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# Problem on HCF | SMO, 2013 | Problem 35

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2013 based on HCF. You may use sequential hints to solve the problem.

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

## Problem on HCF | SMO, 2013 | Problem 35

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

### Key Concepts

HCF and GCD

Number Theory

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

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………………

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………..

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)