XYZ=4000

How many ordered triples (x,y,z) of positive integers satisfy xyz=4000?
Source: NCTM Mathematics Teacher 2006

SOLUTION
This problem is similar to the problem “Triple ABC” posted on 9/14/2012, nevertheless a different solution is offered here.

Given that 4000=2^5\!\cdot 5^3, the positive integers x,y,z must take the form x=2^a\!\cdot 5^d, y=2^b\!\cdot 5^e, and z=2^c\!\cdot 5^f where a,b,c,d,e,f are non-negative integers.
xyz=2^a\!\cdot 2^b\!\cdot 2^c\!\cdot 5^e\!\cdot 5^e\!\cdot 5^f
=2^{a+b+c}\!\cdot 5^{d+e+f}
The more difficult problem of finding the single product xyz=4000 turns into the much simpler problem of finding two products, the first being 2^a\!\cdot 2^b\!\cdot 2^c such that a+b+c=5 and the second 5^d\!\cdot 5^e\!\cdot 5^f such that d+e+f=3.
If a=0, a+b+c have 6 possible values
0+0+5
0+1+4
0+2+3
0+3+2
0+4+1
0+5+0
If a=1, a+b+c have 5 possible values
1+0+4
1+1+3
1+2+2
1+3+1
1+4+0
If a=2, a+b+c have 4 possible values
2+0+3
2+1+2
2+2+1
2+3+0
If a=3, a+b+c have 3 possible values
If a=4, a+b+c have 2 possible values
If a=5, a+b+c have 1 possible value
Total = 6+5+4+3+2+1=21 possible ways to choose 2^a\!\cdot 2^b\!\cdot 2^c
Similarly, we want d+e+f=3.
If d=0, d+e+f have 4 possible values
If d=1, d+e+f have 3 possible values
If d=2, d+e+f have 2 possible values
If d=3, d+e+f have 1 possible value
Total = 4+3+2+1=10 possible ways to choose 5^d\!\cdot 5^e\!\cdot 5^f
There are 21\cdot 10=210 ways to choose 2^a\!\cdot 2^b\!\cdot 2^c\!\cdot 5^d\!\cdot 5^e\!\cdot 5^f; there are 210 ordered triples (x,y,z) such that xyz=4000.

Answer: 210

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.

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