Prove that 29|2⁹⁰ + 5⁹⁰.
I am not sure how to go about proving this divisiblity problem. Can anyone help me on how to go about the process to solve it? Thanks so much.
since we're dealing with sum of cubes,
2^90 + 5^90 = (2^30 + 5^30)(2^60-10^30+5^60)
2^30 + 5^30 = (2^10 + 5^10)(2^20-10^10+5^20)
2^10 + 5^10 = 29*336781
To prove that 29 divides 2⁹⁰ + 5⁹⁰, we can use the concept of modular arithmetic. We will show that when 2⁹⁰ + 5⁹⁰ is divided by 29, the remainder is 0.
Step 1: Start by calculating the remainders of 2⁹⁰ and 5⁹⁰ when divided by 29.
To do this, apply Euler's theorem, which states that for any number a and modulus m, if a and m are coprime (i.e., they don't share any common factors other than 1), then a^(φ(m)) ≡ 1 (mod m), where φ(m) is the Euler's totient function.
In this case, 2 and 29 are coprime, so we can use Euler's theorem to find the remainder of 2⁹⁰ when divided by 29.
Step 1.1: Find φ(29):
φ(29) = 29 * (1 - 1/29) = 28.
Step 1.2: Apply Euler's theorem:
2⁹⁰ ≡ 2^(28 * 3 + 6) ≡ (2^28)³ * 2^6 ≡ 1³ * 2^6 ≡ 2^6 (mod 29).
Similarly, applying Euler's theorem to 5⁹⁰:
5⁹⁰ ≡ 5^(28 * 3 + 6) ≡ (5^28)³ * 5^6 ≡ 1³ * 5^6 ≡ 5^6 (mod 29).
Step 2: Calculate the values of 2^6 and 5^6.
2^6 = 64 ≡ 6 (mod 29),
5^6 = 15,625 ≡ 25 (mod 29).
Step 3: Substitute the values back into the original expression:
2⁹⁰ + 5⁹⁰ ≡ 6 + 25 (mod 29).
Step 4: Simplify:
2⁹⁰ + 5⁹⁰ ≡ 31 ≡ 2 (mod 29).
Since the remainder is 2 (which is not equal to 0), we can conclude that 29 does not divide 2⁹⁰ + 5⁹⁰.
Therefore, the statement "29|2⁹⁰ + 5⁹⁰" is not true.
To prove that 29 divides 2^90 + 5^90, we need to show that there is no remainder when dividing 2^90 + 5^90 by 29.
First, let's try to simplify the expression. Notice that 2^90 and 5^90 are large numbers that will be time-consuming to work with directly. However, we can use the property of modular arithmetic to simplify the expression.
We can rewrite the expression as (2^5)^18 + (5^2)^45. Now, let's find the remainders of (2^5)^18 and (5^2)^45 when divided by 29.
Using Fermat's little theorem, we know that for any prime number p and any integer a (not divisible by p), a^(p-1) ≡ 1 (mod p). In this case, p = 29.
Therefore, using the theorem, we can say that (2^5)^28 ≡ 1 (mod 29) and (5^2)^28 ≡ 1 (mod 29).
Now, let's find the remainders of (2^5)^18 and (5^2)^45 when divided by 28.
(2^5)^18 ≡ (2^5)^(28/2) ≡ (2^5)^14 ≡ 1 (mod 29)
(5^2)^45 ≡ (5^2)^(28*1 + 17) ≡ (5^2)^17 ≡ 1 (mod 29)
Therefore, we can conclude that (2^5)^18 ≡ 1 (mod 29) and (5^2)^45 ≡ 1 (mod 29).
Next, we can rewrite 2^90 + 5^90 as (2^5)^18 + (5^2)^45 ≡ 1 + 1 ≡ 2 (mod 29).
Since 2 is not divisible by 29, we can conclude that 29 does not divide 2^90 + 5^90.