n = 4 digit even number
n = \(2^{a}\) * b ; where a & b are integers; Also, a>0

Now,
323 = 17 * 19

Observe: -
17*19 is part of "b" in the representation of "n" above.
323 does not have a factor of "2" but n surely does.

Smallest possible value of next divisor = 323 * 2 = 646

Hence, E

343= 17*19

So next will be either any multiple of 17 or 19 but greater than 343

19*18 =340
or consider 17*20= 340

Ans C

If we look at the answer choices, we can see that multiplying any of them by 323 will result in a product that is larger than four digits. That means our answer must share a factor with 323. 323 factorizes into 17 and 19. 323+17=340 would be the smallest option.

Answer choice C.
_________________

Call the 4 digit number, N.

We know 323*some factor K1 = N

We want the minimum number X above 323 that multiplied by its factor K2 also equals N.

So N=323K1 = X*K2

We want to minimize X and maximize K2.

Let K2 then equal K1 minus an increment L. We want to minimize L.

So 323K1 = X(K1-L), so

X= 323K1/(K1-L). 323 = 17*19 so

17*19K1/(K1-L) = X

If we want to assume L=1 for the sake of minimization, K1 can't be divisible by (K1-1) unless K1=2, which it can't be because K1 has to be at least 4 in order to create a 4 digit number.

So that leaves 17 or 19 to be divisible by K1-1. So

K1= 18 or 20. Trying both

19*17*18/17 = 19*18 = 342

19*17*20/19 = 17*20 = 340

So the minimum X, the next factor higher than 323 is 340

Posted from my mobile device