WILL CHOOSE BRAINLIEST 5 STARS AND THANKS...How many 6-digit numbers, formed using each digit from 1 to 6 exactly once, are divisible by: Divisible by 5 Divisible by 12 Divisible 25

( 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.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,