Question

If the number of positive integral solutions of the
form (x, y) of the equation 5x + 4y = 221 is equal to k,
then the sum of digits of k will be equal to​

Answer 1

Given : number of positive integral solutions of the form (x, y) of the equation 5x+  4y= 221 is equal to k

To Find : sum of digits of k will be equal to​

Solution:

5x  + 4y  = 221

=> 5x = 221  - 4y

=> 5x = 220 + 1  - 5y  + y

=> x = (44   - y)  + (y + 1)/5

(y + 1)/5  has to be integer  

Hence  y  + 1  = 5p    as y > 0 hence p > 0

=> y = 5p - 1    

5x  + 4(5p - 1)   = 221

=> 5x + 20p - 4 = 221

=> 5x + 20p = 225

=>  x  + 4p  = 45

=> x + 4p = 1 + 44  

=> x + 4p = 1 + 4(11)  

Hence initial value  of  x = 1   and p = 11   as x has  positive integral solutions

x  + 4p  =  41 +  4

=> x + 4p = 41  + 4(1)

=> x = 41  is last solution  and  p = 1

Values of  p are from 1 to 11

Hence total 11 possible solutions.

=> k = 11

Sum of digits

=  1 + 1 = 2

sum of digits of k will be equal to​ 2

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