We need to find 2 digit numbers (10-99) each of which has a) 8 divisors b) 9 divisors
Here's what I found: For 8 divisors, the number must have a. 3 distinct prime factors, or b. of the form {p^3, q} which are: 2^3*3=24 (1,2,3,4,6,8,12,24) 2*3*5=30 (1,2,3,5,6,10,15,30) 2^3*5=40 (1,2,4,5,8,10,20,40) 2*3*7=42 (1,2,3,6,7,14,21,42) 2*3^3=54 (1,2,3,6,9,18,27,54) 2^3*7=56 (1,2,4,7,8,14,28,56) 2*3*11=66 (1,2,3,6,11,22,33,66) 2^3*11=88 (1,2,4,8,11,22,44,88) 2*3*13=78 (1,2,3,6,13,26,39,78) 2*5*7=70 (1,2,5,7,10,14,35,70)
For 9 divisors, it is necessary that the 2-digit number be a perfect square, and the square-root is composite. For this, we can list only (2*3)^2=36 (1,2,3,4,6,9,12,18,36)
So there are 10 2-digit numbers with 8 factors, and 1 with 9 factors.
Note: looks like 72 has 12, the most factors in two digit numbers: 72 (1,2,3,4,6.8,9,12,18,24,36,72)
For this problem, we can choose to eliminate either the x values or the y values.
In this case, it would be easier to do y.
We alter the first equation by multiplying by -2 on both sides.
-4x+-6y=-16
Now, we have the equations
-4x+-6y=-16 and 5x+6y=17.
We can see that there is -6y in one equation and 6y in the other, meaning that if we add both equations together, the ys will cancel each other out because they are opposites.
