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Fundamental Theorem of Arithmetic:

Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

The correct option is: (C) 555681

Explanation:

987 = 3 x 7 x 47

So, the required number must be divisible by each one of 3,7,47.

553681 => (Sum of digits = 28, not divisible by 3) 

555181 => (Sum of digits = 25, not divisible by 3) 

555681 is divisible by each one of 3, 7, 47.

The correct option is: (C) p divides b 

Explanation:

If p divides b², then p also divides b.

The minimum distance that they all three should walk will be the L.C.M. of 40, 45, 50.

Since  40=2×2×2×5

 45=3×3×5

 50=2×5×5

∴ L.C.M. of 40, 45 and 50 = 2×2×2×3×3×5×5= 1800cm = 18m

The correct option is: (D) 1120

Explanation:

L.C.M. (35, 32, 40) = 2⁵ x 5 x 7 = 1120

So, each should walk 1120 cm so that each can cover the same distance in complete steps.

The correct option is: (B) 25

Explanation:

HCF of (192,240,168) = 2 x 2 x 2 x 3 = 24

Number of rooms for participants in Mathematics, physics and Biology respectively is

= 192/24 = 8, 240/24 = 10 and 168/24 = 7

. ^.. Total minimum number of required rooms = 25

We know that LCM = Product of the greatest power of each prime factor, involved in the numbers.

So,

LCM(a, b) = p³q ²

A number which can be expressed in algebraic form i.e., in p/q form is called a rational number.

E.g.: 3/5, -4/9 etc.

Largest number which divides 438 and 606, leaving remainder 6 is actually the largest number which divides 438 – 6 = 432 and 606 – 6 = 600, leaving remainder 0. 

Therefore, HCF of 432 and 600 gives the largest number. 

Now, prime factors of 432 and 600 are: 

432 = 2⁴ × 3³ 

600 = 2³ × 3 × 5² 

HCF = product of smallest power of each common prime factor in the numbers = 2³ × 3 = 24 

Thus, the largest number which divides 438 and 606, leaving remainder 6 is 24.

(b) 16 

We know that the required number divides 240 (245 – 5) and 1024 (1029 – 5). 

∴ Required number = HCF (240, 1024) 

240 = 2 × 2 × 2 × 2 × 3 × 5 

1024 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 

∴ HCF = 2 × 2 × 2 × 2 = 16

The prime factorization of 196 is as follows:

196 = 2 × 2 × 7 × 7

⇒ 98 = 2² ×7²

We know that the exponent of a number a^m is m.

∴The sum of powers = 2 + 2 = 4

We know that any odd positive integer is of the form 4q+1 or, 4q+3 for some whole number q. 

Now that it’s given a > b 

So, we can choose a= 4q+3 and b= 4q+1. 

∴ \(\frac{(a+b)}{2}\) = \(\frac{[(4q+3) + (4q+1)]}{2}\)

⇒ \(\frac{(a+b)}{2}\) = \(\frac{(8q+4)}{2}\) 

⇒ \(\frac{(a+b)}{2}\) = 4q+2 = 2(2q+1) which is clearly an even number. 

Now, doing \(\frac{(a-b)}{2}\) 

⇒ \(\frac{(a-b)}{2}\) = \(\frac{[(4q+3)-(4q+1)]}{2}\) 

⇒ \(\frac{(a-b)}{2}\) = \(\frac{(4q+3-4q-1)}{2}\) 

⇒ \(\frac{(a-b)}{2}\) = \(\frac{(2)}{2}\)

⇒ \(\frac{(a-b)}{2}\) = 1 which is an odd number. 

Hence, one of the two numbers \(\frac{(a+b)}{2}\) and \(\frac{(a-b)}{2}\) is odd and the other is even.

Given:

a and b are two odd positive integers such that a > b.

To prove:

Out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) one is odd and other is even.

Proof:

a and b both are odd positive integers,

Since,

\((\frac{a+b}{2})\) + \((\frac{a-b}{2})\) = \((\frac{a+b+a-b}{2})\) = a

Which is an odd number as "a" is given to be a odd number.

Hence one of the number out of \(\frac{a+b}{2}\), \(\frac{a-b}{2}\) must be even and other must be odd because adding two even numbers gives an even number and adding two odd numbers gives an even number.

Here,

\(\frac{a+b}{2}\) will give an odd number and \(\frac{a-b}{2}\) will give an even number.

EXAMPLE:

Take a = 7 and b = 3 such that a > b

Now,

\(\frac{a+b}{2}\) = \(\frac{7+3}{2}\) = \(\frac{10}{2}\) = 5 which is odd

And \(\frac{a-b}{2}\) = \(\frac{7-3}{2}\) = \(\frac{4}{2}\) = 2 which is even

Conclusion:

Out of out of the numbers \(\frac{a+b}{2}\), \(\frac{a-b}{2}\), \(\frac{a+b}{2}\) is an odd number and \(\frac{a-b}{2}\) is an even number.

Correct option is (D) Every integer is a rational number

Every natural number is a whole number, every whole number is an integer and every integer is a rational number but every rational number need not be an integer, or a whole number or a natural number.

Hence, only correct option is D.

D) Every integer is a rational number

The correct answer is

a² b²

Correct option is (A) an odd number

If n = 2 then \(n^2-1=2^2-1\) = 4 - 1 = 3 which is not divisible by 8.

Thus, options B, C and D are incorrect.

Hence, only A can be correct.

Alternative :-

Let n = 2m+1, \(m\in N.\)

Then \(n^2-1=(2m+1)^2-1\) \(=4m^2+4m+1-1\)

\(=4m^2+4m\) = 4m (m+1)

\(\because\) m (m+1) is always even number.

Thus m (m+1) = 2p, \(p\in N\)

\(\therefore\) \(n^2-1\) = 4m (m+1) = 8p, \(p\in N\)

\(\therefore\) \(n^2-1\) is divisible by 8 if n is an odd number.

Correct option is (A) an odd number

Let the two odd positive numbers x and y be 2k + 1 and 2p + 1, respectively

i.e., x² + y² = (2k + 1)² +(2p + 1)²

= 4k² + 4k + 1 + 4p² + 4p + 1

= 4k² + 4p² + 4k + 4p + 2

= 4 (k² + p² + k + p) + 2

Thus, the sum of square is even the number is not divisible by 4

Therefore, if x and y are odd positive integer, then x² + y² is even but not divisible by four.

Hence Proved

3³x5=3x3x3x5

3²x5²=3x3x5x5

H.C.F=3x3x5=45

Number of goats = 205
Number of donkey = 140
Number of cows = 175
∴ The largest number of animals in one trip = HCF of 105, 140 and 175
First consider 105 and 140
By applying Euclid’s division lemma
140 = 105 × 1 + 35
105 = 35 × 3 + 0
∴ HCF of 105 and 140 = 35
Now consider 35 and 175
By applying Euclid’s division lemma
175 = 35 × 5 + 0
HCF of 105, 140 and 175 = 35

In order to calculate the time take before they meet again, we must first find out the individual time taken by each cyclist in covering the total distance. 

Number of days a cyclist takes to cover the circular field 

= \(\frac{(Total\, distance\, of\, the\, circular\, field\,) }{ (distance\, covered\, in\, 1\, day\, by\, a\, cyclist\,)}\)

So, for the 1st cyclist, number of days = \(\frac{360}{48}\) = 7.5 which is = 180 hours [∵1 day = 24 hours] 

2nd cyclist, number of days = \(\frac{360}{60}\) = 6 which is = 144 hours 

3rd cyclist, number of days = \(\frac{360}{72}\) = 5 which is 120 hours 

Now, by finding the LCM (180, 144 and 120) we’ll get to know after how many hours the three cyclists meet again. 

By prime factorisation, we get 

180 = 2² x 3² x 5 

144 = 2⁴ x 3² 

120 = 2³ x 3 x 5 

⇒ L.C.M (180, 144 and 120) = 2⁴ x 3² x 5 = 720 

So, this means that after 720 hours the three cyclists meet again. 

⇒ 720 hours = \(\frac{720}{24}\) = 30 days [∵1 day = 24 hours] 

Thus, all the three cyclists will meet again after 30 days.

Therefore the required smallest number would be the LCM of the numbers

Prime factors of 35 = 5 × 7

Prime factors of 56 = 2 × 2 × 2×7

Prime factors of 91 = 7 × 13

LCM of 35, 56 and 91 ⇒ 2 × 2 × 2 × 5 × 7×13 = 3640

Number = LCM + Remainder

⇒ 3640 + 7 = 3647

The required number is determine with the help of LCM of 35, 56 and 91.

Using prime factorization, find the LCM:

35 = 5 x 7

56 = 2 x 2 x 2 x 7

91 = 7 x 13

LCM (35, 56 and 91) = 2³ x 5 x 7 x 13 = 3640

Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91. 

Prime factorization of 35, 56 and 91 is: 

35 = 5 × 7 

56 = 2³ × 7 

91 = 7 × 13 

LCM = product of greatest power of each prime factor involved in the numbers = 2³ × 5 × 7 × 13 = 3640 

Least number which can be divided by 35, 56 and 91 is 3640. 

Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Thus, the required number is 3647.

First LCM of 20, 25, 35 and 40, using prime factorization method

20 = 2×2×5

25 = 5×5

35 = 5×7

40 = 2 x 2×2×5

LCM (20,25,35, 40) = 1400

As per given statement, we have to find the greatest number which when divided by 20, 25, 35 and 40 leaves remainder as 14 , 19, 29 and 34 respectively

i.e. 6 less than the divisor in each case.

Required number = 1400 – 6 = 1394

1394 is the required number.

Given numbers are 43, 91 and 183.

Subtract smallest number from both the highest numbers.

we have three cases:

183 > 43; 183 > 91 and 91 > 43

183 – 43 = 140

183 – 91 = 92 and

91 – 43 = 48

Now, we have three new numbers: 140, 48 and 92.

Find HCF of 140, 48 and 92 using prime factorization method, we get

HCF (140, 48 and 92) = 4

The highest number that divide 183, 91 and 43 and leave the same remainder is 4.

i.  Let x = 9.315315 … = \(9.\overline{315}\) …(i) 

Since, three numbers i.e. 3, 1 and 5 are repeating

after the decimal point. 

Thus, multiplying both sides by 1000, 

1000x = 9315.315315…

∴ 1000x =\(9315.\overline{315}\)...(ii)

Subtracting (i) from (ii)

∴ 1000x - x =\(9315.\overline{315}\) - \(9.\overline{315}\)

∴  999x = 9306

∴  \(x=\frac{9306}{999}\) = \(\frac{9\times1034}{9\times111}\) = \(\frac{1034}{111}\) 

∴  9.315315... = \(\frac{1034}{111}\) 

ii.  Let x = 357.417417…= \(357.\overline{417}\) ...(i)

Since, three numbers i.e. 4, 1 and 7 are repeating after the decimal point. 

Thus, multiplying both sides by 1000, 

1000x = 357417.417417… 

∴ 1000x = 357417.417 …(ii) 

Subtracting (i) from (ii), 

1000x – x =\(357417.\overline{417}\) - \(357.\overline{417}\) 

∴ 999x = 357060

∴ \(x=\frac{357060}{999}\) = \(\frac{3\times119020}{3\times333}\) 

∴ 357.417417... = \(\frac{119020}{333}\)

i. 0.555= \(\frac{0.555\times1000}{1\times1000}\) =\(\frac{555}{1000}\) = \(\frac{5\times111}{5\times200}\)  = \(\frac{111}{200}\)

ii. Let x = \(29.\overline{568}\)…(i) 

x = 29.568568… 

Since, three numbers i.e. 5, 6 and 8 are repeating after the decimal point. 

Thus, multiplying both sides by 1000, 

1000x = \(29568.\overline{568}\)…(ii) 

Subtracting (i) from (ii),

1000x – x =\(29568.\overline{568}\) - \(29.\overline{568}\)

∴ 999x = 29539

∴ \(x=\frac{29539}{999}\) 

∴ \(29.\overline{568}\) \(\frac{29539}{999}\)

Given integers here are 225 and 135. On comparing, we find 225 > 135. 

So, by applying Euclid’s division lemma to 225 and 135, we get 

867 = 225 x 3 + 192 

Since the remainder ≠ 0. So we apply the division lemma to the divisor 135 and remainder 90. 

⇒ 135 = 90 x 1 + 45 

Now we apply the division lemma to the new divisor 90 and remainder 45. 

⇒ 90 = 45 x 2 + 0 

Since the remainder at this stage is 0, the divisor will be the HCF. 

Hence, the H.C.F of 225 and 135 is 45.

Given integers here are 196 and 38220. On comparing, we find 38220 > 196. 

So, by applying Euclid’s division lemma to 38220 and 196. We get, 

38220 = 196 x 195 + 0

Since the remainder at this stage is 0, the divisor will be the HCF. 

Hence, the HCF of 38220 and 196 is 196.

Answer :
(i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

(i) HCF of 135 and 225

Here 225 > 135 we always divide greater number with smaller one.
Divide 225 by 135 we get 1 quotient and 90 as remainder so that
225= 135*1 + 90
Divide 135 by 90 we get 1 quotient and 45 as remainder so that
135= 90*1 + 45
Divide 90 by 45 we get 2 quotient and no remainder so we can write it as
90 = 2*45+ 0
As there are no remainder so deviser 45 is our HCF

(ii) HCF of 196 and 38220 :

38220>196 we always divide greater number with smaller one.
Divide 38220 by 196 then we get quotient 195 and no remainder so we can write it as
38220 = 196 * 195 + 0
As there is no remainder so deviser 196 is our HCF

(iii) HCF of 867 and 255
867>255 we always divide greater number with smaller one.
divide 867 by 255 then we get quotient 3 and remainder is 102
so we can write it as
867 = 255 * 3 + 102
Divide 255 by 102 then we get quotient 2 and remainder is 51
So we can write it as
255 = 102 * 2 + 51
Divide 102 by 51 we get quotient 2 and no remainder
So we can write it as
102 = 51 * 2+ 0
As there is no remainder so deviser 51 is our answer

Since 867 > 255, we apply the division algorithm to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division algorithm to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division algorithm to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.

So, HCF (867,255) = 51

Correct option is (B) xy

\((x+y)^{-1}(x^{-1}+y^{-1})\) \(=\frac1{x+y}(\frac1x+\frac1y)\)

\(=\frac1{x+y}(\frac{y+x}{xy})=\frac1{xy}\)

= reciprocal of xy

Thus, given expression \((x+y)^{-1}(x^{-1}+y^{-1})\) is the reciprocal of xy.

Correct option is C) 1/xy

Correct option is (B) 6

\((x-1)^2+(y-2)^4+(z-3)^6 = 0\)

\(\because\) Square of any real number is always greater or equal to zero.

i.e., \(a^2\geq0\) and sum of positive numbers is always positive.

\(\therefore\) \((x-1)^2+(y-2)^4+(z-3)^6 = 0\) is only possible when

x-1 = 0, y-2 = 0 and z-3 = 0

\(\Rightarrow\) x = 1, y = 2 and z = 3

\(\therefore\) Value of xyz \(=1\times2\times3=6\)

Correct option is B) 6

(i) In order to arrange the books as required, we have to find the largest number that divides 96,240 and 336 exactly, clearly, such a number is their HCF.

We have,   96  =  2⁵ x 3

240  =  2⁴ x 3 x 5

and  336   =   2⁴ x 3 x 7

.'.   HCF of 95,240, and 336 is 2⁴ x 3  =   48

So, there must be 48 books in each stack. 

Number of stacks of English books

=    96 / 48   =  2

Number of stacks of Hindi books

240 / 48  =  5

Number of stacks of Sociology books

= 336 / 48  = 7 

(ii) HCF of numbers. 

(iii) Cleanliness has been discussed in this question, it is a good habit that leads to good health

Given: The lengths of three pieces of timber are 42 m, 49 m and 63 m respectively. We have to divide the timber into equal length of planks.

Greatest possible length of each plank = HCF (42, 49, 63)

The prime factorization of 42, 49 and 63 are:

42 = 2 x 3 x 7

49 = 7 x 7

63 = 3 x 3 x 7

HCF (42, 49, 63 ) = 7

The greatest possible length of each plank is 7 m.

Again, The number of planks can be formed = 22

The three given lengths are 7m (700cm), 3m 85cm (385cm) and 12m 95m (1295cm). (∵1m = 100cm). 

∴ Required length = HCF (700, 385, 1295) 

Prime factorization: 

700 = 2 × 2 × 5 × 5 × 7 = 2² × 5² × 7 

385 = 5 × 7 ×11 

1295 = 5 × 7 × 37 

∴HCF = 5 × 7 = 35 

Hence, 

the greatest possible length is 35 cm.

Total number of English books = 336 

Total number of mathematics books = 240 

Total number of science books = 96 

∴ Number of books stored in each stack = HCF (336, 240, 96) 

Prime factorization: 

336 = 2⁴ × 3 × 7 

240 = 2⁴ × 3 × 5 

96 = 2⁵ × 3 

∴ HCF = Product of the smallest power of each common prime factor involved in the numbers = 2⁴ × 3 = 48 

Hence, 

we made stacks of 48 books each.

∴ Number of stacks = \(\frac{336}{48}\) + \(\frac{240}{48}\) + \(\frac{96}{48}\) = (7+5+2) = 14

The lengths of three pieces of timber are 42m, 49m and 63m respectively. 

We have to divide the timber into equal length of planks. 

∴ Greatest possible length of each plank = HCF (42, 49, 63) 

Prime factorization: 

42 = 2 × 3 × 7 

49 = 7 × 7 

63 = 3 × 3 × 7 

∴HCF = Product of the smallest power of each common prime factor involved in the numbers = 7 

Hence, 

the greatest possible length of each plank is 7m.

The lengths of three pieces of timber are 42m, 49m and 63m respectively. 

We have to divide the timber into equal length of planks. 

∴ Greatest possible length of each plank = HCF (42, 49, 63) 

Prime factorization: 

42 = 2 × 3 × 7 

49 = 7 × 7 

63 = 3 × 3 × 7 

∴HCF = Product of the smallest power of each common prime factor involved in the numbers = 7 

Hence, the greatest possible length of each plank is 7m.

Let’s first choose 136 and 170 to find the HCF by using Euclid’s division lemma. 

Thus, we obtain 

170 = 136 x 1 + 34 

Since the remainder 34 ≠ 0. So we apply the division lemma to the divisor 136 and remainder 34. We get, 

136 = 34 x 4 + 0 

The remainder at this stage is 0, the divisor will be the HCF i.e., 34 for 136 and 170. 

Now, we again use Euclid’s division lemma to find the HCF of 34 and 255. And we get, 

255 = 34 x 7 + 17 

Since the remainder 17 ≠ 0. So we apply the division lemma to the divisor 34 and remainder 17. We get, 

34 = 17 x 2 + 0 

So, this stage has remainder 0. Thus, the HCF of the third number 255 and 34 is 17. 

Hence, the HCF of 136, 170 and 255 is 17.

By applying Euclid’s division lemma on 506 and 1155, we get 

1155 = 506 x 2 + 143…………. (1)

As the remainder ≠ 0, apply division lemma on divisor 506 and remainder 143 

506 = 143 x 3 + 77…………….. (2) 

As the remainder ≠ 0, apply division lemma on divisor 143 and remainder 77 

143 = 77 x 1 + 66……………… (3) 

Since the remainder ≠ 0, apply division lemma on divisor 77 and remainder 66 

77 = 66 x 1 + 11……………….. (4) 

As the remainder ≠ 0, apply division lemma on divisor 66 and remainder 11 

66 = 11 x 6 + 0………………… (5) 

Thus, we can conclude the H.C.F. = 11. 

Now, in order to express the found HCF as a linear combination of 506 and 1155, we perform 

11 = 77 – 66 x 1 [from (4)] 

= 77 – [143 – 77 x 1] x 1 [from (3)] 

= 77 – 143 x 1 + 77 x 1 

= 77 x 2 – 143 x 1 

= [506 – 143 x 3] x 2 – 143 x 1 [from (2)] 

= 506 x 2 – 143 x 6 – 143 x 1 

= 506 x 2 – 143 x 7 

= 506 x 2 – [1155 – 506 x 2] x 7 [from (1)] 

= 506 x 2 – 1155 x 7+ 506 x 14 

= 506 x 16 – 1155 x 7

By applying Euclid’s division lemma on 592 and 252, we get 

592 = 252 x 2 + 88……… (1) 

As the remainder ≠ 0, apply division lemma on divisor 252 and remainder 88 

252 = 88 x 2 + 76………. (2) 

As the remainder ≠ 0, apply division lemma on divisor 88 and remainder 76 

88 = 76 x 1 + 12………… (3) 

As the remainder ≠ 0, apply division lemma on divisor 76 and remainder 12 

76 = 12 x 6 + 4………….. (4) 

Since the remainder ≠ 0, apply division lemma on divisor 12 and remainder 4 

12 = 4 x 3 + 0……………. (5) 

Thus, we can conclude the H.C.F. = 4. 

Now, in order to express the found HCF as a linear combination of 592 and 252, we perform 

4 = 76 – 12 x 6 [from (4)] 

= 76 – [88 – 76 x 1] x 6 [from (3)] 

= 76 – 88 x 6 + 76 x 6 

= 76 x 7 – 88 x 6 

= [252 – 88 x 2] x 7 – 88 x 6 [from (2)] 

= 252 x 7- 88 x 14 - 88 x 6 

= 252 x 7- 88 x 20 

= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]

= 252 x 7 – 592 x 20 + 252 x 40 

= 252 x 47 – 592 x 20 

= 252 x 47 + 592 x (-20)

(i) 963 and 657

By applying Euclid’s division lemma 963 = 657 × 1 + 306 …(i)

Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306

657 = 306 × 2 + 45 ….. (ii)

Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 45

306 = 45 × 6 + 36 …..(iii)

Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36

45 = 36 × 1 + 9 …… (iv)

Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9

36 = 9 × 4 + 0

∴ HCF = 9

For linear combination,

Now 9 = 45 – 36 × 1 [from (iv)]

= 45 – [306 – 45 × 6] × 1 [from (iii)]

= 45 – 306 × 1 + 45 × 6

= 45 × 7 – 306 × 1

= 657 × 7 – 306 × 14 – 306 × 1 [from (ii)]

= 657 × 7 – 306 × 15

= 657 × 7 – [963 – 657 × 1] × 15 [from (i)]

= 657 × 22 – 963 × 15

= 657 x 22 - 963 x 15 which is required linear combination.

(ii) 595 and 252

By applying Euclid’s division lemma

595 = 252 × 2 + 91 ….. (i)

Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 91

252 = 91 × 2 + 70 …. (ii)

Since remainder ≠ 0, apply division lemma on divisor 91 and remainder 70

91 = 70 × 1 + 21 ….(iii)

Since remainder ≠ 0, apply division lemma on divisor 70 and remainder 21

70 = 21 × 3 + 7 …..(iv)

Since remainder ≠ 0, apply division lemma on divisor 21 and remainder 7

21 = 7 × 3 + 0

H.C.F = 7

For linear combination,

Now, 7 = 70 – 21 × 3 [from (iv)]

= 70 – [90 – 70 × 1] × 3 [from (iii)]

= 70 – 91 × 3 + 70 × 3

= 70 × 4 – 91 × 3

= [252 – 91 × 2] × 4 – 91 × 3 [from (ii)]

= 252 × 4 – 91 × 8 – 91 × 3

= 252 × 4 – 91 × 11

= 252 × 4 – [595 – 252 × 2] × 11 [from (i)]

= 252 × 4 – 595 × 11 + 252 × 22

= 252 × 26 – 595 × 11

Which is required linear combination.

(iii) 506 and 1155

By applying Euclid’s division lemma

1155 = 506 × 2 + 143 …. (i)

Since remainder ≠ 0, apply division lemma on division 506 and remainder 143

506 = 143 × 3 + 77 ….(ii)

Since remainder ≠ 0, apply division lemma on division 143 and remainder 77

143 = 77 × 1 + 66 ….(iii)

Since remainder ≠ 0, apply division lemma on division 77 and remainder 66

77 = 66 × 1 + 11 …(iv)

Since remainder ≠ 0, apply division lemma on divisor 66 and remainder 11

66 = 11 × 6 + 0

∴ HCF = 11

For linear combination

Now, 11 = 77 – 6 × 11 [from (iv)]

= 77 – [143 – 77 × 1] × 1 [from (iii)]

= 77 – 143 × 1 – 77 × 1

= 77 × 2 – 143 × 1

= [506 – 143 × 3] × 2 – 143 × 1 [from (ii)]

= 506 × 2 – 143 × 6 – 143 × 1

= 506 × 2 – 143 × 7

= 506 × 2 – [1155 – 506 × 27 × 7] [from (i)]

= 506 × 2 – 1155 × 7 + 506 × 14

= 506 × 16 – 1155 × 7

Which is required linear combination.

(iv) 1288 and 575

By applying Euclid’s division lemma

1288 = 575 × 2 + 138 …(i)

Since remainder ≠ 0, apply division lemma on division 575 and remainder 138

575 = 138 × 4 + 23 …(ii)

Since remainder ≠ 0, apply division lemma on division 138 and remainder 23

138 = 23 x 6 + 0

∴ HCF = 23

For linear combination

Now, 23 = 575 – 138 × 4 [from (ii)]

= 575 – [1288 – 575 × 2] × 4 [from (i)]

= 575 – 1288 × 4 + 575 × 8

= 575 × 9 – 1288 × 4

Which is required linear combination.

By applying Euclid’s division lemma on 963 and 657, we get 

963 = 657 x 1 + 306………. (1) 

As the remainder ≠ 0, apply division lemma on divisor 657 and remainder 306 

657 = 306 x 2 + 45………… (2) 

As the remainder ≠ 0, apply division lemma on divisor 306 and remainder 45 

306 = 45 x 6 + 36…………. (3) 

As the remainder ≠ 0, apply division lemma on divisor 45 and remainder 36 

45 = 36 x 1 + 9…………… (4) 

As the remainder ≠ 0, apply division lemma on divisor 36 and remainder 9 

36 = 9 x 4 + 0……………. (5) 

Thus, we can conclude the H.C.F. = 9. 

Now, in order to express the found HCF as a linear combination of 963 and 657, we perform 

9 = 45 – 36 x 1 [from (5)] 

= 45 – [306 – 45 x 6] x 1 

= 45 – 306 x 1 + 45 x 6 [from (3)] 

= 45 x 7 – 306 x 1 

= [657 -306 x 2] x 7 – 306 x 1 [from (2)] 

= 657 x 7 – 306 x 14 – 306 x 1 

= 657 x 7 – 306 x 15 

= 657 x 7 – [963 – 657 x 1] x 15 [from (1)] 

= 657 x 7 – 963 x 15 + 657 x 15 

= 657 x 22 – 963 x 15.

The correct option is: (d) 4/3

The fraction 2(√2 + √6)/3(√2 + √3) is equal to 4/3.

If 'a' and 'b' are rational numbers and 2 + √3/2 - √3 = a + b + √3, then (a + b)² = 121.

HCF of 3³ x 5 and 3² x 5²

= 3² x 5

= 9 x 5

= 45

HCF (a,b) x LCM (a,b) = a x b

or, 12 x LCM (a,b) = a x b

or , LCM (a,b) = 1,800 / 12 = 150

y = 5 x 65

and x = 3 x 195 =  585