Mixing
Digit Sum of three consecutive number is always 6
$begingroup$
I'm aware that in the series of 3 consecutive numbers, one of those is a multiple of 2,3 and 4. But I’m wondering how I’ll explain this observation:
$$begin{align}
1+2+3 &= 6\
4 + 5 + 6 &= 15
end{align}$$
following this pattern continuously I always end up with the digit sum of $6$. I started with $n + (n+1) + (n+2) = 3n + 3$ and the actual number is lets say $10p + q$ hence digit sum is represented by $p + q$... but I cannot connect $3n + 3$ to $p+q$. How sure I am that this will run indefinitely?
Any idea?
number-theory
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
$endgroup$
add a comment |
$begingroup$
I'm aware that in the series of 3 consecutive numbers, one of those is a multiple of 2,3 and 4. But I’m wondering how I’ll explain this observation:
$$begin{align}
1+2+3 &= 6\
4 + 5 + 6 &= 15
end{align}$$
following this pattern continuously I always end up with the digit sum of $6$. I started with $n + (n+1) + (n+2) = 3n + 3$ and the actual number is lets say $10p + q$ hence digit sum is represented by $p + q$... but I cannot connect $3n + 3$ to $p+q$. How sure I am that this will run indefinitely?
Any idea?
number-theory
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
$endgroup$
4
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47
add a comment |
$begingroup$
I'm aware that in the series of 3 consecutive numbers, one of those is a multiple of 2,3 and 4. But I’m wondering how I’ll explain this observation:
$$begin{align}
1+2+3 &= 6\
4 + 5 + 6 &= 15
end{align}$$
following this pattern continuously I always end up with the digit sum of $6$. I started with $n + (n+1) + (n+2) = 3n + 3$ and the actual number is lets say $10p + q$ hence digit sum is represented by $p + q$... but I cannot connect $3n + 3$ to $p+q$. How sure I am that this will run indefinitely?
Any idea?
number-theory
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
$endgroup$
I'm aware that in the series of 3 consecutive numbers, one of those is a multiple of 2,3 and 4. But I’m wondering how I’ll explain this observation:
$$begin{align}
1+2+3 &= 6\
4 + 5 + 6 &= 15
end{align}$$
following this pattern continuously I always end up with the digit sum of $6$. I started with $n + (n+1) + (n+2) = 3n + 3$ and the actual number is lets say $10p + q$ hence digit sum is represented by $p + q$... but I cannot connect $3n + 3$ to $p+q$. How sure I am that this will run indefinitely?
Any idea?
number-theory
number-theory
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
edited Jan 28 at 3:58
Arturo Magidin
266k34590920
266k34590920
asked Jan 28 at 3:44
rosarosa
604616
asked Jan 28 at 3:44
rosarosa
604616
asked Jan 28 at 3:44
rosarosa
604616
604616
4
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47
add a comment |
4
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47
4
4
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The “digit sum” is actually the same thing as the remainder when dividing by $9$, except that unless you have the number $0$, the digit sum will be $9$ when it is evenly divisible by $9$ (not too hard exercise).
If the first number you add is a multiple of $3$, then you are adding $3k$, $3k+1$, and $3k+2$. The sum is
$$3k + (3k+1) + (3k+2) = 9k + 3$$
which will have a “digit sum” of $3$.
If the first number you add is a multiple of $3$ plus $1$, then you are adding $3k+1$, $3k+2$, and $3k+3$; these are the two examples you have. Here you have
$$(3k+1) + (3k+2) + (3k+3) = 9k+ 6$$
giving your observed total of $6$.
Finally, if the first number you add is a multiple of $3$ plus $2$, then you are ading $3k+2$, $3k+3$, and $3k+4$, which gives you
$$(3k+2) + (3k+3) + (3k+4) = 9k + 9$$
Which will give you a “digit sum” of $9$ (or $0$, if you add $-1$, $0$, and $1$).
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
add a comment |
$begingroup$
Counterexamples:
$2+3+4=9$
$5+6+7=18$ where $1+8=9$
$6+7+8=21$ where $2+1=3$
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
The “digit sum” is actually the same thing as the remainder when dividing by $9$, except that unless you have the number $0$, the digit sum will be $9$ when it is evenly divisible by $9$ (not too hard exercise).
If the first number you add is a multiple of $3$, then you are adding $3k$, $3k+1$, and $3k+2$. The sum is
$$3k + (3k+1) + (3k+2) = 9k + 3$$
which will have a “digit sum” of $3$.
If the first number you add is a multiple of $3$ plus $1$, then you are adding $3k+1$, $3k+2$, and $3k+3$; these are the two examples you have. Here you have
$$(3k+1) + (3k+2) + (3k+3) = 9k+ 6$$
giving your observed total of $6$.
Finally, if the first number you add is a multiple of $3$ plus $2$, then you are ading $3k+2$, $3k+3$, and $3k+4$, which gives you
$$(3k+2) + (3k+3) + (3k+4) = 9k + 9$$
Which will give you a “digit sum” of $9$ (or $0$, if you add $-1$, $0$, and $1$).
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
add a comment |
$begingroup$
The “digit sum” is actually the same thing as the remainder when dividing by $9$, except that unless you have the number $0$, the digit sum will be $9$ when it is evenly divisible by $9$ (not too hard exercise).
If the first number you add is a multiple of $3$, then you are adding $3k$, $3k+1$, and $3k+2$. The sum is
$$3k + (3k+1) + (3k+2) = 9k + 3$$
which will have a “digit sum” of $3$.
If the first number you add is a multiple of $3$ plus $1$, then you are adding $3k+1$, $3k+2$, and $3k+3$; these are the two examples you have. Here you have
$$(3k+1) + (3k+2) + (3k+3) = 9k+ 6$$
giving your observed total of $6$.
Finally, if the first number you add is a multiple of $3$ plus $2$, then you are ading $3k+2$, $3k+3$, and $3k+4$, which gives you
$$(3k+2) + (3k+3) + (3k+4) = 9k + 9$$
Which will give you a “digit sum” of $9$ (or $0$, if you add $-1$, $0$, and $1$).
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
add a comment |
$begingroup$
The “digit sum” is actually the same thing as the remainder when dividing by $9$, except that unless you have the number $0$, the digit sum will be $9$ when it is evenly divisible by $9$ (not too hard exercise).
If the first number you add is a multiple of $3$, then you are adding $3k$, $3k+1$, and $3k+2$. The sum is
$$3k + (3k+1) + (3k+2) = 9k + 3$$
which will have a “digit sum” of $3$.
If the first number you add is a multiple of $3$ plus $1$, then you are adding $3k+1$, $3k+2$, and $3k+3$; these are the two examples you have. Here you have
$$(3k+1) + (3k+2) + (3k+3) = 9k+ 6$$
giving your observed total of $6$.
Finally, if the first number you add is a multiple of $3$ plus $2$, then you are ading $3k+2$, $3k+3$, and $3k+4$, which gives you
$$(3k+2) + (3k+3) + (3k+4) = 9k + 9$$
Which will give you a “digit sum” of $9$ (or $0$, if you add $-1$, $0$, and $1$).
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
$endgroup$
The “digit sum” is actually the same thing as the remainder when dividing by $9$, except that unless you have the number $0$, the digit sum will be $9$ when it is evenly divisible by $9$ (not too hard exercise).
If the first number you add is a multiple of $3$, then you are adding $3k$, $3k+1$, and $3k+2$. The sum is
$$3k + (3k+1) + (3k+2) = 9k + 3$$
which will have a “digit sum” of $3$.
If the first number you add is a multiple of $3$ plus $1$, then you are adding $3k+1$, $3k+2$, and $3k+3$; these are the two examples you have. Here you have
$$(3k+1) + (3k+2) + (3k+3) = 9k+ 6$$
giving your observed total of $6$.
Finally, if the first number you add is a multiple of $3$ plus $2$, then you are ading $3k+2$, $3k+3$, and $3k+4$, which gives you
$$(3k+2) + (3k+3) + (3k+4) = 9k + 9$$
Which will give you a “digit sum” of $9$ (or $0$, if you add $-1$, $0$, and $1$).
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
answered Jan 28 at 3:56
Arturo MagidinArturo Magidin
266k34590920
266k34590920
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
add a comment |
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
$begingroup$
My apologies Mr. Arturo, in my haste to see the mathematics, I ignored the first paragraph. I apologise for the inconvenience.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 6:34
add a comment |
$begingroup$
Counterexamples:
$2+3+4=9$
$5+6+7=18$ where $1+8=9$
$6+7+8=21$ where $2+1=3$
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
add a comment |
$begingroup$
Counterexamples:
$2+3+4=9$
$5+6+7=18$ where $1+8=9$
$6+7+8=21$ where $2+1=3$
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
add a comment |
$begingroup$
Counterexamples:
$2+3+4=9$
$5+6+7=18$ where $1+8=9$
$6+7+8=21$ where $2+1=3$
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
$endgroup$
Counterexamples:
$2+3+4=9$
$5+6+7=18$ where $1+8=9$
$6+7+8=21$ where $2+1=3$
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
answered Jan 28 at 3:47
Mohammad Zuhair KhanMohammad Zuhair Khan
1,6792625
1,6792625
add a comment |
add a comment |
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4
$begingroup$
$5+6+7=18$ with digit sum $9$
$endgroup$
– Daniel Mathias
Jan 28 at 3:47