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Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2013 based on HCF.

What is the smallest positive integer n,where \( n \neq 11\) such that the highest common factor of n-11 and 3n +20 is greater than 1?

- 62
- 65
- 66
- 60

HCF and GCD

Number Theory

But try the problem first...

Answer: 64

Source

Suggested Reading

Singapore Mathematics Olympiad

Challenges and Thrills - Pre - College Mathematics

First hint

If you got stuck in this sum we can start from here:

Let d is the highest common factor of n-11 and 3n +20 which is greater than 1.

So d|(n-11) and d|(3n + 20) .

If we compile this two then d|(3n +20 -3(n-11) when d|53 .

Now one thing is clear that 53 is a prime number and also d >1

so we can consider d = 53.

Now try the rest..................

Second Hint

Now from the previous hint

n-11 = 53 k (let kis the +ve integer)

n = 53 k +11

So for any k 3n +20 is a multiple of 53.

so 3n + 20 = 3(53k +11) +20 = 53(3k+1)

Finish the rest...........

Final Step

Here is the final solution :

After the last hint :

n = 64 (if k = 1) which is the smallest integer. as hcf of (n - 11,3n +20)>1(answer)

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