+0  
 
+9
116
9
avatar+549 

In how many ways can 36 be written as the product \(a \times b \times c \times d\) , where \(a, b, c\) and \(d\) are positive integers such that \(a \leq b \leq c \leq d\)?

 Jun 18, 2020
 #1
avatar
0

There's:

1 * 2 * 3 * 6

 Jun 18, 2020
 #2
avatar+549 
+1

that's not the only way... indecision

I need the full answer and solution

amazingxin777  Jun 18, 2020
edited by amazingxin777  Jun 18, 2020
 #3
avatar+978 
-2

do it yourself!

 

List all the combinations then arrange then to the format!

 

Guest gave a wonderful hint!

hugomimihu  Jun 18, 2020
 #4
avatar+549 
+1

tHe hint guest gave i already knew  but it is a great start! 

amazingxin777  Jun 18, 2020
 #7
avatar+307 
+1

After solving the problem, I'm not really sure how this hint ties into the problem. 

Are you saying there is 1 option for a, 2 options for b, 3 options for c, and 6 options for b? Or did you factorize 36? 

Just curious to see if your solution was maybe faster than casework.

thelizzybeth  Jun 18, 2020
edited by thelizzybeth  Jun 18, 2020
 #8
avatar+1130 
+4

How about complementary?

jimkey17  Jun 18, 2020
 #9
avatar+549 
+4

I'll think about complementary counting, haven't tried that method on this problem yet... 

Thanks for the idea

amazingxin777  Jun 22, 2020
 #5
avatar+307 
+1

There aren't that many cases because of the \(a \le b \le c \le d\) restriction, so we don't have to worry about permutations. 

Using casework we find:

 

Case 1: 3 ones (36 as 1 factor)

1*1*1*36

Case 2: 2 ones (36 as 2 factors)

1*1*2*18

1*1*3*12

1*1*4*9

1*1*6*6

Case 3: 1 one (36 as 3 factors)

1*2*3*6

1*2*2*9

1*3*3*4

Case 4: 0 ones (36 as 4 factors)

2*2*3*3

 

Counting them up, we find that there are 9.

 Jun 18, 2020
 #6
avatar+549 
+3

thank you that was correct!!!!

amazingxin777  Jun 18, 2020

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