Question

Find the value of a+b+c+d if the product of first 10 natural numbers is written as 2a+3b+5c+7d

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

First ten natural numbers are 1,2,3,4,5,6,7,8,9,10

We can write the product as 1*2*3*(2*2)*5*(2*3)*7*(2*2*2)*(3*3)*(2*5)

That is

2 power 7*3 power 4* 5power 2* 7

That is a=7,b=4,c=2,d=1

a+b+c+d= 7+4+2+1=14