Question

what will be the minimum value of X for which the number 3(x+63)+640 is completely divisible by 17​

Answer 1

Answer:

7

Step-by-step explanation:

3(x+63)+640

Multiplying 3 with x and 63 then the value will be

(3x+189)+640

if we will take 0 it will be like

3×0+189+640=829

But 829 is not completely divisible by 17. therefore 0 isn't the answer..

I have tried with 1,2,3,4,5,6, but the result is same they all are not completely divisible by 17.

But by 7.

it will be like,

(3×7+189)+640=850.

on dividing 850 by 17 i.e;

850/17= 50.

It is completely divided.

so the answer u r finding is 7....

Answer 2

Answer:

[3(x+63)+640]÷17=1

(3x+189+640)÷17=1

829÷17=-3x

829÷(17×3=-x)

-829÷51=x

x=-16•25

it is not completely divisible