# Can any one figure this out

• cherry2189

#### cherry2189

Can anyone figure this out!

1. The product of the digits in the number 126 is 12:1x2x6=12
a. How many different three digit numbers have a product equal to 12? Explain
b.How many three digit numbers have a product equal to 18? Explain
c.Compare your answer to parts a and b and explain why they work out that way.
d.How many four digit numbers have a product equal to 12? explain
e.How many four digit numbers have a product equal to 18? explain
f.How many five digit numbers have a product equal to 12? explain
g.Can ypu describe a general procedure for figuring out how many n-digit numbers have a product equal to p, where n and p are counting numbers?

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Greetings !
Originally posted by cherry2189
my teacher gave me this question for homework,
Which means that you should've posted this
in the Homework forum and shown where you
got stuck. I'll give you a hint. Find the "options"
of multiplications in the digits' amount.

Live long and prosper.

Originally posted by cherry2189

1. The product of the digits in the number 126 is 12:1x2x6=12
a. How many different three digit numbers have a product equal to 12? Explain

15

Originally posted by cherry2189

b.How many three digit numbers have a product equal to 18? Explain

15

Originally posted by cherry2189

d.How many four digit numbers have a product equal to 12? explain

36

Originally posted by cherry2189

e.How many four digit numbers have a product equal to 18? explain

36

Originally posted by cherry2189

f.How many five digit numbers have a product equal to 12? explain

70

=========================
I'll leave you for the analysis.
(I wrote a basic program to get these answers, you can just use these as a reference as you try to figure out a general algorithm for this question)

1. The product of the digits in the number 126 is 12:1x2x6=12
a. How many different three digit numbers have a product equal to 12? Explain

12 factors as 2*2*3. Since you are allowed to use "1" You can make your number using the digits "1", "2", "3", "4" (= 2*2), and "6".
How can you combine these so that the product is 12 and the digits are in increasing order? (That's what I thought was the easiest way of looking at it.)
If the first digit is 1, and the second 2, the last must be 6.
126
If the first digit is 1, and the second 3, the last must be 4.
134
We can't make any more with first digit 1 second digit 4 or 6.

If the first digit is 2, and the second 2, the last digit must be 3.
223
We can't make any more with first digit 2 and we can't make any with first digit 3 4 or 6 (not in increasing order).

Of course we could rearrange the the digits in 126. There are 3!= 6 ways to do that (they are: 126, 162, 216, 261, 612, 621).

There are also 3!= 6 way to arrange the digits in 134.

There are only 3!/(2!)= 3 ways to arrange the digits in 223
(they are 223, 232, 322).

There are a total of 6+ 6+ 3= 15 three digit numbers whose digits multiply to 12.

18 factors as 2*3*3. Do you see why the answer here is also 15?