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Grade 7 Math Problem (1 Viewer)

chet

Footballguy
My daughter was given the following math problem.

Complete the following crossword. The letters A,B,C,D,E,F below always represents the same number either 1,2,3,5,7,9

The crossword is a 4*4 grid and I will attempt to type out the grid. O is a blank (to be filled in) and @ is a black square.

OO@O

OO@O

OOO@

O@OO

The clues are as numbered as follows. H1A represents the first clue on the top line. H1B is the second clue. H3 is the only clue for the third horizontal line.

H1A (AA-BC) * (D+A)

H1B (E+C)/D

H2A F + A * E

H2B C - E/E

H3 AC * AC

H4A (B + D*D) / (C*C)

H4B F * C

V1 FB*A*(EE +A)

V2 (BBB - D) * F

V3 F * (C + BA)

V4A (BD - C) * A

V4B C * (A + A - F)

From my initial trials, I don't believe there are any constraints on the answers. i.e. The answers are not limited to 1,2,3,5,7,9. Also, an answer that is two spaces could start with a 0. For example, H2A is a 2-square answer that could be below 10.

Any ideas how best to attack this thing?

 
A has to be 1, 2, or 3...because of the H# clue.

Which means if A=1, then f=2 due to clue v4b - or if A=2, then f=1 or 3 ...or if a=3, then f=1, 2 or 5.

 
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Just so I'm clear, you're not supposed to fill in the crossword with A-F, just any digit? Wtf kind of homework is this.

 
Just so I'm clear, you're not supposed to fill in the crossword with A-F, just any digit? Wtf kind of homework is this.
I think each of those letters A-F always equals the same digit (1, 2, 3, 5, 7 and 9)....so that each of the clues complete the crossword and the numbers in the crossword will be whatever numbers make it fit.

 
Yeah, just compiling clues.

I think that's just got to be the way to tackle it, then. Right down all the logical conclusions and then plug away.

 
Why do you say ABCDEF must be 1,2,3,5,7,9 then the answers are not limited to those numbers?

Any way you can attach a pic of the actual problem?

 
The horizontal and vertical clues align so you can form equations, right?

So the intersection of H1A and V1 is (AA-BC)*(D+A) = FB*A*(EE+A), etc.?

 
Bad assumption below.



Pretty sure C = 5

F*C - 10 = C(A + A - F)

CF - 10 = 2CA - CF

2CF - 10 = 2CA

CF - 5 = CA

C = 5

F = 2 or 3

A = 1 or 2
 
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Edit for spoiler tags.

Ugh. So close. A-F = 215793 got almost all the squares to work.
 
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A can not be 1 because H1A would be negative

Oops tried to quote Matty

 
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Wait, Sinn, am I right? I must have messed up some math somewhere confirming (or misread the clue numbers).

 
It just takes playing around with the numbers a bit, and using some basic logic - i.e if an equation has to equal a single digit, that limits the numbers.

This equation: H4A (B + D*D) / (C*C) gave me those three numbers fairly quickly.

 
It just takes playing around with the numbers a bit, and using some basic logic - i.e if an equation has to equal a single digit, that limits the numbers.

This equation: H4A (B + D*D) / (C*C) gave me those three numbers fairly quickly.
There are lots of single digit answers to that one. 9+16/25, 7+9/4, 1+49/25, 3+4/1, etc.

 
It just takes playing around with the numbers a bit, and using some basic logic - i.e if an equation has to equal a single digit, that limits the numbers.

This equation: H4A (B + D*D) / (C*C) gave me those three numbers fairly quickly.
C was 5, aha!

 
It just takes playing around with the numbers a bit, and using some basic logic - i.e if an equation has to equal a single digit, that limits the numbers.

This equation: H4A (B + D*D) / (C*C) gave me those three numbers fairly quickly.
Damn, I had B, C, D, all correct but when doing H1B I put E as 2, which messed everything up and I didn't even consider E as 9, or I would have had the rest :( So close, there goes a good 30 minutes wasted.

 

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