Maths-
General
Easy

Question

If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2t4s2, then the number of ordered pair (p, q) is –

  1. 224    
  2. 225    
  3. 252    
  4. 256    

Hint:

Natural numbers known as prime numbers can only be divided by one (1) and by the number itself. In other terms, prime numbers are positive integers greater than one that only have the number itself and the number's first digit as factors. Here we have given  r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q is r2t4s2, then what is the number of ordered pair (p, q).

The correct answer is: 225


    An ordered pair is made up of the ordinate and the abscissa of the x coordinate, with two values given in parenthesis in a certain sequence.
    Now we have given the LCM as: r2t4s2
    Consider following cases:
    Case 1: if p contains r2 then q will have rk, for the value k=0,1.
    So 2 ways.
    Case 2: if q contains r2 then p will have rk, for the value k=0,1.
    So 2 ways.
    Case 3:Both p and q contain r2
    So 1way.
    So after this we can say that:
    exponent of r=2+2+1 = 5 ways.
    Similarly
    exponent of t=4+4+1=9 ways.
    exponent of s=2+2+1 = 5 ways.
    So total ways will be:
    5 x 5 x 9 = 225 ways.

    Finding the smallest common multiple between any two or more numbers is done using the least common multiple (LCM) approach. A number that is a multiple of two or more other numbers is said to be a common multiple. Here we understood the concept of LCM and the pairs, so the total pairs can be 225.

    Book A Free Demo

    +91

    Grade*

    Related Questions to study

    General
    Maths-

    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

    Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
    There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
    Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
    Then the number of rectangles will be:
    (1+3+5+......+(2m1))(1+3+5+......+(2n1))=m2n2
    So it is m2n2.

    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

    Maths-General
    Now we have given that a rectangle has sides of (2m – 1) & (2n – 1) units as shown in the figure composed of squares having an edge length of one unit.
    There are 2n horizontal lines and 1 2 m vertical lines (numbered 1,2.......2n).
    Two horizontal lines, one with an even number and one with an odd number, as well as two vertical lines must be chosen in order to create the necessary rectangle.
    Then the number of rectangles will be:
    (1+3+5+......+(2m1))(1+3+5+......+(2n1))=m2n2
    So it is m2n2.
    General
    maths-

    nCr + 2nCr+1 + nCr+2 is equal to (2  less or equal thanless or equal than n)

    nCr + 2nCr+1 + nCr+2 is equal to (2  less or equal thanless or equal than n)

    maths-General
    General
    maths-

    The coefficient of x to the power of n in fraction numerator x plus 1 over denominator left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis end fraction is

    The coefficient of x to the power of n in fraction numerator x plus 1 over denominator left parenthesis x minus 1 right parenthesis squared left parenthesis x minus 2 right parenthesis end fraction is

    maths-General
    General
    Maths-

    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 ?

    Maths-General
    General
    Maths-

    The value of 50C4 + not stretchy sum subscript r equals 1 end subscript superscript 6 end superscript 56 minus r C subscript 3 end subscriptis -

    The value of 50C4 + not stretchy sum subscript r equals 1 end subscript superscript 6 end superscript 56 minus r C subscript 3 end subscriptis -

    Maths-General
    General
    Maths-

    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-

    Maths-General
    General
    physics-

    A mass of 100 blank g strikes the wall with speed 5 blank m divided by s at an angle as shown in figure and it rebounds with the same speed. If the contact time is 2 cross times 10 to the power of negative 3 end exponent s e c, what is the force applied on the mass by the wall

    Force = Rate of change of momentum
    Initial momentum stack P with rightwards arrow on top subscript 1 end subscript equals m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top
    Final momentum stack P with rightwards arrow on top subscript 2 end subscript equals negative m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top
    therefore stack F with rightwards arrow on top equals fraction numerator increment stack P with rightwards arrow on top over denominator increment t end fraction equals fraction numerator negative 2 m v sin invisible function application theta over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction
    Substituting m equals 0.1 blank k g comma blank v equals 5 blank m divided by s comma blank theta equals 60 degree
    Force on the ball stack F with rightwards arrow on top equals negative 250 square root of 3 N
    Negative sign indicates direction of the force

    A mass of 100 blank g strikes the wall with speed 5 blank m divided by s at an angle as shown in figure and it rebounds with the same speed. If the contact time is 2 cross times 10 to the power of negative 3 end exponent s e c, what is the force applied on the mass by the wall

    physics-General
    Force = Rate of change of momentum
    Initial momentum stack P with rightwards arrow on top subscript 1 end subscript equals m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top
    Final momentum stack P with rightwards arrow on top subscript 2 end subscript equals negative m v sin invisible function application theta stack i with hat on top plus m v cos invisible function application theta blank stack j with hat on top
    therefore stack F with rightwards arrow on top equals fraction numerator increment stack P with rightwards arrow on top over denominator increment t end fraction equals fraction numerator negative 2 m v sin invisible function application theta over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction
    Substituting m equals 0.1 blank k g comma blank v equals 5 blank m divided by s comma blank theta equals 60 degree
    Force on the ball stack F with rightwards arrow on top equals negative 250 square root of 3 N
    Negative sign indicates direction of the force
    General
    maths-

    If the locus of the mid points of the chords of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, drawn parallel to y equals m subscript 1 end subscript x is y equals m subscript 2 end subscript x then m subscript 1 end subscript m subscript 2 end subscript equals

    If the locus of the mid points of the chords of the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1, drawn parallel to y equals m subscript 1 end subscript x is y equals m subscript 2 end subscript x then m subscript 1 end subscript m subscript 2 end subscript equals

    maths-General
    General
    Maths-

    If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr+1 + nCr –1 + 2 × nCr equals-

    If nCr denotes the number of combinations of n things taken r at a time, then the expression nCr+1 + nCr –1 + 2 × nCr equals-

    Maths-General
    General
    maths-

    The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -

    The number of ways is which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question is -

    maths-General
    General
    physics-

    An intense stream of water of cross-sectional area A strikes a wall at an angle theta with the normal to the wall and returns back elastically. If the density of water is rho and its velocity is v,then the force exerted in the wall will be

    Linear momentum of water striking per second to the wall P subscript 1 end subscript equals m v equals A v rho blank v equals A v to the power of 2 end exponent blank rho, similarly linear momentum of reflected water per second P subscript r end subscript equals A v to the power of 2 end exponent rho

    Now making components of momentum along x- axes and y-axes. Change in momentum of water per second
    equals P subscript i end subscript cos invisible function application theta plus P subscript r end subscript cos invisible function application theta
    equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta
    By definition of force, force exerted on the Wall equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta

    An intense stream of water of cross-sectional area A strikes a wall at an angle theta with the normal to the wall and returns back elastically. If the density of water is rho and its velocity is v,then the force exerted in the wall will be

    physics-General
    Linear momentum of water striking per second to the wall P subscript 1 end subscript equals m v equals A v rho blank v equals A v to the power of 2 end exponent blank rho, similarly linear momentum of reflected water per second P subscript r end subscript equals A v to the power of 2 end exponent rho

    Now making components of momentum along x- axes and y-axes. Change in momentum of water per second
    equals P subscript i end subscript cos invisible function application theta plus P subscript r end subscript cos invisible function application theta
    equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta
    By definition of force, force exerted on the Wall equals 2 A v to the power of 2 end exponent blank rho cos invisible function application theta
    General
    physics-

    The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with above spring of half length, the line O A will

    When the length of spring is halved, its spring constant will becomes double
    open square brackets B e c a u s e blank k proportional to fraction numerator 1 over denominator x end fraction proportional to fraction numerator 1 over denominator L end fraction therefore k proportional to fraction numerator 1 over denominator L end fraction close square brackets
    Slope of force displacement graph gives the spring constant left parenthesis k right parenthesis of spring
    If k becomes double then slope of the graph increases i. e. graph shifts towards force- axis

    The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with above spring of half length, the line O A will

    physics-General
    When the length of spring is halved, its spring constant will becomes double
    open square brackets B e c a u s e blank k proportional to fraction numerator 1 over denominator x end fraction proportional to fraction numerator 1 over denominator L end fraction therefore k proportional to fraction numerator 1 over denominator L end fraction close square brackets
    Slope of force displacement graph gives the spring constant left parenthesis k right parenthesis of spring
    If k becomes double then slope of the graph increases i. e. graph shifts towards force- axis
    General
    physics-

    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are vand 2 v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A

    Let initially particle x is moving in anticlockwise direction and y in clockwise direction
    As the ratio of velocities of xand y particles are fraction numerator v subscript x end subscript over denominator v subscript y end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction, therefore ratio of their distance covered will be in the ratio of 2 blank colon 1. It means they collide at point B

    After first collision at B, velocities of particles get interchanged, i. e., x will move with 2 v and particle y with v
    Second collision will take place at point C. Again at this point velocities get interchanged and third collision take place at point A
    So, after two collision these two particles will again reach the point A

    Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are vand 2 v, respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A

    physics-General
    Let initially particle x is moving in anticlockwise direction and y in clockwise direction
    As the ratio of velocities of xand y particles are fraction numerator v subscript x end subscript over denominator v subscript y end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction, therefore ratio of their distance covered will be in the ratio of 2 blank colon 1. It means they collide at point B

    After first collision at B, velocities of particles get interchanged, i. e., x will move with 2 v and particle y with v
    Second collision will take place at point C. Again at this point velocities get interchanged and third collision take place at point A
    So, after two collision these two particles will again reach the point A
    General
    maths-

    In a model, it is shown that an arch of abridge is semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal, the best approximation of the height of the arch, 2 m from the centre of the base is

    In a model, it is shown that an arch of abridge is semi-elliptical with major axis horizontal. If the length of the base is 9 m and the highest part of the bridge is 3 m from the horizontal, the best approximation of the height of the arch, 2 m from the centre of the base is

    maths-General
    General
    maths-

    The number of non-negative integral solutions of x + y + z  n, where n  N is -

    The number of non-negative integral solutions of x + y + z  n, where n  N is -

    maths-General