This statement is a direct consequence of a fundamental property in number theory known as the "Fundamental Theorem of Arithmetic" and some basic properties of prime numbers.

The Fundamental Theorem of Arithmetic states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers, and this representation is unique, up to the order of the factors. In other words, any integer greater than 1 can be expressed as a unique product of prime numbers.

Now, let's consider the given statement:

"If n is any prime number and a^2 is divisible by n, then n will also divide a."

Proof:

1. Let's assume that n is a prime number, and a2a2 is divisible by n. This implies that a2=kna2=kn, where k is some integer.

2. According to the Fundamental Theorem of Arithmetic, a2a2 can be expressed as the product of prime factors. Since n is prime, it must be one of the prime factors of a2a2.

3. If n is a factor of a2a2, then n must also be a factor of a (this follows from the uniqueness of prime factorization). This is because if a2=kna2=kn, then a must contain at least one factor of n, as otherwise, a2a2 would not be divisible by n.

4. Therefore, n divides a.

So, the statement is justified by the properties of prime numbers and the Fundamental Theorem of Arithmetic.

To find the smallest number that is divisible by both 90 and 144 when increased by 20, we need to find the least common multiple (LCM) of 90 and 144. Then, we'll add 20 to that LCM to get our answer.

First, let's find the LCM of 90 and 144.

The prime factorization of 90 is 2×32×52×32×5.

The prime factorization of 144 is 24×3224×32.

To find the LCM, we take the highest power of each prime factor that appears in either number:

LCM=24×32×5=720LCM=24×32×5=720

Now, we add 20 to 720:

720+20=740720+20=740

So, the smallest number that, when increased by 20, is exactly divisible by both 90 and 144 is 740.

To find the smallest number that leaves remainders of 8 when divided by 28 and 12 when divided by 32, we can use the Chinese Remainder Theorem (CRT).

The CRT states that if we have a system of congruences x≡ai(modmi)x≡ai(modmi) for i=1,2,…,ni=1,2,…,n, where the mimi are pairwise coprime, then there exists a unique solution xx modulo M=m1⋅m2⋅…⋅mnM=m1⋅m2⋅…⋅mn.

In our case, we have:

1. x≡8(mod28)x≡8(mod28)
2. x≡12(mod32)x≡12(mod32)

First, let's find the value of M=28×32=896M=28×32=896.

Next, we find the multiplicative inverses of 32 modulo 28 and of 28 modulo 32. Let's call these inverses y1y1 and y2y2 respectively.

y1y1 is such that 32×y1≡1(mod28)32×y1≡1(mod28). y2y2 is such that 28×y2≡1(mod32)28×y2≡1(mod32).

Using the Extended Euclidean Algorithm or observation, we find y1=22y1=22 and y2=9y2=9.

Now, we can use these inverses to find the solution:

x=(8×32×9+12×28×22)(mod896)x=(8×32×9+12×28×22)(mod896)

Let's compute this:

x=(2304+7392)(mod896)x=(2304+7392)(mod896) x=9696(mod896)x=9696(mod896) x=48x=48

So, the smallest number that satisfies the conditions is 48.

are irrational.