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

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by sriramramu

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by sriramramu

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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

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