Evaluate gcd(19!+19,20!+19).
Details and assumptions
The number n!, read as n factorial, is equal to the product of all positive integers less than or equal to n. For example, 7!=7×6×5×4×3×2×1.
361
I have no idea
To evaluate gcd(19!+19, 20!+19), we need to find the greatest common divisor of the two given numbers.
Let's start by understanding the properties of the gcd.
1. gcd(a, b) = gcd(b, a): This property states that the order of the arguments does not matter. So, we can rearrange the arguments.
2. gcd(a, 0) = a: When one argument is 0, the gcd is the non-zero argument itself.
Now, let's compute 19! and 20!
19! = 19 × 18 × 17 × ... × 2 × 1
20! = 20 × 19 × 18 × ... × 2 × 1
We know that 20! is divisible by both 19! and 19 (since it contains all the numbers from 19! and 19 itself).
Next, let's evaluate gcd(19!+19, 20!+19) using the properties of the gcd:
gcd(19!+19, 20!+19) = gcd(19, 20!+19)
Based on property 1, we rearranged the arguments. Now let's consider 20!+19.
20!+19 = 19! × 20 + 19 = 19!(20 + 1) = 19! × 21
Since 19! is a factor of 20!+19, we can rewrite the equation as:
gcd(19!+19, 20!+19) = gcd(19, 19! × 21)
Since 19 is a prime number and any common divisor of two numbers must be a factor of both numbers, the greatest common divisor will be 19.
Therefore, gcd(19!+19, 20!+19) = 19.
361
soo easy