Determine the largest 3 - digit prime factor of 2000 C 1000.

n C r denotes the number of combinations of n things taking r at a time.

The solution to this puzzle is as follows.<o:p> </o:p>

Let us define [x] as the greatest integer contained in x. For example, [4.5] is 4.

Suppose, r be any 3 digit prime.

Accordingly, r^2 >2000.Thus the highest power of r that divides 2000! is [2000/r] and the highest power of r that divides 1000! is [1000/r].

Now, 2000 C 1000 =(2000!)/(1000!)^2.

Consequently, the highest power of r that divides

2000 C 1000 is [2000/r] – 2*[1000/r].

If r>666, then [2000/r] – 2*[1000/r] =2-2*1=0, so that, r does not divide 2000 C 1000.

Therefore, the desired prime is the largest one such that it is less than 666.

When r=661, we observe that [2000/r] – 2*[1000/r]

= 3 – 2*1 =1.

Consequently, it follows that 661 is the largest 3-digit prime that divides 2000 C 1000.

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