × 1) / (2 × 1) = 12
Hence, option 2 is correct.
Important Questions from Divisibility and Remainder
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Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
select the correct answer using the code given below:
