Not that give you a single digit - with no decimal places after it. (1 + 7*7)/(5*5) = 2
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Grade 7 Math Problem (1 Viewer)
- Thread starter chet
- Start date
bostonfred
Footballguy
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.
I probably should have added that I had eliminated 3,5,7,9 as possible B's from other clues
bostonfred
Footballguy
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.
bostonfred
Footballguy
Your (and sarnoff's) answer of 215793 used that assumption, and it's wrong. Clues H3 and V3 don't work with those numbers.
Shoddy work, imo.
bostonfred
Footballguy
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.
V3: That means that F * (C + BA) - or 3(5+2*1) - would have to begin with a zero. But it doesn't.
Actually, doesn't H1A =2, H1B =4, but V4A =2?
=22-15*(7+2)
=7*9
=63
H1B = (e+c)/d
=9+5/7
=2
V4A = (BD -C)*A
=(17-5)*2
=12*2
=24
The 2 from H1B is the same 2 as in V4A
bostonfred
Footballguy
Ah,.that makes more sense. If AC is 25 instead of 2*5, that changes all of my answers around.
Apologies to Sinn. Still shoddy not to post the work imo.
Sarnoff
Footballguy
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.
hagmania
Footballguy
Actually, doesn't H1A =2, H1B =4, but V4A =2?
=22-15*(7+2)
=7*9
=63
H1B = (e+c)/d
=9+5/7
=2
V4A = (BD -C)*A
=(17-5)*2
=12*2
=24
The 2 from H1B is the same 2 as in V4A
Yeah I used characters for all the other calcs then multiplied the letters here. DERP!
Sarnoff
Footballguy
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
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
bostonfred
Footballguy
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
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:
Controller
Footballguy
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.
Stupid homework, if you ask me.
