Question

Find the least possible value of a+b; where 'a' & 'b' are positive integers such that 11 divides a+13b and 13 divides a+11b.

Answer 1

Answer:

Step-by-step explanation:

a+13b/11=a+11b/13

13(a+13b) =11(a+11b)

13a+169b=11a+121b

13a-11a=121b-169b

2a=-48b

a=-48b/2

a=-24b

Now put the value of a in a+13b/11

=>-24b+13b=11

=>-11b=11

=>-b=11/11

b=-1

Now put the value of b in a=-24b

a=(-24) (-1)

a=24

Answer 2

Answer:

Given

a' & 'b' are positive integers such that 11 divides a+13b and 13 divides a+11b.

We have 13/a+11b

⇒13/a−2b and hence 13/6a−12b this implies 13/6a+b

Similarly,

11/a+13b

⇒11/a+2b

⇒11/6a+12b

⇒11/6a+b

Since gcd(11,13)=1

We conclude 143/6a+b

Thus we may write 6a+b=143k for some integer k

Hence, 6a+6b=143k+5b=144k+6b−(k+b)

This shows that 6/k+b and hence k+b≥6

We therefore obtain 6(a+b)=143k+5b=138k+5(k+b)≥138+(5×6)=168

It follows that a+b≥28

Taking a=23 and b=5

we see that the conditions of the problem satisfied. Thus the minimum value of a+b is 28