10
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When I was last in school, I seem to remember things working differently, but i guess this is the "New Math" I hear them talking about.

I'd better work quickly to understand this or I'll be left behind by the kids and their new-fangled electric slide rules!


$1$ is larger than exactly one of $\{2, 3, 4, 5\}$

$2$ doubled $= 7$

$9$ is prime

$2$, $7$, $12$, $17$, and $22$ are even

$4$, $9$, $14$, $19$ and $24$ are odd

$16 > 21 > 11 > 6 > 1$

$16 > 17$

$1 + 2 + 3 + 4 = 5$

$3 + 8 + 13 + 18 = 23$

$20 + 24 = 25$

$8 × 12 = 13$

$18 × 20 = 19$

The difference between $15$ and $13$ is $14$

The difference between the largest and smallest item in $\{11, 12, 13, 14, 15\}$ is $11$

There are $8$ even numbers in $\{6, 7, 8, 9, 10\}$

There are $20$ even numbers in $\{5, 10, 15, 20, 25\}$

There are a total of $12$ copies of the digit $12$ in $\{2, 7, 12, 17, 22\}$


Where was this New Math invented? RM VS CD LA VO

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  • $\begingroup$ Does the last sentence purposefully says 'digit' 12? $\endgroup$
    – wanderer
    Nov 3, 2017 at 7:15
  • $\begingroup$ @JyotishRobin that probably means 12 could be in between 0 to 9. $\endgroup$
    – Oray
    Nov 3, 2017 at 7:36
  • $\begingroup$ @JyotishRobin Yes it does. 12 is a digit just as 9 is prime. New math! $\endgroup$
    – Aranlyde
    Nov 3, 2017 at 7:42
  • $\begingroup$ the numbers could be in between 1 to 25 only? $\endgroup$
    – Oray
    Nov 3, 2017 at 8:06
  • $\begingroup$ Are Cipher or pattern too applicable tags for this problem? Also, as calculations are involved(perhaps), mathematics is also a suitable candidate. Pls. check. $\endgroup$ Nov 3, 2017 at 12:04

1 Answer 1

5
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Assuming we are looking for a permutation of the numbers 1-25:

Replacing the "new numbers" with letters A-Y the equation is $4+6+13+19=D+F+M+S=1+\{14,16\}+8+15=\{38,40\}$.

Each new number has the value:

$$\begin{array}{r|r|r} \text{new number} & \text{letter} & \text{number} \\ \hline 1 & A & 7 \\ \hline 2 & B & 10 \\ \hline 3 & C & 6 \\ \hline 4 & D & 1 \\ \hline 5 & E & 24 \\ \hline 6 & F & \{14,16\} \\ \hline 7 & G & 20 \\ \hline 8 & H & 4 \\ \hline 9 & I & 13 \\ \hline 10 & J & 22 \\ \hline 11 & K & 17 \\ \hline 12 & L & 2 \\ \hline 13 & M & 8 \\ \hline 14 & N & 11 \\ \hline 15 & O & 19 \\ \hline 16 & P & 25 \\ \hline 17 & Q & \{14,16,18\} \\ \hline 18 & R & 5 \\ \hline 19 & S & 15 \\ \hline 20 & T & 3 \\ \hline 21 & U & 21 \\ \hline 22 & V & \{14,16,18\} \\ \hline 23 & W & 23 \\ \hline 24 & X & 9 \\ \hline 25 & Y & 12 \\ \end{array}$$

Solve path:

"There are a total of L copies of the digit L in (B,G,L,Q,V)" together with "L is even" gives $L=2$.
T and H are the numbers of even numbers in sets with 5 numbers, so they are less than 5.
$H \times L=M$ and $R \times T=S$ gives $T=3$, $R=5$, $H=4$, $M=8$, $S=15$.
The 2 sum condition prune a lot of possibilities, K being the difference in the set together with the previous condition as well. Together with the inequality chain this solves most of the puzzle.
$F, Q, V$ can not be deduced uniquely.

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  • $\begingroup$ did u use computer for this? $\endgroup$
    – Oray
    Nov 3, 2017 at 10:29
  • $\begingroup$ @Oray I made a 25x25 table and crossed out stuff that was impossible. As mentioned in the solve path, a few of the conditions get you started fairly quickly. $\endgroup$
    – w l
    Nov 3, 2017 at 10:31
  • $\begingroup$ I'll mark this as correct. I changed the wording of the first clue late in development and didn't resolve to make sure it wasn't ambiguous. $\endgroup$
    – Aranlyde
    Nov 3, 2017 at 15:59
  • $\begingroup$ I made the logic a while ago using the wording "1 > 2, 1 > 3, 1 > 4, or 1 > 5", and forgot when converting it here that I used linguistic or (exactly 1) vs logical OR (1 or more), making the solution ambiguous. (It all resolves with $1$ = 5 with that stipulation in place) $\endgroup$
    – Aranlyde
    Nov 3, 2017 at 16:06

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