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
If all positive divisors of 132 are arranged in descending order, then what digit will be at unit place of first divisor ?
If 3 2019 is divided by 10, then what is the remainder?
The number 3798125P369 is divisible by 7. What is the value of the digit P?
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:
What is the remainder when 91 × 92 × 93 × 94 × 95 × 96 × 97 × 98 × 99 is divided by 1261?
