Find the missing number 4x4 grid
$begingroup$
I’ve tried everything I can think of that actually uses all the data
mathematics logical-deduction
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I’ve tried everything I can think of that actually uses all the data
mathematics logical-deduction
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
I’ve tried everything I can think of that actually uses all the data
mathematics logical-deduction
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I’ve tried everything I can think of that actually uses all the data
mathematics logical-deduction
mathematics logical-deduction
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
HoldenHolden
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
HoldenHolden
11
asked yesterday
HoldenHolden
11
11
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Holden is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
I thing it is
52
This is because
It seems like each row has a even increment.
And
The three rows increases by one each time, then the final row increase by ten each time. This uses all of the data by row?
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
$endgroup$
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
add a comment |
$begingroup$
These types of puzzles often have lots of arguably correct answers... One (that uses all numbers) is:
The top row is the first digit of the bottom row, and the second row minus the third row gives the second digit.
Or to put it more mathematically... Given the column:
A
B
C
D
Then:
$D = 10 times A + (B - C)$
Which makes the $?$:
$10 times 5 + (9 - 7) = 52$
answered yesterday
AlconjaAlconja
22.4k1294144
$endgroup$
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
I think that the answer is:
52, although how you get there is undefined
because:
if we assume the answer is $52$ (because although it probably isn't the answer, it looks like it should be!), write a solution out as a linear combination of the entries vertically above the baseline, then solutions require that $2A+6B+4C=22to A+3B+2C=11$ (from the first column) and $A+B+C=10$ (from the first differences across columns). This implies that $2B+C=1$ defines all solutions (for example $A=10, B=1, C=-1$, or $A=12, B=3, C=-5$ for another).
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
$endgroup$
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
add a comment |
$begingroup$
I think the answer is 52, since the pattern (from left to right) is adding a certain number. For example...
2,3,4,5
2 adds 1 to become 3. 3 adds 1 to become 4, etc.
So the last line...
22,32,42,?
22 adds 10 to become 32. 32 adds 10 to become 42. And 42 adds 10 to become...
Yes! 52!
So the answer is 52!
Hope this can help you :)
edited 13 hours ago
w l
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
$endgroup$
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
add a comment |
$begingroup$
I believe it's a combination of a pattern and math puzzle.
The answer is:
52
Because, for each column ABCD:
B-C = the second number of D, the answer then becomes AD
So for the first column, 6-4 = 2
So the 2nd number of D is 2
A = 2
AD (Non multiplicative) = 22
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I thing it is
52
This is because
It seems like each row has a even increment.
And
The three rows increases by one each time, then the final row increase by ten each time. This uses all of the data by row?
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
$endgroup$
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
add a comment |
$begingroup$
I thing it is
52
This is because
It seems like each row has a even increment.
And
The three rows increases by one each time, then the final row increase by ten each time. This uses all of the data by row?
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
$endgroup$
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
add a comment |
$begingroup$
I thing it is
52
This is because
It seems like each row has a even increment.
And
The three rows increases by one each time, then the final row increase by ten each time. This uses all of the data by row?
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
$endgroup$
I thing it is
52
This is because
It seems like each row has a even increment.
And
The three rows increases by one each time, then the final row increase by ten each time. This uses all of the data by row?
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
answered yesterday
QuantumTwinkieQuantumTwinkie
14.8k22392
14.8k22392
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
add a comment |
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
1
1
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Not convinced that really uses "all the data" given you could come to the same conclusion ignoring the first three rows.
$endgroup$
– Alconja
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
$begingroup$
Yeah true, your answer definitely covers the data part.
$endgroup$
– QuantumTwinkie
yesterday
add a comment |
$begingroup$
These types of puzzles often have lots of arguably correct answers... One (that uses all numbers) is:
The top row is the first digit of the bottom row, and the second row minus the third row gives the second digit.
Or to put it more mathematically... Given the column:
A
B
C
D
Then:
$D = 10 times A + (B - C)$
Which makes the $?$:
$10 times 5 + (9 - 7) = 52$
answered yesterday
AlconjaAlconja
22.4k1294144
$endgroup$
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
These types of puzzles often have lots of arguably correct answers... One (that uses all numbers) is:
The top row is the first digit of the bottom row, and the second row minus the third row gives the second digit.
Or to put it more mathematically... Given the column:
A
B
C
D
Then:
$D = 10 times A + (B - C)$
Which makes the $?$:
$10 times 5 + (9 - 7) = 52$
answered yesterday
AlconjaAlconja
22.4k1294144
$endgroup$
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
These types of puzzles often have lots of arguably correct answers... One (that uses all numbers) is:
The top row is the first digit of the bottom row, and the second row minus the third row gives the second digit.
Or to put it more mathematically... Given the column:
A
B
C
D
Then:
$D = 10 times A + (B - C)$
Which makes the $?$:
$10 times 5 + (9 - 7) = 52$
answered yesterday
AlconjaAlconja
22.4k1294144
$endgroup$
These types of puzzles often have lots of arguably correct answers... One (that uses all numbers) is:
The top row is the first digit of the bottom row, and the second row minus the third row gives the second digit.
Or to put it more mathematically... Given the column:
A
B
C
D
Then:
$D = 10 times A + (B - C)$
Which makes the $?$:
$10 times 5 + (9 - 7) = 52$
answered yesterday
AlconjaAlconja
22.4k1294144
answered yesterday
AlconjaAlconja
22.4k1294144
answered yesterday
AlconjaAlconja
22.4k1294144
answered yesterday
AlconjaAlconja
22.4k1294144
22.4k1294144
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
...Regarding your first sentence. I wonder if one can find an equally logical solution which isn't 52...
$endgroup$
– BmyGuest
20 hours ago
add a comment |
$begingroup$
I think that the answer is:
52, although how you get there is undefined
because:
if we assume the answer is $52$ (because although it probably isn't the answer, it looks like it should be!), write a solution out as a linear combination of the entries vertically above the baseline, then solutions require that $2A+6B+4C=22to A+3B+2C=11$ (from the first column) and $A+B+C=10$ (from the first differences across columns). This implies that $2B+C=1$ defines all solutions (for example $A=10, B=1, C=-1$, or $A=12, B=3, C=-5$ for another).
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
$endgroup$
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
add a comment |
$begingroup$
I think that the answer is:
52, although how you get there is undefined
because:
if we assume the answer is $52$ (because although it probably isn't the answer, it looks like it should be!), write a solution out as a linear combination of the entries vertically above the baseline, then solutions require that $2A+6B+4C=22to A+3B+2C=11$ (from the first column) and $A+B+C=10$ (from the first differences across columns). This implies that $2B+C=1$ defines all solutions (for example $A=10, B=1, C=-1$, or $A=12, B=3, C=-5$ for another).
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
$endgroup$
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
add a comment |
$begingroup$
I think that the answer is:
52, although how you get there is undefined
because:
if we assume the answer is $52$ (because although it probably isn't the answer, it looks like it should be!), write a solution out as a linear combination of the entries vertically above the baseline, then solutions require that $2A+6B+4C=22to A+3B+2C=11$ (from the first column) and $A+B+C=10$ (from the first differences across columns). This implies that $2B+C=1$ defines all solutions (for example $A=10, B=1, C=-1$, or $A=12, B=3, C=-5$ for another).
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
$endgroup$
I think that the answer is:
52, although how you get there is undefined
because:
if we assume the answer is $52$ (because although it probably isn't the answer, it looks like it should be!), write a solution out as a linear combination of the entries vertically above the baseline, then solutions require that $2A+6B+4C=22to A+3B+2C=11$ (from the first column) and $A+B+C=10$ (from the first differences across columns). This implies that $2B+C=1$ defines all solutions (for example $A=10, B=1, C=-1$, or $A=12, B=3, C=-5$ for another).
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
edited 20 hours ago
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
answered 20 hours ago
JonMark PerryJonMark Perry
19.2k63991
19.2k63991
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
add a comment |
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
But aren't all these linear combinations adding up the the same final answer - i.e. 52 ? At least your two given ones do.
$endgroup$
– BmyGuest
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
$begingroup$
@BmyGuest; thanks, fixed
$endgroup$
– JonMark Perry
20 hours ago
add a comment |
$begingroup$
I think the answer is 52, since the pattern (from left to right) is adding a certain number. For example...
2,3,4,5
2 adds 1 to become 3. 3 adds 1 to become 4, etc.
So the last line...
22,32,42,?
22 adds 10 to become 32. 32 adds 10 to become 42. And 42 adds 10 to become...
Yes! 52!
So the answer is 52!
Hope this can help you :)
edited 13 hours ago
w l
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
$endgroup$
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
add a comment |
$begingroup$
I think the answer is 52, since the pattern (from left to right) is adding a certain number. For example...
2,3,4,5
2 adds 1 to become 3. 3 adds 1 to become 4, etc.
So the last line...
22,32,42,?
22 adds 10 to become 32. 32 adds 10 to become 42. And 42 adds 10 to become...
Yes! 52!
So the answer is 52!
Hope this can help you :)
edited 13 hours ago
w l
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
$endgroup$
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
add a comment |
$begingroup$
I think the answer is 52, since the pattern (from left to right) is adding a certain number. For example...
2,3,4,5
2 adds 1 to become 3. 3 adds 1 to become 4, etc.
So the last line...
22,32,42,?
22 adds 10 to become 32. 32 adds 10 to become 42. And 42 adds 10 to become...
Yes! 52!
So the answer is 52!
Hope this can help you :)
edited 13 hours ago
w l
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
$endgroup$
I think the answer is 52, since the pattern (from left to right) is adding a certain number. For example...
2,3,4,5
2 adds 1 to become 3. 3 adds 1 to become 4, etc.
So the last line...
22,32,42,?
22 adds 10 to become 32. 32 adds 10 to become 42. And 42 adds 10 to become...
Yes! 52!
So the answer is 52!
Hope this can help you :)
edited 13 hours ago
w l
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
edited 13 hours ago
w l
3,8361229
edited 13 hours ago
w l
3,8361229
edited 13 hours ago
w l
3,8361229
3,8361229
answered 16 hours ago
K SharingK Sharing
1289
answered 16 hours ago
K SharingK Sharing
1289
answered 16 hours ago
K SharingK Sharing
1289
1289
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
add a comment |
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
$begingroup$
Thanks @ΩKr for helping me edit the answer!
$endgroup$
– K Sharing
15 hours ago
2
2
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
$begingroup$
Isn't this the exact same as my answer?
$endgroup$
– QuantumTwinkie
14 hours ago
add a comment |
$begingroup$
I believe it's a combination of a pattern and math puzzle.
The answer is:
52
Because, for each column ABCD:
B-C = the second number of D, the answer then becomes AD
So for the first column, 6-4 = 2
So the 2nd number of D is 2
A = 2
AD (Non multiplicative) = 22
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
$endgroup$
add a comment |
$begingroup$
I believe it's a combination of a pattern and math puzzle.
The answer is:
52
Because, for each column ABCD:
B-C = the second number of D, the answer then becomes AD
So for the first column, 6-4 = 2
So the 2nd number of D is 2
A = 2
AD (Non multiplicative) = 22
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
$endgroup$
add a comment |
$begingroup$
I believe it's a combination of a pattern and math puzzle.
The answer is:
52
Because, for each column ABCD:
B-C = the second number of D, the answer then becomes AD
So for the first column, 6-4 = 2
So the 2nd number of D is 2
A = 2
AD (Non multiplicative) = 22
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
$endgroup$
I believe it's a combination of a pattern and math puzzle.
The answer is:
52
Because, for each column ABCD:
B-C = the second number of D, the answer then becomes AD
So for the first column, 6-4 = 2
So the 2nd number of D is 2
A = 2
AD (Non multiplicative) = 22
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
answered 11 hours ago
visualnotsobasicvisualnotsobasic
1317
1317
add a comment |
add a comment |
Holden is a new contributor. Be nice, and check out our Code of Conduct.
Holden is a new contributor. Be nice, and check out our Code of Conduct.
Holden is a new contributor. Be nice, and check out our Code of Conduct.
Holden is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Hi and welcome to this site. Don't be disheartened about the negative votes on your post.They just reflect what people think of this particular puzzle, and Alicona's answer below gives you a nice explanation for the "why". This particular puzzle is not liked, because "These types of puzzles often have lots of arguably correct answers." and, indeed, the two answers given below demonstarte this nicely. But you posted that as "question" and not as "self-made puzzle for others", so it is not a bad posting in my eyes. however, please edit to give credits where the puzzle is from.
$endgroup$
– BmyGuest
20 hours ago