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$22+22=4444$

$43+46=618191$

$77+77=?$

What should come in place of $?$

I cannot see any logic in $43+46=618191$. Is there any?

epimorphic
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Mathematics
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2 Answers2

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If you had $43+46=618198$ then I can see the following pattern: $$\color{red}{4}\color{blue}{3}+\color{red}{4}\color{blue}{6}=\color{red}{\text{reverse(4*4)}}\color{blue}{\text{reverse(3*6)}}\color{green}{\text{reverse(43+46)}}=\color{red}{61}\color{blue}{81}\color{green}{98}$$

Mufasa
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    But this is not a true answer; this should be a comment. – Arbuja Jan 31 '16 at 17:20
  • The same logic can be used to show $22+22=4444$ and hence also be used to calculate $77+77$. The OP didn't clearly state that my assumption of $43+46=618198$ was incorrect. – Mufasa Jan 31 '16 at 17:21
  • It seems convincing. – Mathematics Jan 31 '16 at 17:25
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    'It's convincing' - but gives the incorrect answer for $43+46$? Is there a mistake in your question, if so then it would be convincing. – fosho Jan 31 '16 at 17:29
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Please check your puzzle again if $43+46$ is actually $618198$

If not then expanding on @Daniel's comment

$$22+22=(\text{reverse digits}(2*2))(\text{reverse digits}(2*2))[22+22+(2*2-2*2)]=4444$$

$$43+46=(\text{reverse digits}(4*4))(\text{reverse digits}(3*6))[(43+46+\left(6*3-4*4\right))]=618191$$

So the number of $77+77=\text{reverse}(7*7)\text{Reverse}(7*7)\text{reverse}(77+77+\left(7*7-7*7\right))=9494154$

Arbuja
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