58. Let GCF (x,y) represent the greatest common factor of x any y. If p is a positive even integer less than 11, for what value of p does GCF (p^2,81) have the greatest value?
hmmm. Is it 4?
Is 4^2 = 36?
Does 4 have 3 as a factor?
Better read what I wrote, and this time pay attention.
To find the value of p that makes the GCF(p^2, 81) the greatest, we need to examine the factors of both p^2 and 81.
Let's start by finding the factors of p^2. Since p is a positive even integer less than 11, we can consider the possible values of p: 2, 4, 6, 8, and 10. Squaring these values gives us p^2: 4, 16, 36, 64, and 100.
Now let's look at the factors of 81. The factors of 81 are 1, 3, 9, 27, and 81.
To find the greatest common factor (GCF), we need to identify the highest common factor shared by both sets of factors.
For p = 2, the factors of p^2 = 4 are 1 and 4. The factors of 81 do not include 4 as a common factor, so GCF(4, 81) = 1.
For p = 4, the factors of p^2 = 16 are 1 and 16. The factors of 81 do not include 16 as a common factor, so GCF(16, 81) = 1.
For p = 6, the factors of p^2 = 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. The factors of 81 include 9 as a common factor, so GCF(36, 81) = 9.
For p = 8, the factors of p^2 = 64 are 1, 2, 4, 8, 16, 32, 64. The factors of 81 do not include 64 as a common factor, so GCF(64, 81) = 1.
For p = 10, the factors of p^2 = 100 are 1, 2, 4, 5, 10, 20, 25, 50, 100. The factors of 81 do not include 100 as a common factor, so GCF(100, 81) = 1.
Therefore, the value of p that makes GCF(p^2, 81) the greatest is p = 6.
I'd make that 6, since 81 = 3^4.
p^2 must have some powers of 3 in it, else GCF(p^2,81) = 1.
GCF(36,81) = 9
9 would be better, but we needed an even integer.