# How many 3 digit integers…?

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?

• First thing: How many ways are there to factor 24? How many of those are three single digit integer. How many ways are there to arrange those factors into three digit numbers. – fleablood Oct 29 '18 at 16:36
• Finding the largest is easier then counting them. What is the largest single digit factor of $24$. That will be the first digit. If you factor what is left what is the largest single digit factor of that. That will be the second digit. – fleablood Oct 29 '18 at 16:50

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.

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?

$$xyz= 2^3 .3$$

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 $$8 3 1$$