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Grade 7 Math Problem (1 Viewer)

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.
Not that give you a single digit - with no decimal places after it. (1 + 7*7)/(5*5) = 2

 
I was working under the assumption that an answer like 049 wouldn't be allowed, so a three digit number would have to be 100 or more. But I think that assumption leads to a dead end.

 
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.
Not that give you a single digit - with no decimal places after it. (1 + 7*7)/(5*5) = 2
I probably should have added that I had eliminated 3,5,7,9 as possible B's from other clues

 
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.
Not that give you a single digit - with no decimal places after it. (1 + 7*7)/(5*5) = 2
Maybe you're using other assumptions, but if B is 9, D is 4 and C is 5, the answer is a single digit (1). If B is 7, D is 3 and C is 2, the answer is 4. Etc.

You can plug numbers in to guess at it, but it doesn't mean the first three that works are right.

 
H3: If A is 2 and C is 5 then AC*AC is 100.

V3: That means that F * (C + BA) - or 3(5+2*1) - would have to begin with a zero. But it doesn't.

 
H3: If A is 2 and C is 5 then AC*AC is 100.

V3: That means that F * (C + BA) - or 3(5+2*1) - would have to begin with a zero. But it doesn't.
When two letters are next to each other, they represent a two-digit number. AC=25. When numbers are meant to be multiplied together, the * appears between them.

 
Here's the whole thing:

A=2, B=1, C=5, D=7, E=9, F=3

6 3 ◼ 2

2 1 ◼ 4

6 2 5 ◼

2 ◼ 1 5

H1A (AA-BC) * (D+A) = (22-15) * (7+2) = 7*9 = 63

H1B (E+C)/D = (9+5)/7 = 14/7 = 2

H2A F + A * E = 3 + 2 * 9 = 3 + 18 = 21

H2B C - E/E = 5-9/9 = 5-1 = 4

H3 AC * AC = 25 * 25 = 625

H4A (B + D*D) / (C*C) = (1 + 7*7) / (5 * 5) = (1+49) / 25 = 50/25 = 2

H4B F * C = 3 * 5 = 15

V1 FB*A*(EE +A) = 31 * 2 * (99+2) = 31 * 2 * 101 = 6262

V2 (BBB - D) * F = (111 - 7) * 3 = 104 * 3 = 312

V3 F * (C + BA) = 3 * (5 + 12) = 3 * 17 = 51

V4A (BD - C) * A = (17 - 5) * 2 = 12 * 2 = 24

V4B C * (A + A - F) = 5 * (2 + 2 - 3) = 5 * 1 = 5
 
With work shown:

1) H4B is F*C, which is two digits. Neither C nor F can be 1, since F*C wouldn't be two digits.

3) V4B is one digit. C* (A+A-F). C can't be 1, which means that 2A-F can't be greater than 4. A can't be 1, either. So if A is 2, F must be 1, 2 or 3. But it can't be 1, as I pointed out above. So it would have to be 2 or 3. If A is 3, F must be 2 or 3. If A is 5, then F must be 3. A can't be 7 or higher and still have V4B be one digit. So we know that A has to be 2, 3 or 5, and that F must be 2 or 3, and we can narrow it down further once we know what A is.

4) H4B is F*C, which is two digits. And since we know that F is either 2 or 3, C must be 5, 7 or 9.

5) H4A is (B+D*D)/C*C and is a single digit integer. DD can be either 1, 4, 9, 25, 49 or 81. CC can be any of those except 1. That makes it really hard for C to be a big number, which narrows us down significantly. Something like (3 + 25)/4 = 7 would fit the bill, but we know that C must be 5, 7 or 9. For B+DD/25, 49 or 81 to be a single digit integer, there's a very short list of possible answers. 1 + 49/.25 is possible. So would 9 + 16/25. I think those are the only two possible answers. In both cases, we know that C is 5. B can be either 1 or 9, and D can be either 4 or 7.

A

B

C 5

D

E

F

6) If C is 5, and F is either 2 or 3, then F*C is either 10 or 15. So we can fill in the first digit in our puzzle.

OO@O

OO@O

OOO@

O@1O

7) V3 is equal to F(C+BA). If the last digit is 1, then F must be an odd number. As I showed above, F must be 2 or 3. Which means that F is 3.

So now we know

A

B

C 5

D

E

F 3

which means we can also figure out H4B: FC =15

OO@O

OO@O

OOO@

O@15

8) That makes V4B pretty easy. C * (A + A - F) = 5. 5* (A+A-3)=5. A+A = 4. A=2.

A 2

B

C 5

D

E

F 3

9) That makes H3 simple math. AC is 25. 25*25 = 625.

OO@O

OO@O

625@

O@15

10) So V3 equals 51. F * (C + BA), or 3 * (5+_2) = 51. 5 +_2 = 17. Which means B=1

A 2

B 1

C 5

D

E

F 3

11) H2B is C-E/E. C is 5. E/E is 1. That means H2B is 4.

OO@O

OO@4

625@

O@15

12) V4A is (BD - C) * A. That's (1_-5)*2. So D is either 2 or 7. Going back to H4A, (B + D*D) / (C*C), if we know B is 1 and C is 5, then D must be 2 or 7. V2 is (BBB-D)*F Based on the crossword, we know it ends with a 2. 111-D*3 to be an even number, 111-D must be an even number. Thus D must be an odd number, which means D is 7. 104*3 = 412.

A 2

B 1

C 5

D 7

E

F 3

O3@O

O1@4

625@

O@15

13) H1A is (AA-BC)*(D+A). 22-15*7+2 = 7*9 = 63

63@O

O1@4

625@

O@15

14) V4A is (BD - C) * A, or 17-5*2 = 24

63@2

O1@4

625@

O@15

15) H1B is (E+C)/D, and must equal 2 according to the crossword. (E + 5)/7 = 2. E+5=14. E=9.

A 2

B 1

C 5

D 7

E 9

F 3

16) V1 is FB*A*(EE +A). 31*2*(99+2) = 62*101 = 6262

63@2

21@4

625@

2@15
 
Last edited by a moderator:
Tough problem, unless someone explained to the kids that two letters next to each other do not infer muliplication. Because in algebra and all math beyond, multiplication is always inferred. This problem, then, temporarily suspends a rule that the kids will need to know.

Stupid homework, if you ask me.

 

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