Maths-
General
Easy

Question

The maximum value of P such that 3P divides 99 × 97 × 95 × .... × 51 is-

  1. 11
  2. 14
  3. 13
  4. 12

The correct answer is: 14


    99 × 97 × 95 ×....× 51
    equals times fraction numerator 100 factorial over denominator left parenthesis 100 cross times 98 cross times 96 cross times horizontal ellipsis cross times 52 right parenthesis end fraction cross times fraction numerator 1 over denominator 50 factorial end fraction times fraction numerator 100 factorial over denominator 2 to the power of 25 cross times 50 factorial cross times 50 factorial end fraction
    maximum power of 3 in 100 !
    equals times open square brackets 100 over 3 close square brackets plus open square brackets 100 over 9 close square brackets plus open square brackets 100 over 27 close square brackets plus open square brackets 100 over 81 close square brackets
    = 33 + 11 + 3 + 1 = 48.
    maximum power of 3 in 50! = open ceil 50 over 3 close ceil plus open ceil 50 over 9 close ceil plus open ceil 50 over 27 close ceil
    = 16 + 5 + 1 = 22
    maximum power of 3 in 25! = open square brackets 25 over 3 close square brackets plus open ceil 25 over 9 close square brackets
    = 8 + 2 = 10
    thereforeexponent of 3 = 48 + 10 –
    (22 × 2) = 14

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