Answer:
( edited )
1. divisible by 5 : 120
2. divisible by 12 : 192
3. divisible by 25 : 24
See explanations below.
Step-by-step explanation:
Given:
6 digit number
digits all unique between 1 to 6
divisible by 5, 12 and 25
Find how many such numbers
solution:
1. Divisible by 5.
The last digit must be 5. That leaves five digits (1,2,3,4,6) for the rest of the numbers. So there are 5! = 120 such numbers.
2. Divisible by 12 (edited again)
All numbers formed from unique digits 1,2,3,4,5,6 are divisible by 3, since 1+2+3+4+5+6 = 21 which is divisible by 3.
For the 6 digit numbers divisible by 12, the two end digits must be divisible by 4, for which there are 8 ( 12,16,24,32,36, 52, 56, 64)
For each of the 8 numbers, we have 4 more digits to tag on for which there are 4! = 24 permutations. Thus there are 8*24 = 192 numbers divisible by 12.
3. divisible by 25
All numbers ending in 00, 25, 50 or 75 are divisible by 25.
Using digits 1 to 6, only those ending in 25 qualify.
Thus that leaves us with 4 digits to make 4! = 24 variations.
Thus there are 24 such numbers.
