How many 3 digit integers...?

Question

How many different 3-digit integers have the product of their digits equal to 4!? What is the largest of these integers?

I know 4! Is 24 but still confusing to do this. How do I find the largest let alone how many?

Answer 1

How many ways can you write $24$ as a product of one digit whole numbers?

$24 = 1 \cdot 4 \cdot 6$ is one way, giving you the numbers $146$, $164$, $416$, $461$, $614$, and $641$.

Try to find the others.

Answer 2

Just do it.

$4! = 24$ how many ways are there to factor $24$ into three factors.

There's $1*1*24$

$1*2*12$

$1*3*8$

..... etc.

How many of those have only single digit factors?

There's

$1*3*8$

$1*4*6$

$2*2*6$

... etc.

How many ways are there to arrange $1,3,8$ into different orders? There's $138, 183, 318, ...$. How many?

Do that for all the other ways of factoring. How many are there? And what is the largest one?

Hint: if $8$ is the largest single digit factor than doesn't it make sense that the largest such three digit number would be in the $8$ hundreds.

Now, that was the hard way. Were there any handy mathematical observations you might have used to make this easier? Would knowing that $24 = 2^3*3$ is the unique prime factorization of $24$ have helped you find the ways to find three term factorizations? Would knowing there are $k!$ ways to arrange $k$ objects have helped? What if some of the objects were indistinguishable?

Answer 3

$xyz= 2^3 .3 $

powers of $2$ can be divided among $x,y,z$ in

$^{3+(3-1)} \ C _{3-1}= 10 $ ways

and power of $3$ can be divided into $^{1+(3-1)} \ C _{3-1}= 3 $ ways

thus giving total $10\times 3 =30 $ways

but here we also counted arrangements like $2 \times 1\times 12 $ and $1\times 1\times 24$ which violates 3 digit number policy so, by subtracting these i.e, $\left(3! + \dfrac{3!}{2!}\right)=9 $ ways

we get

total number of $3$ digit numbers having product $4 != (30-9)=21 $

and out of all such $3$ digit numbers largest is $8 3 1$