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#### The number of ordered pairs (m, n), m, n {1, 2, … 100} such that 7^{m} + 7^{n} is divisible by 5 is -

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- 1250
- 2000
- 2500
- 5000

#### Answer:The correct answer is: 2500Note that 7^{r} (r N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).

Thus, 7^{m} + 7^{n} cannot end in 5 for any values of m, n N. In other words, for 7^{m} + 7^{n} to be divisible by 5, it should end in 0.

For 7^{m} + 7^{n} to end in 0, the forms of m and n should be as follows :

m n

1 4r 4s + 2

2 4r + 1 4s + 3

3 4r + 2 4s

4 4r + 3 4s + 1

Thus, for a given value of m there are just 25 values of n for which 7^{m} + 7^{n} ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]

There are 100 × 25 = 2500 ordered pairs (m, n) for which 7^{m} + 7^{n} is divisible by 5.

Hence

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### Related Questions to study

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#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

#### Consider the following statements:

1. The number of ways of arranging m different things taken all at a time in which p m particular things are never together is m! – (m – p + 1)! p!.

2. A pack of 52 cards can be divided equally among four players in order in ways.

Which of these is/are correct?

maths-General

(1) Total number of ways of arranging m things = m!.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

To find the number of ways in which p particular things are together, we consider p particular things as a group.

Number of ways in which p particular things are together = (m – p + 1)! p!

So, number of ways in which p particular things are not together

= m! – (m – p + 1)! p!

Total number of ways =

Hence, both of statements are correct.

maths-

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

#### The total number of function ‘ƒ’ from the set {1, 2, 3} into the set {1, 2, 3, 4, 5} such that ƒ(i) ƒ(j), i < j, is equal to-

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Let ‘l’ is associated with ‘r’ ,

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

r {1, 2, 3, 4, 5} then ‘2’ can be associated with r, r + 1, ….., 5.

Let ‘2’ is associated with ‘j’ then 3 can be associated with j, j + 1, …., 5. Thus required number of functions

= =

=

= = 35

Hence (a) is correct answer.

maths-

#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).#### The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x| k, |y| k, |x – y| k ; is-

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|x| k –k x k ….(1)

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

= (3k

& |y| k –k y k ….(2)

& |x – y| k |y – x| k ….(3)

– k y – x k x – k y x + k

Number of points having integral coordinates

= (2k + 1)

^{2}– 2[k + (k – 1) + …. + 2 + 1]= (3k

^{2}+ 3k + 1).maths-

#### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.#### The numbers of integers between 1 and 10^{6} have the sum of their digit equal to K(where 0 < K < 18) is -

maths-General

The required no. of ways = no. of solution of the equation (x

Where 0 x

= Coefficient of x

= Coefficient of x

= Coefficient of x

(1 + 6 C

=

=

_{1}+ x_{2}+ x_{3}+ x_{4}+ x_{5}+ x_{6}= K)Where 0 x

_{i} 9, i = 1, 2, …6, where 0 < K < 18= Coefficient of x

^{K}in (1 + x + x^{2}+….. + x^{9})^{6}= Coefficient of x

^{K}in= Coefficient of x

^{k}in (1 – 6x^{10}+ 15 x^{20}– ….)(1 + 6 C

_{1}x + 7 C_{2}x^{2}+ …. +(7 – K – 10 – 1) C_{K–10}x^{K–10}+ ….+(7 + K – 1) C_{K }x^{K}+ …)=

^{k + 6}C_{K}– 6.^{K–4}C_{K–10}=

^{k + 6}C_{6}– 6.^{K–4}C_{6 }.maths-

#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.#### The straight lines I_{1}, I_{2}, I_{3} are parallel and lie in the same plane. A total number of m points are taken on I_{1} ; n points on I_{2 }, k points on I_{3}. The maximum number of triangles formed with vertices at these points are -

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Total number of points = m +n + k. Therefore the total number of triangles formed by these points is

^{m + n + k}C_{3}. But out of these m + n + k points, m points lie on I_{1}, n points lie on I_{2}and k points lie on I_{3}and by joining three points on the same line we do not obtain a triangle. Hence the total number of triangles is^{m + n +}^{k}C_{3}–^{m}C_{3}–^{n}C_{3}–^{k}C_{3}.maths-

#### If the line is a normal to the hyperbola then

#### If the line is a normal to the hyperbola then

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#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

#### If the tangents drawn from a point on the hyperbola to the ellipse make angles α and β with the transverse axis of the hyperbola, then

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#### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

#### The eccentricity of the hyperbola whose latus rectum subtends a right angle at centre is

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#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

#### If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r^{2}t^{4}s^{2}, then the number of ordered pair (p, q) is –

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#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

#### A rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having edge length one unit then no. of rectangles which have odd unit length

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^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

^{n}C_{r} + ^{2n}C_{r+1} + ^{n}C^{r+2} is equal to (2 r n)

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The coefficient of in is

The coefficient of in is

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#### How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?

#### How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which not two S are adjacent ?

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#### The value of ^{50}C_{4} + is -

#### The value of ^{50}C_{4} + is -

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#### The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-

#### The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is-

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