Question

4. Find least positive value of a+b where a,b are positive integers such that |a+13 and 13a+11b.​

Answer 1

Step-by-step explanation:

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