Its your turn, divide it

Let the 8-digit number formed from digits 0 to 7 be $n = 100a + b$ , where $a$ is a 6-digit integer and $b$ , 2-digit integer. We note that $n \equiv 100a+b \equiv b \pmod{25}$ . Therefore for $25|n$ , $\implies b = 25$ , $50$ , and $75$ (it cannot end in 00 because repetition is not allowed).

• The number of 8-digit numbers end with 25 $= \color{#D61F06}{5} \times 5 \times 4 \times 3 \times 2 \times 1 = 600$ ( $\color{#D61F06}{\text{Note: }}$ 0 cannot be the first digit)
• The number of 8-digit numbers end with 50 $= 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
• The number of 8-digit numbers end with 75 $= \color{#D61F06}{5} \times 5 \times 4 \times 3 \times 2 \times 1 = 600$ ( $\color{#D61F06}{\text{Note: }}$ 0 cannot be the first digit)

Therefore, the total number of 8-digit numbers divisible by 25 is $600+720+600 = \boxed{1920}$ .

From the question we need eight digit numbers and are given eight different numbers that can't be repeated. The numbers divisible by 25 always end in 25, 50, 75 or 100 but since we can't repeat numbers, we can't have a number ending in 100. Starting with all the numbers ending in 50 we have already placed two of the numbers that have to be used, so we now only need to rearrange the remaining numbers, there are 6 of them so we have $6! = 720$ rearrangements. Now we must consider the 75 and 25 endings, choosing which we have means 2 different options. To have an eight digit number there can't be a leading 0 so we must put the 0 in to a digit opening first that isn't the first openings. There are 5 choices there and then we must rearrange the remaining 5 numbers so $5! = 120$ ways for this meaning the total for this is $2 \cdot 5 \cdot 5! = 1200$ This leads us to the grand total of $720 + 1200 = 1920$