Problem:
Find the number of 7-tuples of positive integers (a,b,c,d,e,f,g) that satisfy the following systems of equations:
abccdeefg=70=71=72
Solution:
We begin by observing that the variable c appears in both the first and second equations. Therefore, it must divide both abc=70 and cde=71. Since gcd(70,71)=1, the only common positive integer divisor of 70 and 71 is 1. Thus, the only possible value for c is 1.
Substituting c=1 into the equations, we get:
- ab=70
- de=71
- efg=72
Now, we will count the number of positive integer solutions to this system.
1. For ab=70:
The number of positive integer solutions (a,b) is equal to the number of positive divisors of 70=2⋅5⋅7, which is (1+1)(1+1)(1+1)=8.
2. For de=71:
Since 71 is a prime number, the only factor pairs are (1,71) and (71,1). So there are 2 choices for (d,e).
3. For efg=72:
This depends on the value of e. From above, the only valid values of e are 1 and 71.
- If e=1, then fg=72. The number of positive integer solutions (f,g) is equal to the number of positive divisors of 72=23⋅32, which is (3+1)(2+1)=12.
- If e=71, then fg=7172, which is not an integer, so this case is invalid.
Therefore, only the case e=1 contributes valid solutions.
Thus, the total number of valid tuples is: 8⋅1⋅12=96.
The problems on this page are the property of the MAA's American Mathematics Competitions