Question

There exist integers a and b
for
which
42a +7b=1

​

Answer 1

Answer:

11-Oct-2017 · 1 answer

Proof: For the sake of contradiction, there exists integers a and b for which 21a+30b=1, then, 21a+30b=1 3(7a+10b)=1 3n=1 for some integer

Answer 2

Given: 42a +7b=1

To find: integers a and b

Solution:

We need the values of a and b which are integers and they satisfy the equation 42a+7b=1.

We can take a as 0 and b as 1/7 which are integers

when we put these values in the equation then it is satisfied.

42a+7b=1

LHS                                        RHS

42 into 0+ 7 into 1/7             1

0+1= 1

LHS=RHS

The value for integers a and b is 0 and 1/7.