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Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2012 based on Prime numbers.

Let A be a 4 - digit integer. When both the first digit (leftmost) and the third digit are increased by n, and the second digit and the fourth digit are decreased by n, the new number is n times A. Find the value of A.

- 1201
- 1551
- 1818
- 2000

Algebra

Prime Number

But try the problem first...

Answer: 1818

Source

Suggested Reading

Singapore Mathematics Olympiad

Challenges and Thrills - Pre - College Mathematics

First hint

If you got stuck you can follow this hint:

We can assume the 4 digit number to be A = \(\overline {abcd}\)

If we expand it into the equation

1000(a+n) + 100(b - n) + 10(c+n) + (d-n) = nA

Try the rest of the sum ...........

Second Hint

After the previous hint :

If we compare the equation it gives :

A + 909 n = nA or

(n-1)A = 909 n

Now one thing we can understand that n and (n-1) are relatively prime and 101 is a prime number . So n= 2 or n= 4.

We have almost got the answer .So try to do the rest now ..........

Final Step

If n = 4 then A = 1212, which is impossible right?

as b<n given .so

n=2 and A = \( 909 \times 2\) = 1818

- https://www.cheenta.com/problem-on-series-smo-2009-problem-no-25/
- https://www.youtube.com/watch?v=5fWkdSs5PZk&t=63s

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