Learn UNIT I: Number systems with Free Lessons & Tips

To find the Highest Common Factor (HCF) of 52 and 117 and express it in the form 52x + 117y, we can use the Euclidean algorithm.

1. Step 1: Find the Remainder

   Divide the larger number by the smaller number and find the remainder.

   117=2×52+13117=2×52+13

   So, the remainder is 13.

2. Step 2: Replace Numbers

   Now, replace the divisor with the previous remainder, and the dividend with the divisor.

   52=4×13+052=4×13+0

   Here, the remainder is 0, so we stop. The divisor at this step, which is 13, is the HCF.

3. Expressing in the given form

   Now, we backtrack to express the HCF, which is 13, in the form 52x + 117y.

   Using the reverse steps of the Euclidean algorithm, we can express the HCF as a linear combination of 52 and 117:

   13=117−2×5213=117−2×52

   Hence, the HCF of 52 and 117 expressed in the form 52x + 117y is 13=117−2×5213=117−2×52.

Sure, let's prove this statement.

Let's start with m=2k+1m=2k+1 and n=2l+1n=2l+1, where kk and ll are integers.

Now, let's square both mm and nn:

m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1m2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1 n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1n2=(2l+1)2=4l2+4l+1=2(2l2+2l)+1

Now, let's sum them:

m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2m2+n2=2(2k2+2k)+1+2(2l2+2l)+1=2(2k2+2k+2l2+2l)+2

Since 2k2+2k+2l2+2l2k2+2k+2l2+2l is an integer, let's denote it as qq. Then:

m2+n2=2q+2m2+n2=2q+2

This clearly shows that m2+n2m2+n2 is even, as it is divisible by 22.

To prove that m2+n2m2+n2 is not divisible by 44, let's consider the possible remainders when dividing by 44:

• If kk and ll are both even, then m2+n2m2+n2 will leave a remainder of 22 when divided by 44.
• If kk and ll are both odd, then m2+n2m2+n2 will also leave a remainder of 22 when divided by 44.

Thus, m2+n2m2+n2 is even but not divisible by 44 when both mm and nn are odd positive integers.

To find aa, we can use the relationship between the highest common factor (HCF) and the least common multiple (LCM).

Given: HCF(6,a)=2HCF(6,a)=2 LCM(6,a)=60LCM(6,a)=60

We know that the product of the HCF and LCM of two numbers is equal to the product of the numbers themselves. So, we have:

HCF(6,a)×LCM(6,a)=6×aHCF(6,a)×LCM(6,a)=6×a

Substituting the given values:

2×60=6×a2×60=6×a 120=6a120=6a

Now, we solve for aa:

a=1206a=6120 a=20a=20

Therefore, a=20a=20.

To find the size of tiles required and how many tiles are needed, we first need to determine the largest possible size of the square marble tiles that can fit evenly into the room's dimensions.

Given:

• Room dimensions: 10m x 7m

To find the largest possible size of the square tiles, we need to find the greatest common divisor (GCD) of the dimensions (length and width) of the room.

GCD(10, 7) = 1

This means that there's no integer length that evenly divides both 10 and 7. So, we cannot find a single tile size that perfectly fits the room without cutting.

However, we can approximate the largest possible tile size by using factors of the GCD, which in this case is 1. So, theoretically, we could use a tile size of 1m x 1m, but that wouldn't be practical.

In real-world scenarios, the tile size is usually chosen for convenience and aesthetics. A common approach is to use a tile size that evenly divides the room dimensions, even if it involves some cutting.

Let's say we decide to use a 0.5m x 0.5m tile size. Then, we can calculate how many tiles are needed:

For the length of the room (10m), we would need 10m / 0.5m = 20 tiles. For the width of the room (7m), we would need 7m / 0.5m = 14 tiles.

So, in total, we would need 20 tiles lengthwise and 14 tiles widthwise, resulting in 20 x 14 = 280 tiles.

However, keep in mind that some tiles will need to be cut to fit the edges of the room, especially along one of the dimensions (either length or width). The number of tiles that need to be cut will depend on the exact layout of the tiles and the dimensions of the room.

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.

Sure, let's prove this by induction.

Base Case:
When n=2n=2, a2=a×aa2=a×a, where aa is a positive rational number. The product of two rational numbers is rational, so a2a2 is rational.

Inductive Step:
Assume that anan is rational for some positive integer n>1n>1.

Consider an+1an+1: an+1=an×aan+1=an×a

By the inductive hypothesis, anan is rational. And since aa is rational (given in the problem), the product of two rational numbers is rational. Therefore, an+1an+1 is rational.

By mathematical induction, anan is rational for all positive integers n>1n>1.

). Therefore, this example demonstrates that the product of a rational number and an irrational number can indeed be rational.

are irrational.