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I am looking at an old IBM PAT (circa 1965) and am stumped by a number sequence question. This is it:

Identify the pattern in the following sequences and select the option which indicates the next element in the sequence.

6 10 7 14 14 18

Possible answers:

  • a) 14
  • b) 15
  • c) 16
  • d) 18
  • e) 21

If anyone can identify the pattern and the correct answer, I would be very interested to hear.

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3 Answers 3

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I believe the correct answer is b) because:

6 + 4 = 10
10 - 3 = 7
7 * 2 = 14
14 / 1 = 14
14 + 4 = 18
18 - 3 = 15

So the sequence is:

First add 4 then subtract 3 after that multiply with 2 and divide by 1

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    $\begingroup$ I think I understand but some people might not. Could you please add what that sequence is? $\endgroup$
    – Yout Ried
    Commented Mar 11, 2019 at 13:45
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    $\begingroup$ At first this answer seemed like a stretch to me, but the more I look at it, the more is makes sense. Even though there are four different steps before they begin repeating, the steps use the four basic operations in the canonical order, and the other numbers involved also have a simple pattern to them. Very clever. $\endgroup$ Commented Mar 11, 2019 at 14:49
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I saw this as a merge of two series, and agree with @Coder97 about the answer being b) with different reasoning. I saw the odd indexed elements of the series following this pattern:

next = current + 1 (if even), next = current * 2 (if odd)

and the even indexed elements of the series following this pattern:

next = current + 4

It might be interesting to try and prove @Coder97 and my solutions are mathematically the same.

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  • $\begingroup$ Your odd-index rule matches the other solution indefinitely, but your even-index rule differs as soon as you extend the sequence to the next number. $\endgroup$ Commented Mar 11, 2019 at 23:45
  • $\begingroup$ @JaapScherphuis yeah! Found that after right after I posted. Now I'm curious who designed this... $\endgroup$ Commented Mar 12, 2019 at 14:36
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Sorry if i’m not really good at explaining,i got 18,i made groups of two Numbers (6,10),(,7,14),(14,18),(b,a) We observe that 6x2*2=14,and 10x2-2=18 And 6+10=16;14+18=32 so we do the same with 7,14;7+14=21 then a+b=42;7x2+2=b;14x2-2=a and b=18

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    $\begingroup$ This puzzle already has an accepted answer? Do you think your answer adds value? If not, you can delete it. First check the accepted answer before posting to see if your answer would be a good addition to the site. $\endgroup$
    – JLee
    Commented Jun 9, 2022 at 21:59
  • $\begingroup$ I think there could be multiple solutions to number sequence puzzles,i think my answer adds some value bcz it’s a different method of solving this number sequence. $\endgroup$ Commented Jun 10, 2022 at 6:47
  • $\begingroup$ Ok, just letting you know. Welcome to Puzzling! $\endgroup$
    – JLee
    Commented Jun 10, 2022 at 9:24

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