Not that Obvious: Missing Numbers In A Table

[Ilikebugs]

Ive tried to make each variable relate to a but it hasn't worked.View attachment 6511

 

[MarkFL]

Re: Not that Obvious

Suppose we let the sum of any row be $S_R$ and the sum of any column be $S_C$, where $S_R\ne S_C$. Using the table, we get the equations:

\(\displaystyle a+157=S_R\)

\(\displaystyle b+c+113=S_R\)

\(\displaystyle d+e+120=S_R\)

\(\displaystyle b+110=S_C\)

\(\displaystyle d+108=S_C\)

\(\displaystyle a+118=S_C\)

\(\displaystyle c+e+54=S_C\)

You now have 7 equations in 7 unknowns...can you proceed?

 

[Evgeny.Makarov]

Re: Not that Obvious

You have five equations and five unknowns. Solving this system of equation using any standard technique should work.

Ive tried to make each variable relate to a but it hasn't worked.

Please describe in more detail what you tried.

Edit: By the five equations I mean equating the sum of the first and second row as well as the sum of the second and third row, and similarly for columns.

 

[Ilikebugs]

Re: Not that Obvious

a+157=b+c+113=d+e+120
b+110=d+108=a+118=c+e+54

a=c+b-44=d+e-37
b=a-c+44
c=a-b+44
d=a-e+37
e=a-d+37

a=b+8=d+10=c+e-64

I tried to change each variable to a+ something.

 

[MarkFL]

Re: Not that Obvious

This is the kind of strategy I would use:

Subtract the second equation from the first to get:

\(\displaystyle a-b=c-44\)

Subtract the 4th from the 6th to get:

\(\displaystyle a-b=-8\)

And so we conclude:

\(\displaystyle c=32\)

Next, observe that the 1st and 3rd equations involve $a,\,d,\,e$ and the 5th and 6th equations involve $a,\,d$...so can you use a similar technique to determine $e$?

 

[Ilikebugs]

Re: Not that Obvious

e=47?

 

[MarkFL]

Re: Not that Obvious

e=47?

It would help if you show your work...:D

 

[Ilikebugs]

Re: Not that Obvious

a+157=SRa+157=SR

b+c+113=SRb+c+113=SR

d+e+120=SRd+e+120=SR

b+110=SCb+110=SC

d+108=SCd+108=SC

a+118=SCa+118=SC

c+e+54=SC

1-3=
a-d-e+37=0
a-d=e-37

6-5=
a-d=10

10=e-37

e=47

 

[MarkFL]

Re: Not that Obvious

When I subtract the 5th equation from the 6th, I get:

\(\displaystyle a-d=-10\)

Which means:

\(\displaystyle e=27\)

What would you do next?

 

[Ilikebugs]

Re: Not that Obvious

32+27+54=SC

=113

a=-5
b=3
d=5

 

[MarkFL]

Re: Not that Obvious

32+27+54=SC

=113

a=-5
b=3
d=5

Yes, that's exactly what I would have done. Good work! (Star)

edit: We've made an error somewhere...the numbers don't add up...however, I have to run now...check the work and see if you can find the error. :D

 

[MarkFL]

Re: Not that Obvious

According to an online calculator, we should find:

\(\displaystyle (a,b,c,d,e)=(-1,7,36,9,27)\)

Let's see if we can get that...(1) - (2):

\(\displaystyle a-b=c-44\)

(6) - (4):

\(\displaystyle a-b=-8\)

Yes, we find:

\(\displaystyle c=36\)

The rest should now fall into place. :D