How many ways can you arrange the digits 1, 2, 4, 5, 6, 7, to get a six-digit number that is divisible by 25?

Guest May 7, 2020

#1**0 **

for a number to be divisible by 25, it must end in either 00 (making it a multiple of 100), 25 (a multiple of 100 + 25), 50 (a multiple of 50), or 75 (a multiple of 100 + 75). since 0 isn't in our digits, the digits must be in the form

\(\times\times\times\times25\)

or

\(\times\times\times\times75\)

where \(\times\) represents any digit not used. there are \(4!\) ways to arrange the digits that fit the first form and \(4!\) ways for the second, so we have a total of

\(4! + 4! = 24 + 24 =\boxed{48}\)

ways to arrange the digits.

chrissy May 7, 2020