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I was given this question today as part of a technical analysis, which I failed to answer. I am not sure what is expected?

Is it really just a case of substituting the values at the bottom in place of ...? It seems unlikely. Can someone please explain the question to me?

1. Add up the numbers in the following series

17 + 34 + 51 + … + 71,417


a) 150047117
b) 170047117
c) 130047117
d) 160047117
e) 190047117
JL.
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2 Answers2

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$$17(1 + 2 \ldots 4201) = 17 (4201)(4202)/2 = 150047117$$

Note: It is making use of Gauss's formula for the sum of the 4201 numbers.

Approach

  • You can see that there is a common factor of $17$ in the increasing number.
  • You can now factor (divide) the last number by $17$ and it gives $4201$.
  • Now, you need a formula to sum the first $4201$ numbers.
  • Thanks to Gauss, we have the sum of the first $n$ numbers as $\dfrac{n(n+1)}{2}$.
Amzoti
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  • Can you please explain a bit more about the theory and the steps you took to get to the answer? – JL. Sep 17 '13 at 12:19
  • That's awesome, thank you. – JL. Sep 17 '13 at 12:32
  • You are very welcome. Note that one of the main branches of math is the search for pattern and sometimes when working problems, see if you cand find a pattern or modify the problem to expose a pattern! Glad I could help. Regards – Amzoti Sep 17 '13 at 12:33
  • Nicely done, and nice feedback, too! +1 – amWhy Sep 17 '13 at 13:00
  • The "crowd" here at MSE is getting to be a difficult crowd to please lately! – amWhy Sep 17 '13 at 13:57
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The problem wants you to add up all the multiples of $17$ from $17\cdot1$ to $17\cdot4201=71,417$. If you're a Gauss, you can see in an instant how to do this. If you're not a Gauss, you might still realize that the sum must also be a multiple of $17$. You might also notice that at most one of the offered answers can be a multiple of $17$, since they differ from one another by small multiples of $10,000,000$, which is not divisible by $17$. So it suffices to see which one $17$ actually divides.

My pocket calculator won't allow $9$-digit numbers, which stymied me at first, until I noticed the $17$ at the end of each number. So instead of looking at $150047117/17$, for example, it suffices to look at $1500471/17$. Etc. As soon as you get an integer quotient, you're done.

Barry Cipra
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  • There's a little bit of subtlety in the fact that if $150047100$ is a multiple of $17$ then $1500471$ is (certainly this doesn't work so well if you choose, say $25$ instead of $17$!) – Ben Millwood Oct 11 '13 at 23:47