Question

(68-a)(68-b)(68-c)(68-d)(68-e) = 725 ? find a+b+c+d=? solve

Answer 1

5*5*29*1*1=725
68-63*68-63*68-39*68-67*68-67=725
so a=63,b=63,c=39,d=67,e=67
first of all factorize 725- >> 29*5*5*1*1
now try to manipulate all the factors in the form of the equation 
(68-63)*(68-63)*(68-39)*(68-67)*(68-67)
a =63
b=63
c=39
d=67
e=67
now a+b+c+d+e = 299

Answer 2

Answer:

a + b + c + d =  232

Step-by-step explanation:

Prime factorisation of 725:

= 5 * 145

= 5 * 5 * 29

= 5 * 5 * 29 * 1

= 5 * 5 * 29 * 1 * 1

Now the given equation can be written as :

( 68 - a ) ( 68 - b ) ( 68 - c ) ( 68 - d ) ( 68 - e ) = 725

( 68 - a ) ( 68 - b ) ( 68 - c ) ( 68 - d ) ( 68 - e ) = 5 * 5 * 29 * 1 * 1

So,

The given equation can also be written as :

( 68 - 63 ) ( 68 - 63 ) ( 68 - 39 ) ( 68 - 67 ) ( 68 - 67 ) = 725

As,

68 - 63 = 5

68 - 63 = 5

68 - 39 = 29

68 - 67 = 1

68 - 67 = 1

On comparing the equation we get ,

a = 63,

b = 63,

c = 39,

d = 67,

e = 67

Now,

a + b + c + d = 63 + 63 + 39 + 67 = 232

Hence,

a + b + c + d =  232