What is her real age? [closed]
$begingroup$
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
$endgroup$
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
|
show 1 more comment
$begingroup$
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
$endgroup$
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
1
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
3
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
1
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34
|
show 1 more comment
$begingroup$
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
$endgroup$
He : What is your age?
She : 35 years old, ignoring the intervening Saturdays and Sundays.
What is her real age?
logical-deduction calculation-puzzle
logical-deduction calculation-puzzle
edited Aug 29 '18 at 10:09
JonMark Perry
20.7k64199
20.7k64199
asked Aug 29 '18 at 9:26
SmartSmart
2,63351838
2,63351838
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
closed as off-topic by Quintec, JonMark Perry, El-Guest, Rubio♦ Aug 29 '18 at 15:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Quintec, JonMark Perry, El-Guest, Rubio
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
1
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
3
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
1
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34
|
show 1 more comment
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
1
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
3
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
1
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
1
1
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
3
3
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
1
1
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34
|
show 1 more comment
5 Answers
5
active
oldest
votes
$begingroup$
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
$endgroup$
add a comment |
$begingroup$
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
$endgroup$
add a comment |
$begingroup$
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
$endgroup$
add a comment |
$begingroup$
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
$endgroup$
add a comment |
$begingroup$
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
$endgroup$
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags>!
as opposed to>
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
$endgroup$
– user477343
Aug 29 '18 at 11:38
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
$endgroup$
add a comment |
$begingroup$
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
$endgroup$
add a comment |
$begingroup$
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
$endgroup$
Based on my initial intuition, I'm guessing it's
49
Since
It's as if she only lived for 5 days a week instead of 7, so she actually lived $frac75$ longer than her reported age.
answered Aug 29 '18 at 9:41
sedricksedrick
1,686514
1,686514
add a comment |
add a comment |
$begingroup$
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
$endgroup$
add a comment |
$begingroup$
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
$endgroup$
add a comment |
$begingroup$
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
$endgroup$
Her real age is
35.
If you ignore all the Saturdays and Sundays,
the number of days lived is 5/7 of what it would normally be, but the number of days in a year is also 5/7 of the norm.
answered Aug 29 '18 at 11:43
BassBass
31k473190
31k473190
add a comment |
add a comment |
$begingroup$
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
$endgroup$
add a comment |
$begingroup$
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
$endgroup$
add a comment |
$begingroup$
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
$endgroup$
(fun answer)
We do not know. Because women are known to lie about their ages all the time, even in riddles.
answered Aug 29 '18 at 13:24
CashbeeCashbee
1,69524
1,69524
add a comment |
add a comment |
$begingroup$
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
$endgroup$
add a comment |
$begingroup$
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
$endgroup$
add a comment |
$begingroup$
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
$endgroup$
Ok, I've got a weird solution for this but here goes. If we go with the assumption that in this scenario your recorded age only increases on the date of your birthday each year and only then when that date falls on a weekday. If it was a weekend (Saturday or Sunday) in a particular year then you wouldn't increase your age.
Taking that I ran some code to work out for each date from today's date to the start of the year how old you would be using these mechanics.
The results came up that, depending on your birthday you could be anywhere between 47 (26 dates), 48 (110 dates) 49 (96 dates) and 50 (9 dates) years old in reality and still have the age of 35. The only outlier to this would be if you were born on February 29th (a Leap Year). For these lucky people if you could theoretically live that long they would have an age of 197.
The code I used is below
Sub CalcAge()
Dim datBDay As Date
Dim datTempDate As Date
Dim iActAge As Integer
Dim iCnt As Integer
datBDay = Date
While datBDay > #7/18/2018#
iCnt = 1
iActAge = 35
While iCnt < iActAge
datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
If Weekday(datTempDate, vbMonday) > 5 Then
'falls on weekend therefore increases age
iActAge = iActAge + 1
End If
iCnt = iCnt + 1
Wend
'Leap Year Calculations
' While iCnt < iActAge
' datTempDate = DateAdd("yyyy", 0 - iCnt, datBDay)
' If isLeapYear(Year(datTempDate)) Then
' If Weekday(datTempDate, vbMonday) > 5 Then
' 'falls on weekend therefore increases age
' iActAge = iActAge + 1
' End If
' Else
' iActAge = iActAge + 1
' End If
'
'
' iCnt = iCnt + 1
' Wend
'
Debug.Print "Actual Age if born on " & datBDay & " is :"; iActAge
datBDay = datBDay - 1
Wend
End Sub
Public Function isLeapYear(yr As Integer) As Boolean
isLeapYear = (Month(DateSerial(yr, 2, 29)) = 2)
End Function
There might be some innacuracies as worked this up quickly but just wanted to look at this from a different angle.
edited Aug 29 '18 at 15:18
answered Aug 29 '18 at 14:57
FrazzleUKFrazzleUK
1415
1415
add a comment |
add a comment |
$begingroup$
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
$endgroup$
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags>!
as opposed to>
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
$endgroup$
– user477343
Aug 29 '18 at 11:38
add a comment |
$begingroup$
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
$endgroup$
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags>!
as opposed to>
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
$endgroup$
– user477343
Aug 29 '18 at 11:38
add a comment |
$begingroup$
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
$endgroup$
I think
$49$
Calculation:
Let $X =$ her actual age.
There are $260.7$ week days in a year ( $365div 7 = 52.14$ and $52.14 times 5 = 260.7$)
So...
$35$ is to $X$ as $260.7$ is to $365$
$35div X = 260.7div 365$
Solving for X by cross multiplying
$260.7X = 35 times 365$
$260.7X = 12,775$
$$X = 49.00$$
edited Aug 29 '18 at 11:41
user477343
3,01711061
3,01711061
answered Aug 29 '18 at 10:08
JamesJames
25628
25628
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags>!
as opposed to>
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
$endgroup$
– user477343
Aug 29 '18 at 11:38
add a comment |
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags>!
as opposed to>
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)
$endgroup$
– user477343
Aug 29 '18 at 11:38
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags
>!
as opposed to >
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)$endgroup$
– user477343
Aug 29 '18 at 11:38
$begingroup$
Always make sure to hide your answer(s) in spoiler quotes/tags
>!
as opposed to >
, in order to not spoil the answer for users attempting to solve the puzzle. I have proposed such an edit :)$endgroup$
– user477343
Aug 29 '18 at 11:38
add a comment |
$begingroup$
Today 35 years old, correct?
$endgroup$
– Doomenik
Aug 29 '18 at 9:31
$begingroup$
@Doomenik Yes...
$endgroup$
– Smart
Aug 29 '18 at 9:32
1
$begingroup$
@rsp But the year is 52 weeks (including Sat and Sun) plus a day or two. What does a year mean in the context of your question? 365 days? How about the leap years? Saturdays or Sunday on 29th February? Etc.
$endgroup$
– rhsquared
Aug 29 '18 at 9:40
3
$begingroup$
How is this a puzzle and not merely an arithmetic story-problem?
$endgroup$
– Rubio♦
Aug 29 '18 at 13:32
1
$begingroup$
@Rubio to me it seemed like a pretty decent trick question puzzle all the way up to the appearance of the tick..
$endgroup$
– Bass
Aug 29 '18 at 15:34