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-1 1/8= -9/8

8/9* (-9/8) = -1

therefore -9/8 is not a multiplicative inverse of 8/9 because product is -1

a * 1/a =1

then 1/a is multiplicative inverse of a

(15)ⁿ can end with the digit 0 only if (15)ⁿ is divisible by 2 and 5 But prime factors of (15)ⁿ are 3ⁿ x 5ⁿ 

By Fundamental theorem of Arithmetic, there is no natural number n for which (15)ⁿ ends with the digit zero.

(d) 15/16

1095/ 1168 = 1095 ÷ 73/ 1168 ÷ 73 = 15/16 

Hence, HCF of 1095 and 1168 is 73.

(i) 1/√2 is a rational number

we find two integers b such as 1/√2 = a/b

where a and b co-prime integers (b ≠ 0)

Square both sides

(1/√2)² = (a/b)² 

⇒ 1/2 = a²/b²

b² = 2a²

So, b² divides by 2.

.’. So, b also divides by 2.

Now let b = 2c, where c is any integer

b² = (2c)²

⇒ b² = 4c²

⇒ 2a² = 4c² (∴ b² = 2a²)

⇒ a² = 2c²

Hence, a², divides by 2.

a also divides by 2.

Hence, 2 is a common  factor of a and b.

But, this contradicts the fact that a and b have no common factor other than 1.

This contradiction arises by assuming that 1/√2 is rational,

Hence 1/√2 is a irrational number.

(ii) Let 6 + √2 is a rational number

we can find two integers a and b (b ≠ 0)

Such as

6 + √2 = a/b

√2 = a/b - 6

√2 = (a - 6b)/b

a, b and 6 all are integers.

(a - 6b)/b is a rational numbers

√2 is also a rational number

But this contradicts the fact that √2 is an irrational number.

Our hypothesis is wrong.

Hence, 6 + √2 is an irrational number.

(iii) Let 3√2 is a rational number we find two integers a and b such as

3√2 = a/b (where a and b co-prime integers)

⇒ √2 = a/3b

⇒ a, b and 3 are integers.

a/3b is a rational number

√2 is a rational number

3√2 will be also a rational numbers.

But this contradicts the fact that √2 is an irrational number.

So, our hypothesis is wrong.

So, 3√2 is an irrational number.

Minimum number of rooms required = (Total number of participants) 

HCF (60, 84,108) ……..(1)

Find HCF of 60, 84 and 108

60 = 2² x 3 x 5

84 = 2² x 3 x 7

108 = 2² x 3³

HCF = 2 x 2 x 3 = 12

Form given: Total number of participants = 60 + 84 + 108 = 252

From Equation (1),

Minimum number of rooms required = 252/12 = 21

Let 5 – √3 is a rational number.

we have to find out two integers a and b such as

5 - √3 = a/b

- √3 = a/b - 5

√3 = 5 - a/b

√3 = (5b - a)/b

a, b and 5 all are integers

(5b - a)/b is a rational number.

√3 will be also rational number.

But this contradicts the fact that √3 is an irrational number.

So our hypothesis is wrong.

Hence, 5 – √3 is an irrational number.

The smallest number which when increased by 17 is exactly divisible by both 468 and 520 is obtained by subtracting 17 from the LCM of 468 and 520. 

Prime factorization of 468 and 520 is: 

468 = 2² × 3² × 13 

520 = 2³ × 5 × 13 

LCM = product of greatest power of each prime factor involved in the numbers = 2² × 3² × 5 × 13 = 4680 

The required number is 4680 – 17 = 4663. 

Hence, the smallest number which when increased by 17 is exactly divisible by both 468 and 520 is 4663.

Prime factorization: 

15 = 3 × 5 

24 = 2³ × 3 

36 = 2² × 3² 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5 = 360 

Now, the greatest four digit number is 9999. 

On dividing 9999 by 360 we get 279 as remainder. 

Thus, 

9999 – 279 = 9720 is exactly divisible by 360. 

Hence, 

the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

Prime factorization: 

15 = 3 × 5 

24 = 2³ × 3 

36 = 2² × 3² 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 3² × 5 = 360 

Now, the greatest four digit number is 9999. 

On dividing 9999 by 360 we get 279 as remainder. 

Thus, 9999 – 279 = 9720 is exactly divisible by 360. 

Hence, the greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

The greatest four digit number = 9999

We can get the required number by following steps:

Step 1: Divide 9999 by LCM of 15, 24 and 36

Step 2: Subtract result of Step 1 from 9999.

Prime factorization:

15 = 3 × 5

24 = 23 × 3

36 = 22 × 32

LCM(15, 24, 36) = 23 × 32 × 5 = 360

Step 1: Dividing 9999 by 360, we get 279 as remainder.

Step 2: Subtract 279 from 9999

9999 – 279 = 9720

The greatest number of four digits which is exactly divisible by 15, 24 and 36 is 9720.

First, let’s find the L.C.M of 8, 15 and 21. 

By prime factorisation, we have 

8 = 2 × 2 × 2 

15 = 3 × 5 

21 = 3 × 7 

⇒ L.C.M (8, 15 and 21) = 2³ × 3 × 5 × 7 = 840 

When 110000 is divided by 840, the remainder that is obtained is 800. 

So, 110000 – 800 = 109200 should be divisible by each of 8, 15 and 21. 

Also, we have 110000 + 40 = 110040 is also divisible by each of 8, 15 and 21. 

⇒ 109200 and 110040 both are greater than 100000 but 110040 is greater than 110000. 

Hence, 109200 is the number nearest to 110000 and greater than 100000 which is exactly divisible by each of 8, 15 and 21.

Correct option is (A) 46

Correct option is (C) 40

Correct option is A) distributive law

Correct option is (D) ab

Correct option is (A) log₈ 2 = 1/3

\(\sqrt[3]8\) = 2

i.e., \((8)^\frac13\) = 2

\(\Rightarrow\) \(log_8\,(8)^\frac13\) \(=log_8\,2\)   (By taking \(log_8\) both sides)

\(\Rightarrow\) \(\frac13\,log_8\,8\) \(=log_8\,2\)

\(\Rightarrow\) \(log_8\,2\) \(=\frac13\)    \((\because log_8\,8=1)\)

Correct option is log ₈ 2 = 1/3

Correct option is (A) log₁₀1000 = 3 

Correct option is (B) Irrational

a - b \(=(2+\sqrt3)\) \(-(2-\sqrt3)\)

\(=2+\sqrt3-2+\sqrt3\)

\(=2\sqrt3\) which is an irrational number.

Correct option is (B) Irrational

Correct option is (A) 1

For 6a53435 to divisible by 3, sum of its digits must be divisible by 3.

\(\therefore\) 6+a+5+3+4+3+5 is divisible by 3.

\(\Rightarrow\) 26+a is divisible by 3.

Therefore, possible values for a are 1, 4 & 7.

\(\therefore\) Least possible value of a is 1.

Correct option is (A) 1

Correct option is (A) 1/2

\(log\,_{25}5=\frac1{log\,_{5}25}\)   \((\because log_ab=\frac1{log_ba})\)

\(=\frac1{log\,_{5}5^2}\)

\(=\frac1{2\,log\,_{5}5}\)   \((\because log\,a^n=n\,log\,a)\)

\(=\frac12\)   \((\because log_aa=1)\)

Correct option is A) 1/2

Correct option is (B) 12

\(log_\sqrt2\,64=log_\sqrt2\,2^6=6\,log_\sqrt2\,2\)

\(=\frac6{log_2\,\sqrt2}\) \(=\frac6{log_2\,2^\frac12}\) \(=\frac6{\frac12\,log_2\,2}\)

\(=\frac{6\times2}1\) = 12    \((\because log_2\,2=1)\)

Correct option is B) 12

11/2³ x 3

We know either 2 or 3 is not a factor of 11, so it is in its simplest form. 

Moreover, (2³× 3) ≠ (2^m × 5ⁿ ) 

Hence, the given rational is non – terminating repeating decimal.

Correct option is (C) 2 or 5 only

If the prime factor of the denominator of a rational number are 2 or 5 only then the rational number has a repeating decimals.

Correct option is C) 2 or 5 only

Correct option is (D) 1/m

\(a^x=(\frac {a}{k})^y=k^m\)

\(\Rightarrow\) \(log\,(a^x)=log\,(\frac {a}{k})^y=log\,(k^m)\)    (By taking log)

\(\Rightarrow\) x log a = y log \(\frac ak\) = m log k

\(\Rightarrow\) x log a = m log k and y log \(\frac ak\) = m log k

\(\Rightarrow\) \(x=\frac{m\;log\,k}{log\,a}\) \(and\,y=\frac{m\;log\,k}{log\,\frac ak}\) \(=\frac{m\;log\,k}{log\,a-log\,k}\)

\((\because\) \(log\,\frac AB\) = log A - log B)

Now, \(\frac{1}{x}-\frac{1}{y}\) \(=\frac{log\,a}{m\;log\,k}\) \(-\frac{log\,a-log\,k}{m\;log\,k}\)

\(=\frac{log\,a-log\,a+log\,k}{m\;log\,k}\)

\(=\frac{log\,k}{m\;log\,k}\) \(=\frac{1}{m}\)

Correct option is D) 1/m

Correct option is (D) 3 (log 7 – log 5)

\(log\,\frac{343}{125}=log\,\frac{7^3}{5^3}\)

\(=log\,(\frac75)^3=3\,log\,\frac75\)

= 3 (log 7 – log 5)

Correct option is D) 3 (log 7 – log 5)

657 and 963
By applying Euclid’s division lemma
963 = 657 × 1 + 306
Since remainder ≠ 0, apply division lemma on division 657 and remainder 306
657 = 306 × 2 + 45
Since remainder ≠ 0, apply division lemma on division 306 and remainder 45
306 = 45 × 6 + 36
Since remainder ≠ 0, apply division lemma on division 45 and remainder 36
45 = 36 × 1 + 19
Since remainder ≠ 0, apply division lemma on division 36 and remainder 19
36 = 19 × 4 + 0
∴ HCF = 657
Given HCF = 657 + 963 × (-15)
⇒ 9 = 657 × −1445
⇒ 9 + 14445 = 657 x
⇒ 657x = 1445y
⇒ x = 1445y/657

⇒ x = 22

The require number when divides 285 and 1249, leaves remainder 9 and 7, 

this means 285 – 9 = 276 and 1249 – 7 = 1242 are completely divisible by the number
∴ The required number = HCF of 276 and 1242
By applying Euclid’s division lemma
1242 = 276 × 4 + 138
276 = 138 × 2 + 0

∴ HCF = 138
Hence remainder is = 0
Hence required number is 138

For the question it’s understood that, 

2011 – 9 = 2002 and 2623 – 5 = 2618 has to be exactly divisible by the number. 

Thus, the required number should be the H.C.F. of 2002 and 2618 

Applying Euclid’s division lemma, we get 

2618 = 2002 x 1 + 616 

2002 = 616 x 3 + 154 

616 = 154 x 4 + 0. (Here the remainder becomes 0) 

And hence the H.C.F. (2002, 2618) = 154 

∴ The required number is 154.

From the question, it’s given that 

Number of chocolates of 1st brand in a pack = 24 

Number of chocolates of 2nd brand in a pack = 15. 

So, the least number of both brands of chocolates need to be purchased is given by their LCM. 

L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5 = 120 

Hence, the number of packets of 1st brand to be bought = 120 / 24 = 5 

And, the number of packets of 2nd brand to be bought = 120 / 15 = 8

From the question it’s understood that, 

1251 – 1 = 1250, 9377 – 2 = 9375 and 15628 – 3 = 15625 has to be exactly divisible by the number. 

Thus, the required number should be the H.C.F of 1250, 9375 and 15625. 

First, consider 1250 and 9375 and apply Euclid’s division lemma 

9375 = 1250 x 7 + 625 

1250 = 625 x 2 + 0 

∴ H.C.F (1250, 9375) = 625 

Next, consider 625 and the third number 15625 to apply Euclid’s division lemma 

15625 = 625 x 25 + 0 

We get, the HCF of 625 and 12625 to be 625. 

∴ H.C.F. (1250, 9375, 15625) = 625 

So, the required number is 625.

The new numbers after subtracting remainders are:

626 - 1 = 625

3127 - 2 = 3125

15628 - 3 = 15625

Prime factors of 625 = 5 × 5 × 5 × 5

Prime factors of 3125 = 5 × 5 × 5 × 5 × 5

Prime factors of 15625 = 5 × 5 × 5 × 5 × 5 × 5

Therefore HCF of 625, 3125 and 15625 is:

5 × 5 × 5 × 5 = 625

Hence the largest number which divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively is 625

(a) 13 

We know the required number divides 65 (70 – 5) and 117 (125 – 8) 

∴ Required number = HCF (65, 117) 

We know, 

65 = 13 × 5 

117 = 13 × 3 × 3 

∴ HCF = 13

We know the required number divides 540 (546 – 6) and 756 (764 – 8), respectively. 

∴ Required largest number = HCF (540, 756) 

Prime factorization: 

540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3² × 5 

756 = 2 × 2 × 3 × 3 × 3 × 7 = 2² × 3² × 7 

∴ HCF = 2² × 3³ = 108 

Hence, the largest number is 108.

We know the required number divides 540 (546 – 6) and 756 (764 – 8), respectively. 

∴ Required largest number = HCF (540, 756) 

Prime factorization: 

 540 = 2 × 2 × 3 × 3 × 3 × 5 = 2² × 3² × 5 

756 = 2 × 2 × 3 × 3 × 3 × 7 = 2² × 3² × 7 

∴ HCF = 2² × 3³ = 108 

Hence, the largest number is 108

Euclid’s division algorithm states that for any two positive integers a and b, there exit unique integers q and r, such that a = bq + r. where 0 ≤ r ≤ b. 

We know, Dividend = Divisor × Quotient + Remainder 

Given: Divisor = 61, Quotient = 27, Remainder = 32 

Let the Dividend be x. 

∴ x = 61 × 27 + 32 = 1679 

Hence, the required number is 1679.

Given: Dividend = 1365, Quotient = 31, Remainder = 32 

Let the divisor be x. 

Dividend = Divisor × Quotient + Remainder 

1365 = x × 31 + 32 

⇒ 1365 – 32 = 31 x 

⇒ 1333 = 31 x 

⇒ x = 1333/31 = 43 

Hence, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder. 

We know, Dividend = Divisor × Quotient + Remainder 

Given: Divisor = 61, Quotient = 27, Remainder = 32 

Let the Dividend be x. 

∴ x = 61 × 27 + 32 

= 1679 

Hence, the required number is 1679.

Given: Dividend = 1365, Quotient = 31 and Remainder = 32

Let us consider, Divisor as a number y.

To Find: The value of y

As per Euclid’s lemma, we have

Dividend = (divisor  \times  quotient) + remainder

After substituting the values, we get

1365 = (y × 31) + 32

1365 – 32 = 31y

1333 = 31y

or y = 1333/31 = 43

So, 1365 should be divided by 43 to get 31 as quotient and 32 as remainder.

Let = x = 571

√x  =  √571

Now 571 lies between the perfect squares of (23)² and(24)²

Prime numbers less than 24 are 2,3,5,7, 1,1, 1,3, 17,19,23.

Since 571 is not divisible by any of the above numbers.

So, 571  is a prime number.

The approximate square root of 571 = 24. 

Prime number < 24 are 2, 3, 5, 7, 11, 13, 17, 19, & 23. But 571 is not divisible by any of these prime numbers 

so 571 is a prime number.

(7 x 13 x 11) + 11 =11 x (7 x 13 + 1)

=11 x (91+1)

= 11x 92=11 x 2 x 2 x 23 

and (7 x 6 x 5 x 4 x 3 x 2 x 1)+3

=3(7 x  6 x 5 x 4 x 2 x 1+1)

=3 x(1581) = 3 x 47 x 41.

Given numbers have more than two prime factors.

Hence both number are composite

3 x 12 x 101 + = 4(3 x 3 x 101 + 1)

= 4 (909 + 1)

=4(910)

= 2 x 2 x 2 x 5 x 7 x 13

= a composite number

:. [Product of more than two prime factors]

b = 1,001 / 143 = 7

143 = 11 x 13, 

so c = 11 or 13 

and d = 13 or 11

No, because 6ⁿ = (2 × 3)n = 2n × 3n , so the only primes in the factorisation of 6ⁿ are 2 and 3, and not 5. 

Hence, it cannot end with the digit 5.

No

According to Euclid’s division lemma, 

a = 3q + r, where 0 ≤ r < 3 

and r is an integer. Therefore, the values of r can be 0, 1 or 2.

Let x = \(\frac{p}{q}\) be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non - negative integers.

Then 'x' has a decimal expansion which terminates after n or m places, whichever is maximum

The maximum power of 2 or 5 in the given rational number is 4.

So, it will terminate after 4 places of decimals.

The decimal expansion of the given rational number will terminate after 4 places of decimals.

Let x = p/q be a rational number, such that the prime factorization of q is of the form 2ⁿ 5^m, where n, m are non - negative integers.

Then x has a decimal expansion which terminates.

The denominator 1250 can also be written as 2 × 5⁴.

The maximum power of 2 or 5 in 1250 is 4.

So, it will terminate after 4 places of decimals.

Answer is (1) 30√210

(2) \(6\sqrt[3]{3}\)