Question #275903

Find the last three digits of the number 3×7×11×· · ·×2003. [Hint: Chinese

remainder theorem.]


1
Expert's answer
2021-12-07T13:59:00-0500

Let  n=0500(4n+3)\prod_{n=0}^{500} (4n + 3), then the answer is x mod 1000.


Using Chinese remainder theorem we can calculate modulo 125 and 8.

Since 125 is a divisor of x, there is the congruence for modulo 125:

x0mod125.x\equiv0 mod 125.


And for modulo 8 we have two cases : 

1) 4n13mod84n-1\equiv3 mod 8 if n even (occurs 251 times).

2) 4n31mod84n-3\equiv-1 mod 8 if n odd (occurs 250 times).


To calculate xmod8x mod 8 we can use 32=91mod8.3^2=9\equiv1 mod 8.

That is why x3251(1)25032513mod8.x\equiv3^{251}(-1)^{250}\equiv3^{251}\equiv3 mod 8.


The last thing we need to do is to check the multiples of 125 until they are sufficient to match the above congruence.

Multiples of 125: 125, 250, 375, 500, 625, 750, 875, 1000, 1125, 1250 and so on.

The first value that matches our congruence (x3mod8)(x\equiv3mod8) is 875, because the remainder of the division 875 by 8 is equal to 3.


Answer: 875.


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