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Here's a cross-number puzzle I came across recently.

 

\(\displaystyle \hspace{50pt}\text{Clues} \\ \underline{\text{Across}} \hspace{70pt} \underline{\text{Down}} \\ 1\quad a^{2}\hspace{75pt}1\quad e \\ 5 \quad b^{3} \hspace{75pt}2 \quad f^{2} \\ 6 \quad c^{4} \hspace {75pt} 3 \quad g \\ 7 \quad d^{4} \hspace{75pt} 4 \quad h^{4} \)

 

The letters a, b, c, d, e, f, g and h denote distinct positive integers. 

 

\(\begin{array}{|l|l|l|l|} \hline 1 \hspace{20pt} & 2 \hspace{20pt} & 3 \hspace{20pt} & 4 \hspace{20pt} \\ & & & \\ \hline 5 & & & \\ & & & \\ \hline 6 & & & \\ & & & \\ \hline 7 & & & \\ & & & \\ \hline \end{array}\)

 

The answer to each clue is a 4-digit number, the first digit of which is not zero.

Show that your answer to the complete puzzle is not unique, but that the sum e + g is unique, and calculate its value.

 Apr 5, 2023
 #1
avatar+118627 
+1

Hi Tiggsy,

 

Thanks that was fun.

 

I won't give away the whole answer but the     e+g = 11132

 Apr 5, 2023
 #2
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+1

Thanks Melody.

 

Your answer is correct, but then you would know that.

Sadly, no one else seems to be interested.

 

Best wishes,

Tiggsy.

Guest Apr 6, 2023
 #3
avatar+118627 
0

Hi Tiggsy,

 

Maybe other people have had a go. 

 

 

Anyway I enjoyed it   :)

It was enjoyable puzzle, and it was good that I could post an answer without giving anything away.

 

------------------------

 

If anyone wants hints you should ask for them.    laugh

We always like seeing interest in our posts even if they seem too hard !

Melody  Apr 6, 2023

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