Problem on Prime Numbers | SMO, 2012 | Problem 20

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.

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 ...........

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 ..........

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

as b<n given .so

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

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