6
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Here's a simple maths puzzle I've made up:

$$1+0=1\\11+5=6\\5+2=3\\51+7=10\\56+13=\,?$$

Can you find the value of the question mark?

Hints:

Hint 1: Starred questions are tremendous.

Hint 2: The value of the question mark has four factors.

Hint 3: Pythagoras.

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3 Answers 3

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I think the answer is

$15$

Reasoning

ans $ = \sqrt{a+b^2}$ where first term is $a$ and second term is $b$ so $56+13^2 = 225$ and its root is equal to $15$.

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    $\begingroup$ Welcome to Puzzling.SE! I really like this answer! It fits so well, I bet it's right. $\endgroup$
    – Chowzen
    Commented Mar 27, 2018 at 7:45
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I think the answer is

$-6$

Reasoning

In each equation the replacement rule is
$a + b = gpf(a) - gpf(b)$

where $gpf$ is the Greatest Prime Factor function and we use the convention
$gpf(0) = 0 \,\,,\,\, gpf(1) = 1$

So, $56 + 13 = gpf(56) - gpf(13) = 7-13 = -6$

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  • $\begingroup$ Similarly, if $a + b = c$ in the OP's puzzle, then $\text{gpf}(a) = b + c$. $\endgroup$
    – Mr Pie
    Commented Mar 25, 2018 at 0:58
  • $\begingroup$ Good try! However it is not the one I thought of. Note that the Greatest Prime Factor function is only defined for integers greater than $1$ so the first equation $1+0=1$ wouldn't apply. $\endgroup$
    – TheSimpliFire
    Commented Mar 25, 2018 at 7:45
  • $\begingroup$ a/gpf(a) + b? So for first three that leaves just 1 on the left, then 51/17 + 7 = 3 + 7 = 10... $\endgroup$
    – Will Crawford
    Commented Mar 26, 2018 at 5:55
  • $\begingroup$ @WillCrawford You should post this as an answer. Though valid (if you include $1$ in the definition of the GPF), there is an answer that relies only on the most basic mathematical operations. $\endgroup$
    – TheSimpliFire
    Commented Mar 26, 2018 at 6:44
  • $\begingroup$ @WillCrawford Try to link your ideas with Hint 1 :) $\endgroup$
    – TheSimpliFire
    Commented Mar 26, 2018 at 6:45
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All the numbers $1, 6, 3, 10$ are the triangular numbers but just rearranged. We can make a pattern like so to generate these numbers:

$$0 + 1 = 1\tag*{$T_1$}$$ $$0 + 1 + 2 = 3\tag*{$T_2$}$$ $$0 + 1 + 2 + 3 = 6\tag*{$T_3$}$$ $$0 + 1 + 2 + 3 + 4 = 10\tag*{$T_4$}$$

By this reason, I believe that

the next number is also a triangular number.

Now the sequence $1, 6, 3, 10$ is ordered as $T_1, T_3, T_2, T_4$ which has the pattern

$T_{(n)}, T_{(n+2)}, T_{(n)+1}, T_{(n+2)+1}$ for $n = 1$.

Therefore, the next answer is

$T_{(n)+2}$ which when $n = 1$, we have that $T_{(n)+2} = T_{1+2} = T_3 = 6$ which has exactly four factors, namely $1$, $2$, $3$ and $6$.

If this pattern continued, the next triangular number would be $T_{(n+2)+2} = 21$, however my reasoning does not explain why we have to add certain numbers to form this pattern.

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    $\begingroup$ Good attempt. The actual formula is not based on triangular numbers, but somehow all the answers are though! $\endgroup$
    – TheSimpliFire
    Commented Mar 26, 2018 at 6:41
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    $\begingroup$ @TheSimpliFire S, Q, A, T and Star, Ques, Are, Tre. I still don't get it. I'm not wise enough. $\endgroup$
    – Mr Pie
    Commented Mar 27, 2018 at 9:46
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    $\begingroup$ "are= R. Combining gives SQRT. See @Abhinavsharma's answer. $\endgroup$
    – TheSimpliFire
    Commented Mar 27, 2018 at 12:00
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    $\begingroup$ @TheSimpliFire SQRT for square root. Good God, you are a genius! I guess I should have looked at Hint 3: Pythagoras. That one is the easiest because $a^2 + b^2 = c^2\Leftrightarrow c = \sqrt{a^2 + b^2}$ and now let $c =\,?$ and the rest is clear. $\endgroup$
    – Mr Pie
    Commented Mar 27, 2018 at 12:05
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    $\begingroup$ I will :) ${}{}$ $\endgroup$
    – TheSimpliFire
    Commented Mar 27, 2018 at 12:09

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