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

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  1. 1250    
  2. 2000    
  3. 2500    
  4. 5000    

    Answer:The correct answer is: 2500Note that 7r (r  N) ends in 7, 9, 3 or 1 (corresponding to r = 1, 2, 3 and 4 respectively).
    Thus, 7m + 7n cannot end in 5 for any values of m, n  N. In other words, for 7m + 7n to be divisible by 5, it should end in 0.
    For 7m + 7n 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 7m + 7n ends in 0. [For instance, if m = 4r, then = 2, 6, 10,….. , 98]
     There are 100 × 25 = 2500 ordered pairs (m, n) for which 7m + 7n is divisible by 5.
    Hence

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    Related Questions to study

    General
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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 less or equal than 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 fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
    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 = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
    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 less or equal than 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 fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fractionways.
    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 = fraction numerator 52 factorial over denominator left parenthesis 13 factorial right parenthesis to the power of 4 end exponent end fraction
    Hence, both of statements are correct.
    General
    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) less or equal than ƒ(j), straight for all i < j, is equal to-

    Let ‘l’ is associated with ‘r’ ,
    element of {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
    = not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
    = fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
    = fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 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) less or equal than ƒ(j), straight for all i < j, is equal to-

    maths-General
    Let ‘l’ is associated with ‘r’ ,
    element of {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
    = not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript open parentheses not stretchy sum subscript j equals r end subscript superscript 5 end superscript left parenthesis 6 minus j right parenthesis close parentheses= not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript fraction numerator left parenthesis 6 minus r right parenthesis left parenthesis 7 minus r right parenthesis over denominator 2 end fraction
    = fraction numerator 1 over denominator 2 end fraction open parentheses not stretchy sum subscript r equals 1 end subscript superscript 5 end superscript left parenthesis 42 minus 13 r plus r to the power of 2 end exponent right parenthesis close parentheses
    = fraction numerator 1 over denominator 2 end fraction open parentheses 42.5 minus 13. fraction numerator 6.5 over denominator 2 end fraction plus fraction numerator 5.6.11 over denominator 6 end fraction close parentheses= 35
    Hence (a) is correct answer.
    General
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    The number of points in the Cartesian plane with integral co-ordinates satisfying the inequalities |x|  less or equal than k, |y| less or equal than k, |x – y| less or equal than k ; is-

    |x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
    & |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

    not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
    thereforeNumber of points having integral coordinates
    = (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
    = (3k2 + 3k + 1).

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

    maths-General
    |x| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(1)
    & |y| less or equal thannot stretchy rightwards double arrow –k less or equal thanless or equal than k ….(2)
    & |x – y| less or equal thannot stretchy rightwards double arrow |y – x| less or equal than k ….(3)

    not stretchy rightwards double arrow – k less or equal than y – x less or equal than k not stretchy rightwards double arrow x – k less or equal than y less or equal than x + k
    thereforeNumber of points having integral coordinates
    = (2k + 1)2 – 2[k + (k – 1) + …. + 2 + 1]
    = (3k2 + 3k + 1).
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    maths-

    The numbers of integers between 1 and 106 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 (x1 + x2 + x3 + x4 + x5 + x6 = K)
    Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
    = Coefficient of xK in (1 + x + x2 +….. + x9)6
    = Coefficient of xK in open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
    = Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
    (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
    = k + 6CK – 6. K–4CK–10
    = k + 6C6 – 6. K–4C6 .

    The numbers of integers between 1 and 106 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 (x1 + x2 + x3 + x4 + x5 + x6 = K)
    Where 0 xi 9, i = 1, 2, …6, where 0 < K < 18
    = Coefficient of xK in (1 + x + x2 +….. + x9)6
    = Coefficient of xK in open parentheses fraction numerator 1 minus x to the power of 10 end exponent over denominator 1 minus x end fraction close parentheses to the power of 6 end exponent
    = Coefficient of xk in (1 – 6x10 + 15 x20 – ….)
    (1 + 6 C1x + 7 C2 x2 + …. +(7 – K – 10 – 1) CK–10 xK–10 + ….+(7 + K – 1) CK xK + …)
    = k + 6CK – 6. K–4CK–10
    = k + 6C6 – 6. K–4C6 .
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    The straight lines I1, I2, I3 are parallel and lie in the same plane. A total number of m points are taken on I1 ; n points on I2 , k points on I3. The maximum number of triangles formed with vertices at these points are -

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    m + n + kC3mC3nC3kC3.
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