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

SOLUTION

GIVEN

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

TO DETERMINE

The sum of digits of k

EVALUATION

LINEAR DIOPHANTINE EQUATION

Here the given equation is

$\sf{5x + 4y = 221} \: \: \: ........(1)$

The gcd of 5 & 4 is 1 and 1 divides 221

Hence 5x + 4y = 221 has integral solutions

Now

$\sf{1 = 5 - 4 = 5 + ( - 4)}$

$\therefore \: \: \sf{221 = 5.221 + 4.( - 221)}$

Thus one integral solution of the given equation is

$\sf{x_0 = 221 \: \: \: \: and \: \: \: \: y_0 = - 221}$

Hence all integral solutions of the given equation are given by

$\sf{x = 221 + 4n \: , \:y = - 221 - 5n \: \: for \: all \: integer \: n }$

If all the solutions are positive

$\sf{221 + 4n > 0 \: \: and \: \: - 221 - 5n > 0}$

Now

$\sf{221 + 4n > 0 \: \: \: gives}$

$\sf{4n \: > - 221}$

$\displaystyle \sf{ \implies \: n > - \frac{221}{4} }$

$\displaystyle \sf{ \implies \: n > - 55.25 }$

Again

$\displaystyle \sf{ \: - 221 - 5n \: > 0 \: \: \: gives}$

$\displaystyle \sf{ - 5n > 221}$

$\displaystyle \sf{ \implies \:n < - 44.2 }$

From above

$\sf{ - 55.25 \: < \: n \: < \: - 44.2}$

Now n must be an integer

$\sf{ \therefore \: n = - 55, - 54, - 53, - 52, - 51, - 50, - 49 , - 48, - 47, - 46, - 45\: }$

So there are 11 positive integral solutions of the form (x, y) of the equation 5x+ 4y= 221

∴ k = 11

Hence The sum of digits of k = 1 + 1 = 2

FINAL ANSWER

The sum of digits of k = 2

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Answer 2

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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