find least positive value of a+b where a,b are positive integer such that |a+13 and 13a+11b
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Answer:
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
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