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Excellent Bike Riders. All under 100. Find the missing members marked x. They are all members of special club.

97 X 59

29 X 23

X 7 X

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  • $\begingroup$ No calculations needed..bike riders is a clue $\endgroup$ – Uvc May 7 at 10:10
  • $\begingroup$ I’m guessing it’s something to do with rot13(ahzoref gung plpyr) $\endgroup$ – MichaelMaggs May 7 at 11:10
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I'm going to guess:

97 61 59
29 47 23
19 7 17

These are all the:

Cyclic numbers under 100
7, 17, 19, 23, 29, 47, 59, 61, 97 shown here

Although the order doesn't quite seem to make 100% sense to me

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  • $\begingroup$ Nor to me. But you must be right in taking all the numbers less than 100 from this sequence: oeis.org/A001913 $\endgroup$ – MichaelMaggs May 7 at 14:36
  • $\begingroup$ wow, every sequence is in there isn't it! I'll add it as a ref to backup the numbers if that's ok. Still not sure if there's a way of ordering these to make more sense though. $\endgroup$ – Smock May 7 at 15:10
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Solution

$\begin{array}{rrr} 97 & 13 & 59 \\ 29 & 13 & 23 \\ 11 & 7 & 3 \end{array}$

Explanation

Let $A = ||a_{ij}||$ be the 3-by-3 matrix in question. All $a_{ij} (i=1,2,3; j=1,2,3)$ must satisfy the following: $$\begin{eqnarray}a_{ij} \mathrm{\ is\ prime}, \\ a_{i2} = \frac{a_{i1} + a_{i3}}{8 - 2i}, i = 1,2,3 \\ a_{i1} > a_{i3} \end{eqnarray}$$ (In simpler words, the number in middle column is the average is a prime (a member of a special club) equals to the average of the numbers to the left and right of it, divided by 6, 4, and 2, respectively. The leftmost number is always greater than the rightmost one.)
So, $$\begin{eqnarray} (97 + 59) / 6 = 13, \\ (29 + 23) / 4 = 13, \\ (11 + 3) / 2 = 7. \end{eqnarray}$$ (3 and 11 are the only distinct primes whose sum is equal to 14)

However

that seems too far-fetched and completely unrelated to bikes.

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  • $\begingroup$ No calculations needed...bike riders is a clue $\endgroup$ – Uvc May 7 at 10:11

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