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How to find the product of $n$ terms of an arithmetic progression where common difference is not unity.

I just want to know the last $3$ digits of $7 \times 23 \times 39 \times \ldots \times 2071$ where common difference is $16$.

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Computers were invented to solve problems like these. –  Gerry Myerson Jul 1 '12 at 11:11
    
It was asked in a prestigious management studies exam. –  Bazinga Jul 1 '12 at 11:12
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By the way, are you aware that users are only allowed to ask 50 questions in a month? I believe you have asked over 30 in a week! –  Gerry Myerson Jul 1 '12 at 11:15
    
Sorry, didn't know that. –  Bazinga Jul 1 '12 at 11:16
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May I suggest that in the future you provide all relevant information with your questions including, in particular, the source of the question. –  Gerry Myerson Jul 2 '12 at 0:17

2 Answers 2

up vote 3 down vote accepted

There are two (unrelated) questions here. An answer to the first question is $$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$ To answer the second question, consider $n=\prod\limits_{k=0}^{129}(7+16k)$. Note that $7+16k=0\pmod{5}$ for every $k=3\pmod{5}$. Using this for $k=3$, $k=8$ and $k=13$ yields $n=0\pmod{125}$, that is, $n=125m$ for some integer $m$.

Likewise, $7+16k=-1\pmod{8}$ and the number of terms in the product which defines $n$ is $130$, which is even, hence $n=1\pmod{8}$. And $125=5\pmod{8}$ hence $n=5m\pmod{8}$, which implies $m=5n\pmod{8}$, that is, $m=5\pmod{8}$.

Finally, $n=125\cdot5\pmod{125\cdot8}$, that is, $n=\underline{\ \ \ \ \ \ \ \ }\pmod{1000}$.

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You are asked for the final three digits. I reckon this is quite easy once you can answer two questions - so this is an extended hint.

  1. What happens when you multiply your product by 8?

  2. Can you establish the value of the product mod 16? (or mod 8)

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