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Any rectangular shape on a calculator numpad when divided by 11 gives an integer. Why?
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I have come across this fact a very long time ago, but I still can't tell why is this happening.
Given the standard calculators numpad:
7 8 9
4 5 6
1 2 3
if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by 11 without a remainder.
Examples of the shapes: 1254, 3179, 2893, 8569, 2871, and so on. It is not allowed to use zero, only digits 1..9.
UPDATE: The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results.
The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:
A B 7 8 9
C D 4 5 6
E F 1 2 3
All valid rectangular shapes, for example, A85C, E25C, 39BF, and so on, being divided by 11 still give an integer result!
Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number 11 and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.
What law is this fun rule based on?
combinatorics number-theory permutations calculator
asked 1 hour ago
noncom
1464
 |Â
show 2 more comments
up vote
6
down vote
favorite
I have come across this fact a very long time ago, but I still can't tell why is this happening.
Given the standard calculators numpad:
7 8 9
4 5 6
1 2 3
if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by 11 without a remainder.
Examples of the shapes: 1254, 3179, 2893, 8569, 2871, and so on. It is not allowed to use zero, only digits 1..9.
UPDATE: The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results.
The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:
A B 7 8 9
C D 4 5 6
E F 1 2 3
All valid rectangular shapes, for example, A85C, E25C, 39BF, and so on, being divided by 11 still give an integer result!
Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number 11 and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.
What law is this fun rule based on?
combinatorics number-theory permutations calculator
asked 1 hour ago
noncom
1464
4
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
1
@m_t_ indeed, all these numbers give0according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!
– noncom
1 hour ago
1
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question
– noncom
21 mins ago
1
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago
 |Â
show 2 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I have come across this fact a very long time ago, but I still can't tell why is this happening.
Given the standard calculators numpad:
7 8 9
4 5 6
1 2 3
if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by 11 without a remainder.
Examples of the shapes: 1254, 3179, 2893, 8569, 2871, and so on. It is not allowed to use zero, only digits 1..9.
UPDATE: The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results.
The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:
A B 7 8 9
C D 4 5 6
E F 1 2 3
All valid rectangular shapes, for example, A85C, E25C, 39BF, and so on, being divided by 11 still give an integer result!
Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number 11 and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.
What law is this fun rule based on?
combinatorics number-theory permutations calculator
asked 1 hour ago
noncom
1464
I have come across this fact a very long time ago, but I still can't tell why is this happening.
Given the standard calculators numpad:
7 8 9
4 5 6
1 2 3
if you dial any rectangular shape, going only in right angles and each shape consisting of 4 points, then the dialed number is always divisible by 11 without a remainder.
Examples of the shapes: 1254, 3179, 2893, 8569, 2871, and so on. It is not allowed to use zero, only digits 1..9.
UPDATE: The part of the question below proved to be an error on my part because I did not double-check what the Programmers calculator was showing, and it turned out that it was rounding the results.
The same rule also works even for the Programmer calculator layout on MS Windows 10 which looks like this:
A B 7 8 9
C D 4 5 6
E F 1 2 3
All valid rectangular shapes, for example, A85C, E25C, 39BF, and so on, being divided by 11 still give an integer result!
Initially I was thinking that it's just somehow tied to picking digits from the triplets and being just another peculiarity of the decimal base and number 11 and started looking this way, but discovering that it works for the hexidecimal base and even with the hex part of the keyboard layout not obeying exactly the pattern of the decimal part layout, I'm lost.
What law is this fun rule based on?
combinatorics number-theory permutations calculator
combinatorics number-theory permutations calculator
asked 1 hour ago
noncom
1464
asked 1 hour ago
noncom
1464
edited 18 mins ago
asked 1 hour ago
noncom
1464
asked 1 hour ago
noncom
1464
asked 1 hour ago
noncom
1464
1464
4
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
1
@m_t_ indeed, all these numbers give0according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!
– noncom
1 hour ago
1
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question
– noncom
21 mins ago
1
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago
 |Â
show 2 more comments
4
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
1
@m_t_ indeed, all these numbers give0according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!
– noncom
1 hour ago
1
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question
– noncom
21 mins ago
1
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago
4
4
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
1
1
@m_t_ indeed, all these numbers give
0 according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!– noncom
1 hour ago
@m_t_ indeed, all these numbers give
0 according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!– noncom
1 hour ago
1
1
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal
11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question– noncom
21 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal
11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question– noncom
21 mins ago
1
1
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago
 |Â
show 2 more comments
3 Answers
3
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votes
up vote
4
down vote
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 leq a,b,c,d < l$. (This represents the $l$-base number $overlineabcd$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l-1) - (a - b + c - d) \ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $overlineabcd$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
add a comment |Â
up vote
2
down vote
Let's denote:
- $d_1$: the left-bottom digit of such a rectangle
- $h$: the horizontal distance (difference) to the next digit to the right
- $v$: the vertical distance (difference) to the digit above $d_1$
Then, the entered number $n$ starting at bottom-left going counterclockwise has the digits
$$n = d_1d_2d_3d_4$$
with
- $d_2 = d_1 + h$
- $d_3 = d_1 + h + v$
- $d_4 = d_1 + v$
Then, the alternating digit sum of the 4 digits is:
$$d_1 - (d_1+h) + (d_1+h+v) - (d_1+v) = 0 Rightarrow n equiv 0 mbox mod 11$$
This does not change with cyclic permutations of the digits.
answered 50 mins ago
trancelocation
5,6551515
add a comment |Â
up vote
1
down vote
Note that any four digit number made from rectangular shape can be expressed in the $2times 2$ matrix form:
$$beginpmatrix a&b\ c&dendpmatrix, \
a=4,5,7,8; b=5,6,8,9;c=1,2,4,5; d=2,3,5,6.$$
Divisibility of $overlinexyzw$ by $11$ is:
$$overlinexyzwequiv (x+z)-(y+w)equiv 0 pmod11.$$
Note that for all suitable numbers:
$$(a+d)-(b+c)=0.$$
The result follows.
answered 15 mins ago
farruhota
15.4k2734
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 leq a,b,c,d < l$. (This represents the $l$-base number $overlineabcd$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l-1) - (a - b + c - d) \ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $overlineabcd$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
add a comment |Â
up vote
4
down vote
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 leq a,b,c,d < l$. (This represents the $l$-base number $overlineabcd$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l-1) - (a - b + c - d) \ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $overlineabcd$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 leq a,b,c,d < l$. (This represents the $l$-base number $overlineabcd$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l-1) - (a - b + c - d) \ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $overlineabcd$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
Since you are concerned only with rectangular patterns, you have four digit numbers in a certain base $l$, and you want to check divisibility by $11$, where $11 = l+1$.
When you have a four digit number in base $l$, write it as $al^3 + bl^2 + cl + d$, where $0 leq a,b,c,d < l$. (This represents the $l$-base number $overlineabcd$). Now, we have a cute fact : $l^3 + 1$ is a multiple of $l+1$, since $l^3+1 = (l^2 - l + 1)(l+1)$. Furthermore, $l^2 - 1 = (l-1)(l+1)$. Therefore, we make the following rewrite :
$$
al^3 + bl^2 + cl + d = a(l^3 + 1) + b(l^2 - 1) + c(l-1) - (a - b + c - d) \ = (l+1)(...) + ((b+d)-(a+c))
$$
where $l+1 = 11$ in base $l$.
Therefore, the remainder when $overlineabcd$ is divided by $11$ is $(b+d) - (a+c)$.
When you consider four numbers ($1 to 9$, since I found problems in the letters for hex) in a rectangle and form a four digit number out of them now, can you see why this number $(b+d) - (a+c)$ is in fact zero, therefore giving your desired result?
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
edited 57 mins ago
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
answered 1 hour ago
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
34k32871
34k32871
add a comment |Â
add a comment |Â
up vote
2
down vote
Let's denote:
- $d_1$: the left-bottom digit of such a rectangle
- $h$: the horizontal distance (difference) to the next digit to the right
- $v$: the vertical distance (difference) to the digit above $d_1$
Then, the entered number $n$ starting at bottom-left going counterclockwise has the digits
$$n = d_1d_2d_3d_4$$
with
- $d_2 = d_1 + h$
- $d_3 = d_1 + h + v$
- $d_4 = d_1 + v$
Then, the alternating digit sum of the 4 digits is:
$$d_1 - (d_1+h) + (d_1+h+v) - (d_1+v) = 0 Rightarrow n equiv 0 mbox mod 11$$
This does not change with cyclic permutations of the digits.
answered 50 mins ago
trancelocation
5,6551515
add a comment |Â
up vote
2
down vote
Let's denote:
- $d_1$: the left-bottom digit of such a rectangle
- $h$: the horizontal distance (difference) to the next digit to the right
- $v$: the vertical distance (difference) to the digit above $d_1$
Then, the entered number $n$ starting at bottom-left going counterclockwise has the digits
$$n = d_1d_2d_3d_4$$
with
- $d_2 = d_1 + h$
- $d_3 = d_1 + h + v$
- $d_4 = d_1 + v$
Then, the alternating digit sum of the 4 digits is:
$$d_1 - (d_1+h) + (d_1+h+v) - (d_1+v) = 0 Rightarrow n equiv 0 mbox mod 11$$
This does not change with cyclic permutations of the digits.
answered 50 mins ago
trancelocation
5,6551515
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let's denote:
- $d_1$: the left-bottom digit of such a rectangle
- $h$: the horizontal distance (difference) to the next digit to the right
- $v$: the vertical distance (difference) to the digit above $d_1$
Then, the entered number $n$ starting at bottom-left going counterclockwise has the digits
$$n = d_1d_2d_3d_4$$
with
- $d_2 = d_1 + h$
- $d_3 = d_1 + h + v$
- $d_4 = d_1 + v$
Then, the alternating digit sum of the 4 digits is:
$$d_1 - (d_1+h) + (d_1+h+v) - (d_1+v) = 0 Rightarrow n equiv 0 mbox mod 11$$
This does not change with cyclic permutations of the digits.
answered 50 mins ago
trancelocation
5,6551515
Let's denote:
- $d_1$: the left-bottom digit of such a rectangle
- $h$: the horizontal distance (difference) to the next digit to the right
- $v$: the vertical distance (difference) to the digit above $d_1$
Then, the entered number $n$ starting at bottom-left going counterclockwise has the digits
$$n = d_1d_2d_3d_4$$
with
- $d_2 = d_1 + h$
- $d_3 = d_1 + h + v$
- $d_4 = d_1 + v$
Then, the alternating digit sum of the 4 digits is:
$$d_1 - (d_1+h) + (d_1+h+v) - (d_1+v) = 0 Rightarrow n equiv 0 mbox mod 11$$
This does not change with cyclic permutations of the digits.
answered 50 mins ago
trancelocation
5,6551515
edited 29 mins ago
answered 50 mins ago
trancelocation
5,6551515
answered 50 mins ago
trancelocation
5,6551515
answered 50 mins ago
trancelocation
5,6551515
5,6551515
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that any four digit number made from rectangular shape can be expressed in the $2times 2$ matrix form:
$$beginpmatrix a&b\ c&dendpmatrix, \
a=4,5,7,8; b=5,6,8,9;c=1,2,4,5; d=2,3,5,6.$$
Divisibility of $overlinexyzw$ by $11$ is:
$$overlinexyzwequiv (x+z)-(y+w)equiv 0 pmod11.$$
Note that for all suitable numbers:
$$(a+d)-(b+c)=0.$$
The result follows.
answered 15 mins ago
farruhota
15.4k2734
add a comment |Â
up vote
1
down vote
Note that any four digit number made from rectangular shape can be expressed in the $2times 2$ matrix form:
$$beginpmatrix a&b\ c&dendpmatrix, \
a=4,5,7,8; b=5,6,8,9;c=1,2,4,5; d=2,3,5,6.$$
Divisibility of $overlinexyzw$ by $11$ is:
$$overlinexyzwequiv (x+z)-(y+w)equiv 0 pmod11.$$
Note that for all suitable numbers:
$$(a+d)-(b+c)=0.$$
The result follows.
answered 15 mins ago
farruhota
15.4k2734
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that any four digit number made from rectangular shape can be expressed in the $2times 2$ matrix form:
$$beginpmatrix a&b\ c&dendpmatrix, \
a=4,5,7,8; b=5,6,8,9;c=1,2,4,5; d=2,3,5,6.$$
Divisibility of $overlinexyzw$ by $11$ is:
$$overlinexyzwequiv (x+z)-(y+w)equiv 0 pmod11.$$
Note that for all suitable numbers:
$$(a+d)-(b+c)=0.$$
The result follows.
answered 15 mins ago
farruhota
15.4k2734
Note that any four digit number made from rectangular shape can be expressed in the $2times 2$ matrix form:
$$beginpmatrix a&b\ c&dendpmatrix, \
a=4,5,7,8; b=5,6,8,9;c=1,2,4,5; d=2,3,5,6.$$
Divisibility of $overlinexyzw$ by $11$ is:
$$overlinexyzwequiv (x+z)-(y+w)equiv 0 pmod11.$$
Note that for all suitable numbers:
$$(a+d)-(b+c)=0.$$
The result follows.
answered 15 mins ago
farruhota
15.4k2734
answered 15 mins ago
farruhota
15.4k2734
answered 15 mins ago
farruhota
15.4k2734
answered 15 mins ago
farruhota
15.4k2734
15.4k2734
add a comment |Â
add a comment |Â
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4
A number is divisible by 11 iff the alternating sum of its digits is divisible by 11. Alternating sum means a-b+c-d+...
– m_t_
1 hour ago
1
@m_t_ indeed, all these numbers give
0according to this rule! So, probably the answer then lies in congruency of how the digits get arranged on the keypad, that following the mentioned shape cancels out the sum, collapsing it to zero!– noncom
1 hour ago
1
One second : $A85C$ is not a multiple of $11$ , and neither is $39BF$.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
59 mins ago
@ðÑÂтþýòіûûðþûþфüÑÂûûñÑÂрó hmm, indeed... neither it is a multiple of decimal or hexidecimal
11. Looks like that's simply the calculator doing the rounding! Gotta make corrections to the question– noncom
21 mins ago
1
@noncom There is a different layout, other than the MS layout, which will lead to your "rectangle property" holding for that layout, for any non-prime base. You can see my edit below.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
8 mins ago