What Are the Possible Three Digit Numbers with a Product of 12?

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Discussion Overview

The discussion revolves around identifying three-digit numbers whose digits multiply to specific products, particularly 12 and 18. Participants explore the mathematical reasoning behind the combinations of digits that yield these products, as well as extending the inquiry to four and five-digit numbers.

Discussion Character

  • Exploratory, Homework-related, Mathematical reasoning

Main Points Raised

  • One participant notes that the product of the digits in the number 126 is 12 and asks for the total number of three-digit numbers with a product of 12.
  • Another participant suggests finding the "options" of multiplications in the digits' amount as a hint for solving the problem.
  • A participant provides specific counts of three-digit numbers with products of 12 and 18, stating there are 15 for both cases.
  • Further, the same participant claims there are 36 four-digit numbers for both products and 70 five-digit numbers for a product of 12.
  • Another participant explains the factorization of 12 as 2*2*3 and discusses how to combine digits to achieve the product of 12, emphasizing the importance of ordering the digits.
  • This participant also calculates the arrangements of the digits for the combinations that yield the product of 12, leading to a total of 15 valid three-digit numbers.
  • They also note that 18 factors as 2*3*3 and imply that the reasoning for the count of three-digit numbers is similar to that for the product of 12.

Areas of Agreement / Disagreement

Participants present various counts for the number of valid three-digit, four-digit, and five-digit numbers, but there is no explicit consensus on the methodology or the correctness of the counts provided. The discussion remains unresolved regarding the general procedure for determining the counts for n-digit numbers.

Contextual Notes

Some participants rely on specific factorization and combinatorial reasoning, but the discussion does not clarify assumptions about the use of digits or the constraints on digit selection.

cherry2189
Can anyone figure this out!

please help me somebody!
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. :wink:

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

=========================
Those are the right answers.
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?
 

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