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The units digit of 3^n follows the following patterns: 3,9,7,1,3,9,7,1,.....

Thus if we know what both x and y are, we can solve it (statement 1).

So how did you find the values of x & y from Stmt 1??

Well we don't know the values of X and Y individually, but all we need to know is how many times a number with 3 in the units digit is multiplied by itself. Since X and Y are both exponents of such numbers, knowing x+y is sufficient.

Of course I might be wrong so the OA would be appreciated.

In this case, since the base number is different i.e 243 & 463, it makes sense to use the various combinations of x & y: 1. 1&6 or 6&1 2. 2&5 or 5&2 3. 3&4 or 4&3

The units digit of 3^n follows the following patterns: 3,9,7,1,3,9,7,1,.....

Substituting n in the above combinations and multiplying the ending unit digits of each of these numbers, we get the same unit digit i.e., 7.

Choice (B), keeps the n open for y, so the unit digit of the resultant number may vary.

If (243)^x(463)^y = n, where x and y are positive integers, what is the units digit of n?

(1) x + y = 7 (2) x = 4

Units digit of \(243^x\) equals to units digit of \(3^x\) and units digit of \(463^y\) equals to units digit of \(3^y\) (general rule). Hence units digit of \(243^x*463^y\) equals to units digit of \(3^x*3^y=3^{x+y}\). So knowing the value of \(x+y\) is sufficient to determine units digit of \(n\).

(1) \(x+y=7\). Sufficient. (As cyclicity of \(3\) is \(4\), units digit of \(3^7\) would be the same as of units digit of \(3^3\) which is \(7\))