1 x 1 x36 = 36
1 x 2 x18 = 36
1 x 3 x12 = 36
1 x 4 x9 = 36
1 x 6 x6 = 36
1 x 9 x4 = 36
1 x 12 x3 = 36
1 x 18 x2 = 36
1 x 36 x1 = 36
2 x 1 x18 = 36
2 x 2 x9 = 36
2 x 3 x6 = 36
2 x 6 x3 = 36
2 x 9 x2 = 36
2 x 18 x1 = 36
3 x 1 x12 = 36
3 x 2 x6 = 36
3 x 3 x4 = 36
3 x 4 x3 = 36
3 x 6 x2 = 36
3 x 12 x1 = 36
4 x 1 x9 = 36
4 x 3 x3 = 36
4 x 9 x1 = 36
6 x 1 x6 = 36
6 x 2 x3 = 36
6 x 3 x2 = 36
6 x 6 x1 = 36
9 x 1 x4 = 36
9 x 2 x2 = 36
9 x 4 x1 = 36
12 x 1 x3 = 36
12 x 3 x1 = 36
18 x 1 x2 = 36
18 x 2 x1 = 36
36 x 1 x1 = 36
2007-12-21 02:16:25
·
answer #1
·
answered by cidyah 7
·
0⤊
2⤋
36 = 4 x 9
36 = 2 x 2 x 3 x 3
OR
36 = 1 x 2 x 18
36 = 1 x 3 x 12
36 = 1 x 4 x 9
36 = 1 x 6 x 6
2007-12-21 01:39:49
·
answer #2
·
answered by Como 7
·
1⤊
0⤋
1 X 36
2 X 18
3 X 12
4 X 9
6 X 6
1 X 1 X 36
1 X 2 X 18
1 X 3 X 12
1 X 4 X 9
1 X 1 X 36
2 X 2 X 9
2 X 3 X 6
3 X 3 X 4
There are lots of them
2007-12-21 01:37:58
·
answer #3
·
answered by Jeƒƒ Lebowski 6
·
3⤊
0⤋
I don't much care for 1 as a factor of a number, it is too open ended. You can have as many ones as you like.
Now if the puzzle were "Find out how many distinct ways of multiplying three integers together there are that give 36 as a product ". Then 1 is much more satisfactory factor. Don't forget the negative integers either.
1 x 1 x 36
-1 x -1 x 36
1 x -1 x -36
1 x 2 x 18 (four ways of assigning sign)
1 x 3 x 12 (ditto)
1 x 4 x 9 (4 ways)
1 x 6 x 6 (only three ways of assigning sign)
2 x 2 x 9 (3 ways)
2 x 3 x 6 (4 ways)
3 x 3 x 4 (3 ways)
I make that 27 distinct integer triples that when multiplied together give 36 as the product.
2007-12-21 02:05:45
·
answer #4
·
answered by frothuk 4
·
0⤊
1⤋
To each of the above you can add 1's as many as you like:
Consider the factors ordered in increasing value
Lowest factor 2
2*2*3*3 (prime factorization)
2*2*9
2*3*6
2*18
Lowest Factor 3
3*3*4
3*12
Lowest factor 4
4*9
Lowest factor 6
6*6
2007-12-21 01:45:16
·
answer #5
·
answered by zazensoto 3
·
0⤊
1⤋
3 x 3 x 4 is the best combination with three numbers
2007-12-21 02:11:54
·
answer #6
·
answered by Anonymous
·
0⤊
2⤋
It would be easy to show with the prime factorization tree. Difficult to do it here though.
Check the first answer then.
Good Luck,
Kempos
2007-12-21 01:40:42
·
answer #7
·
answered by Anonymous
·
0⤊
1⤋
36x1
1x31
2013-09-24 09:02:39
·
answer #8
·
answered by Anonymous
·
0⤊
0⤋