Question

find a+b+c+d if product of 10 natural numbers is 2^a×3^b×5^c×7^d

Answer 1

hey dear,

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let us assume a,b,c, d are natural numbers and more than zero.

d = 1 , c = 1 gives b = 2 , a = (10! - 18 ) / 2,    

      a+b+c+d = 18, 14, 391

a = 3, b = 1 c = 1  gives  d = (10! - 14 ) / 7

      a+b+c+d = 5, 18, 403

 

  the value of a+b+c+d lies in between the above two values.

if a,b,c, d can be 0 , then

  values lie  between    10! / 2 and 10! / 7

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Hope this helps you

Answer 2

sameeeeeeeeeeeeeeeee