-9
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Given

$D$ is the total number of days in a normal year

$a$, $b$ are positive integers

$P$, $Q$, $R$, $S$, $T$ are consecutive integers

Figure them out by combining the following six expressions into 2 equations involving D. Valid solution is the one that gives numerical values for all a b P Q R S T.

X is the missing operator

$ D ^ { a x b }$ x $P^a$ X $Q^a$ X $R^a$ X $S^a$ X $T^a$

You are allowed to use operators

$+$ $-$ $\times$ $/$ $($ $)$ $=$

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closed as unclear what you're asking by PiIsNot3, Omega Krypton, Alconja, Rubio May 8 at 2:57

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Just to clarify do you mean for this puzzle to be asking the solver to sub in those signs in any way that creates valid equation? $\endgroup$ – gabbo1092 May 7 at 19:53
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    $\begingroup$ Excuse me, but what is a valid solution? Can you please post an example which is a valid (but incorrect) solution? $\endgroup$ – Weather Vane May 7 at 20:14
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    $\begingroup$ @Uvc It's bad form to revise the specifications of the original puzzle after an answer has already been given. I'd suggest you roll back to before you added the "2 equations" part. And I also agree with Weather Vane that it's not clear what we're supposed to do. Do you mean to find two equations such that each equation separately contains all those expressions or when combined contain them all? I.e. would $ D^{ab} = P^a + Q^a $ and $ D^{ab} = R^a + S^a + T^a $ be considered valid? $\endgroup$ – PiIsNot3 May 7 at 20:25
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    $\begingroup$ @pilsnot3..I am very sorry to cause this confusion...is there a way to delete it as I should have specified that a be greater than 1 $\endgroup$ – Uvc May 7 at 20:29
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    $\begingroup$ Have you been thinking about this comment? You are becoming a prolific puzzle-maker, and some are very good, but is there some scope for you to review them as a puzzle solver, before posting? $\endgroup$ – Weather Vane May 7 at 20:34
1
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It seems that this can be solved fairly trivially.

We are given that $D = 365$.

If we take $a=1$ and $b=2$, then $D^{ab} = D^2 = 365^2 = 133225$.

Now, since $a=1$, we just need five consecutive integers that sum to $133225$.

We can easily find those by taking $133225 \div 5 = 26645$, and then taking the two preceding and two following integers.

Then we have
$P^a = 26643^1 = 26643$
$Q^a = 26644^1 = 26644$
$R^a = 26645^1 = 26645$
$S^a = 26646^1 = 26646$
$T^a = 26647^1 = 26647$

So the final solution is

$$365^{1\times2} = 26643^1 + 26644^1 + 26645^1 + 26646^1 + 26647^1$$

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  • $\begingroup$ The question asks for two equations. $\endgroup$ – Weather Vane May 7 at 20:17
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    $\begingroup$ @WeatherVane It didn't when I was typing this up. It was just changed. $\endgroup$ – GentlePurpleRain May 7 at 20:17
  • $\begingroup$ You are right, that was slipped in after my last edit. $\endgroup$ – Weather Vane May 7 at 20:19
  • $\begingroup$ Edited to provide clarification and true to my original intent of the puzzle $\endgroup$ – Uvc May 8 at 14:12

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