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This one is really not as difficult as it seems. here is how to approach it :

From the given equation for abcd, we can clearly see that the digits of the number, abcd are the powers to which 3,5,7,11 have been raised to.

so from *m*=(3^r)(5^s)(7^t)(11^u), we know that the four digit number "m" is "rstu" and its value is 1000r+100s+10t+u

*n*=(25)(*m*) = (25)(3^r)(5^s)(7^t)(11^u)
= (3^r)(5^(s+2))(7^t)(11^u)

Thus the four digit number "n" is "r(s+2)tu" and its value is 1000r+100(s+2)+10t+u

Finally n-m = 100(s+2) - 100s = 200



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