Chapter-1, Real numbers

Explanation:

i. 135 and 225

The question shows that 225 is greater than 135 as you can see. Due to Euclid's division process, we have the following:

225 = 135 × 1 + 90

So, applying the division lemma for 90 once more, the remainder is 90 0, and we obtain

135 = 90 × 1 + 45

45 0, repeating the previous process for 45, gives us,

90 = 45 × 2 + 0

Now that the remaining is zero, our approach comes to an end. The divisor in the final step is 45, HCF (225,135) = HCF (135, 90) = HCF (90, 45) = 45.

The HCF of 225 and 135 is therefore 45.

ii. 196 and 38220

Since 38220 > 196 in the given problem, we can solve it by using Euclid's procedure for division and selecting 38220 as the divisor.

38220 = 196 × 195 + 0

The remaining amount is already zero in this case. HCF(196, 38221) = 196 as a result.

As a result, 196 is the HCF of 196 and 38220.

iii. 255 and 867

867 is greater than 255, as we all know. Let's divide 867 using Euclid's technique now to obtain,

867 = 255 × 3 + 102

Remainder 102 0, therefore using 255 as the divisor and the division lemma technique, we have,

255 = 102 × 2 + 51

Again, 51 ≠ 0. The new divisor is 102, so by repeating the previous process, we obtain,

102 = 51 × 2 + 0

Our process ends here because the remaining amount is now zero. Because the divisor in the final step is 51, HCF (867,255) = HCF(255,102) = HCF(102,51) = 51.

The HCF of 867 and 255 is therefore 51.

Explanation:

With b equal to 6, let a be any positive integer. If some number q 0, and r = 0, 1, 2, 3, or 4, then a = 6q + r by Euclid's algorithm, since 0r6.

Thus, replacing r's value yields,

When r = 0, a = 6q.

Similar to this, the values of an are 6q+1, 6q+2, 6q+3, 6q+4, and 6q+5 for r=1, 2, 3, 4, and 5.

If a = 6q, 6q+2, or 6q+4, it may be divided by 2 and is an even integer. If a number is positive, it can either be even or odd. Since q is an integer, each positive odd number has the form 6q+1, 6q+3, or 6q+5.

Explanation:

Given,

Army contingent personnel total: 616

32 members of the army band

The largest common factor between the two groups must be determined whether they must march in a single column. The maximum number of columns they can march in is given by HCF(616, 32).

When we find their HCF using Euclid's technique, we get,

Since 616>32, this means

616 = 32 × 19 + 8

Given that 8 0, and using 32 as the new divisor, we have,

32 = 8 × 4 + 0

As we currently have a residual of 0, HCF (616, 32) equals 8.

The maximum number of columns they can march in is eight.

Explanation:

Let y equal 3 and x be any positive integer.

Therefore, using Euclid's division formula,

For some numbers q>=0 and r = 0, 1, and 2, as well as r>=0 and r<3, x = 3q + r.

Hence, x=3q, 3q+1, and 3q+2.

Now, by squaring both sides in accordance with the question, we obtain,

x2 = (3q)2 = 9q2 = 3 × 3q2

Let 3q2 = m

Therefore, x2 = 3m, and hence. (1)

x2 = (3q + 1)

2 = (3q) (3q)

2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

Replace 3q2+2q = m to obtain

x2= 3m + 1 ……………………………. (2)

x2= (3q + 2)

2 = (3q) (3q)

2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

Once more, substitute 3q2+4q+1 = m to obtain

x2= 3m + 1…………………………… (3)

The square of every positive integer is therefore either of the form 3m or 3m + 1 for some integer m, as shown by equations 1, 2, and 3.

Explanation:

Let y equal 3 and x be any positive integer.

Therefore, using Euclid's division formula,

x = 3q+r, where q0 and r=0, 1, 2, and r>0, r>3, respectively.

As a result, when we input the value of r, we obtain

x = 3q

or

x = 3q + 1

or

x = 3q + 2

So, if we take the cube of each of the three formulas above, we obtain,

Case I If r is zero, then

Where m = 3q3, x2= (3q)3 = 27q3, 9(3q3), and 9m

case (ii): If r equals 1, then

x3 = (3q+1)

3 = (3q)3 +13 + 3q+1 = 27q3+1 + 27q2 + 9q

Using 9 as the common component, we obtain,

x3 = 9(3q3+3q2+q)+1

Using = m, we obtain,

When we use (3q3+3q2+q) = m, we have,

x3 = 9m+1

Case (iii): If r equals 2, then

x3 = (3q+2)

3 = (3q)3+23+3+3+3+3q+2(3q+2) = 27q3+54q2+36q+8

Using 9 as the common component, we obtain,

x3=9(3q3+6q2+4q)+8

When we use (3q3+6q2+4q) = m, we obtain.

x3 = 9m+8

Hence, it is established from all three of the aforementioned situations that the cube of any positive integer has the form 9m, 9m + 1, or 9m + 8.

Explanation:

(i)140

We may obtain the prime factor of 140 by taking the LCM of that number.

Therefore, 140 = 2 × 2 × 5 ×7 × 1 = 22 × 5 × 7

(ii) 156

We may obtain the prime factor of 156 by taking the LCM of that number.

Thus, 156 = 2 × 2 × 13 × 3 × 1 = 22× 13 × 3

(iii) 3825

We may obtain the prime factor of 3825 by taking the LCM of that number.

Therefore, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32×52×17

(iv) 5005

We may obtain the prime factor of 5005 by taking the LCM of that number.

Therefore, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429

We may obtain the prime factor of 7429 by taking the LCM of that number.

Therefore, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23.

Explanation:

(i) 26 and 91

When 26 and 91 are expressed as the product of their prime factors, we obtain,

26 = 2 × 13 × 1

91 = 7 × 13 × 1

LCM (26, 91) = 2 × 7 × 13 × 1 = 182 as a result.

HCF (26, 91) = 13 as well.

Verification

The result of multiplying 26 by 91 is now 2366.

LCM and HCF combined product is 182 x 13 = 2366.

LCM × HCF thus equals the product of 26 and 91.

(ii) 510 and 92

510 and 92 are expressed as the product of its main elements to give us,

510 = 2 × 3 × 17 × 5 × 1

92 = 2 × 2 × 23 × 1

LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460 as a result.

HCF (510, 92) = 2 as well.

Verification

The result of the product of 510 and 92 is now 46920.

Moreover, the LCM and HCF product is 23460 x 2 = 46920.

LCM HCF thus equals the product of 510 and 92.

(iii)(336 and 54)

When 336 and 54 are expressed as the product of their prime components, we obtain,

336 = 2 × 2 × 2 × 2 × 7 × 3 × 1

54 = 2 × 3 × 3 × 3 × 1

Hence, LCM(336, 54) = = 3024

HCF(336, 54) = 23 = 6 as well.

Verification

The result of multiplying 336 by 54 is now 18,144.

LCM and HCF combined products are 3024 x 6 = 18,144.

As a result, LCM HCF equals the product of 336 and 54.

Explanation:

(i) 12,15 and 21

Each of the three integers' prime factors are added up to give us the following results:

12=2×2×3

15=5×3

21=7×3

Therefore,

HCF(12,15,21) = 3

LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Each of the three integers' prime factors are added up to give us the following results:

17=17×1

23=23×1

29=29×1

Therefore,

HCF(17,23,29) = 1

LCM(17,23,29) = 17 × 23 × 29 = 11339

(iii) 8, 9, and 25

Each of the three integers' prime factors are added up to give us the following results:

8=2×2×2×1

9=3×3×1

25=5×5×1

Therefore,

HCF(8,9,25)=1

LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800.

Explanation:

We must first determine the entire forms of LCM and HCF as well as their true meanings in order to answer the issue. Then, we shall represent the sum of the prime numbers 306 and 657 in writing. Then we will choose the numbers that both 306 and 657 have in common. After that, we shall multiply the chosen numbers. These chosen numbers will be added to the other numbers to create the LCM of these two numbers.

As we are aware,

The product of the two supplied numbers is HCF×LCM

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

LCM(306,657) = 22338 as a result.

Explanation:

Given that any integer with a unit place of 0 or 5 is divisible by 5, the number 6n should be divisible by 5 if its final digit is zero (0).

6n's prime factorization is (2×3)n

As a result, the prime number 5 is not included in the prime factorization of 6n.

Hence, it is evident that 6n cannot be divisible by 5 for any natural number n, and this implies that 6n cannot terminate in 0 for any natural number n.

Explanation:

According to the definition of a composite number, if a number is composite, it signifies that it includes components besides 1 and itself. Consequently, for the indicated expression;

7 × 11 × 13 + 13

Using 13 as a common number, we obtain,

=13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13

The composite number is therefore 7 × 11 × 13 + 13

Let us take the second number now.

7×6×5×4×3×2×1+5

When we use 5 as a common component, we obtain,

=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009

Hence, the composite number 7× 6× 5× 4× 3 ×2 ×1 + 5 exists.

Explanation:

The LCM of 18 and 12 can be used to calculate the time when Sonia and Ravi will reassemble at their starting place because they both move in the same direction and at the same time.

Hence, the LCM(18,12) = 2×3×3×2×1=36.

We can thus conclude from this that Sonia and Ravi will meet again at the starting point after 36 minutes.

Explanation:

Assume for the moment that √5 is a rational integer.

i.e. √5 = x/y (where, x and y are co-primes) (where, x and y are co-primes)

y√5= x

When we square both sides, we have,

(y√5)2 = x2

⇒5y2 = x2……………………………….. (1)

As a result, since x2 is divisible by 5, x is as well.

If we substitute the value of x in equation (1) with x = 5k for some value of k, we obtain

5y2 = (5k)2

⇒y2 = 5k2

is divided by five meaning that y can be divided by 5.

X and Y are obviously not co-primes. As a result, we should not assume that 5 is a rational number.

√5 is therefore an irrational number.

Explanation:

Let's assume that 3 + 2 5 makes sense.

Finding co-prime x and y (y 0) such that 3 + 25 = x/y is then possible.

Organizing results in,

Given that x and y are integers,

5 is also a rational number as a result. Yet, this goes against the idea that 5 is unreasonable.

Hence, we deduce that 3 + 25 is an irrational number.

Explanation:

(i)1/√2

Let's assume 1/2 is a rational number.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y

Organizing results in,

√2 = y/x

Despite the fact that 2 is irrational, it is a rational number because x and y are both integers.

As a result, we can say that 1/2 is irrational.

(ii) 7√5

Let's assume that the number 75 is rational.

Thus that 75 = x/y, we can identify coprime a and b (b 0).

Organizing results in,

√5 = x/7y

In spite of the fact that 5 is irrational because x and y are integers, 5 is a rational number.

Hence, we might say that 75 is irrational.

(iii) 6 +√2

Assume that 2 + 6 is a rational number.

Thus that 6 + 2 = x/y, we can obtain co-primes x and y (y 0).

Organizing results in,

√2 = (x/y) – 6

As x and y are integers, (x/y) - 6 is a rational number, which means that 2 is also a rational number. The idea that 2 is an irrational number is refuted by this.

Hence, we might say that 6 + 2 is irrational.

Explanation:

The denominator has terminating decimal expansion if it only contains the numbers 2 and 5, or if it takes the form 2m 5n.

The denominator has a non-terminating decimal expansion if the denominator contains factors than 2 and 5.

(i) 13/3125

When we factor the denominator, we have,

3125 = 5 × 5 × 5 × 5 × 5 is equal to 55

As there is just one element in the denominator, 5, 13/3125 has a terminating decimal expansion.

(ii) 17/8

When we factor the denominator, we have,

8 = 2×2×2 = 23

As there is just one element in the denominator, 2, 17/8 has a terminating decimal expansion.

(iii) 64/455

When we factor the denominator, we have,

455 = 5×7×13

64/455 has a non-terminating decimal expansion because the denominator is not in the form of 2m 5n.

(iv) 15/ 1600

When we factor the denominator, we have,

1600 = 26×52

As 2m 5n is the denominator's form, 15/1600 has a terminating decimal expansion.

(v) 29/343

When we factor the denominator, we have,

343 = 7×7×7 = 73 29/343 has a non-terminating decimal expansion because the denominator is not in the form of 2m 5n.

(vi) 23/(2352)

It is obvious that the denominator takes the shape of 2m 5n.

Hence, 23/ (2352) has a decimal expansion that terminates.

(vii) 129/(225775)

It is clear that the denominator does not have the form 2m 5n.

This results in a non-terminating decimal expansion for 129/ (225775).

(viii) 6/15

6/15 = 2/5

Since the denominator only has the number 5 as a component, the decimal expansion of 6/15 is terminating.

(ix) 35/50

35/50 = 7/10

When we factor the denominator, we have,

10 = 2 × 5

The denominator of 35/50 has a terminating decimal expansion since it is in the form 2m 5n.

(x) 77/210

77/210 = (7× 11)/ (30 × 7) is equal to 11/30

When we factor the denominator, we have,

30 = 2 × 3 × 5

The denominator is not in the form of 2m 5n, as you can see.

As a result, the decimal expansion of 77/210 is non-terminating.

Explanation:

Explanation:

(i) 43.123456789

It is a rational number in the form of p/q since it has a terminating decimal expansion, and q only has factors of 2 and 5.

(ii) 0.120120012000120000…

 It is an irrational number since its decimal expansion is neither terminating nor repeating.

It is a rational number in the form of p/q since it has non-terminating but repeating decimal expansion, and q has factors besides 2 and 5.