6 Digit Number Puzzle

6 Digit Number Puzzle

How many six-digit numbers can be formed using the digits 1 to 6 exactly once, such that the number is divisible by the digit in its unit place (i.e., the last digit)?

In other words, from all possible 6-digit numbers with no repeated digits from 1 to 6, how many are divisible by their last digit?

6 digit number puzzle

Solution: 6 Digit Number

Divisibility Rule Analysis

Divisibility Rule of 1

6 digit number puzzle with solution

Divisibility Rule of 2

6 Digit Number Puzzle solution rule 2

Divisibility Rule of 3

rule 3 for 6 digit puzzle

Divisibility Rule of 4

rule 4 How many numbers are divisible by 4 from digit puzzle

Divisibility Rule of 5

Number Puzzle rule 5

Divisibility Rule of 6

rule 6, numbers divisible by 6

Detailed Analysis

Explanation-
XXXXX1 is always divisible by 1, so we have 5! numbers.
XXXXX2 is always divisible by 2, so we have 5! numbers.
XXXXX3 is always divisible by 3 (sum of digits is always 21), so we have 5! numbers.
XXXXY4 is divisible by 4 only if Y is 2 or 6, so we have 2 \times 4! numbers.
XXXXX5 is always divisible by 5, so we have 5! numbers.
XXXXX6 is always divisible by 6 (even number divisible by 3), so we have 5! numbers.

So total number of numbers with required property

= 5 \times 5! + 2\times 4! = 600 + 48 = 648 numbers.

6 Digit Numbers puzzle

Also Check: