Mixing
Math riddle: reach sum of 100 with numbers $0…9$
2
$begingroup$
This is a riddle my friend gave me and we don't know the answer so we would like some help. The task is to use all of the numbers $0,1,2,...,9$ once each to get a sum of $100$ only using the plus sign.
You can do whatever you want with the numbers so long the plus sign and only that is included. For example, one could concatenate and use the number $12$ or $23$.
Multiplications or exponents or other signs are not allowed.
puzzle
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
$endgroup$
|
show 4 more comments
2
$begingroup$
This is a riddle my friend gave me and we don't know the answer so we would like some help. The task is to use all of the numbers $0,1,2,...,9$ once each to get a sum of $100$ only using the plus sign.
You can do whatever you want with the numbers so long the plus sign and only that is included. For example, one could concatenate and use the number $12$ or $23$.
Multiplications or exponents or other signs are not allowed.
puzzle
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
$endgroup$
$begingroup$
That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:34
$begingroup$
@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
$endgroup$
– Trevor Norton
Feb 11 '16 at 21:36
$begingroup$
Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:36
$begingroup$
@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:38
1
$begingroup$
I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
$endgroup$
– lulu
Feb 11 '16 at 21:50
|
show 4 more comments
2
2
2
2
$begingroup$
This is a riddle my friend gave me and we don't know the answer so we would like some help. The task is to use all of the numbers $0,1,2,...,9$ once each to get a sum of $100$ only using the plus sign.
You can do whatever you want with the numbers so long the plus sign and only that is included. For example, one could concatenate and use the number $12$ or $23$.
Multiplications or exponents or other signs are not allowed.
puzzle
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
$endgroup$
This is a riddle my friend gave me and we don't know the answer so we would like some help. The task is to use all of the numbers $0,1,2,...,9$ once each to get a sum of $100$ only using the plus sign.
You can do whatever you want with the numbers so long the plus sign and only that is included. For example, one could concatenate and use the number $12$ or $23$.
Multiplications or exponents or other signs are not allowed.
puzzle
puzzle
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
edited Feb 11 '16 at 21:48
Bobson Dugnutt
8,53831939
8,53831939
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
asked Feb 11 '16 at 21:31
Themistoklis HarisThemistoklis Haris
348
348
$begingroup$
That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:34
$begingroup$
@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
$endgroup$
– Trevor Norton
Feb 11 '16 at 21:36
$begingroup$
Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:36
$begingroup$
@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:38
1
$begingroup$
I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
$endgroup$
– lulu
Feb 11 '16 at 21:50
|
show 4 more comments
$begingroup$
That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:34
$begingroup$
@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
$endgroup$
– Trevor Norton
Feb 11 '16 at 21:36
$begingroup$
Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:36
$begingroup$
@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:38
1
$begingroup$
I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
$endgroup$
– lulu
Feb 11 '16 at 21:50
$begingroup$
That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:34
$begingroup$
That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:34
$begingroup$
@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
$endgroup$
– Trevor Norton
Feb 11 '16 at 21:36
$begingroup$
@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
$endgroup$
– Trevor Norton
Feb 11 '16 at 21:36
$begingroup$
Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:36
$begingroup$
Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:36
$begingroup$
@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:38
$begingroup$
@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
$endgroup$
– Bobson Dugnutt
Feb 11 '16 at 21:38
1
1
$begingroup$
I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
$endgroup$
– lulu
Feb 11 '16 at 21:50
$begingroup$
I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
$endgroup$
– lulu
Feb 11 '16 at 21:50
|
show 4 more comments
2 Answers
2
active
oldest
votes
17
$begingroup$
It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+cdots+9=45$ is divisble by $9$, and so it can never be $100$.
To comment on the problem where we can use a decimal point - we have the same problem. Multiply both sides by a power of $10^k$ that makes the solution $x$ integers only - and $10^kxequiv 1^kxequiv xequiv 100equiv 1mod 9$. But still the solution multiplied by $10^k$ only adds numbers with digits $1,cdots,9$ and a number of $0$'s, so $10^kxequiv 0mod 9$, which is again a contradiction.
A good website to visit on related problems is cut-the-knot.
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
$endgroup$
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
add a comment |
-1
$begingroup$
I think it is possible if we write $49 + 50+ 1/2 + 38/76$, which sums up to $100$.
edited Jan 4 at 11:00
YiFan
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
17
$begingroup$
It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+cdots+9=45$ is divisble by $9$, and so it can never be $100$.
To comment on the problem where we can use a decimal point - we have the same problem. Multiply both sides by a power of $10^k$ that makes the solution $x$ integers only - and $10^kxequiv 1^kxequiv xequiv 100equiv 1mod 9$. But still the solution multiplied by $10^k$ only adds numbers with digits $1,cdots,9$ and a number of $0$'s, so $10^kxequiv 0mod 9$, which is again a contradiction.
A good website to visit on related problems is cut-the-knot.
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
$endgroup$
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
add a comment |
17
$begingroup$
It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+cdots+9=45$ is divisble by $9$, and so it can never be $100$.
To comment on the problem where we can use a decimal point - we have the same problem. Multiply both sides by a power of $10^k$ that makes the solution $x$ integers only - and $10^kxequiv 1^kxequiv xequiv 100equiv 1mod 9$. But still the solution multiplied by $10^k$ only adds numbers with digits $1,cdots,9$ and a number of $0$'s, so $10^kxequiv 0mod 9$, which is again a contradiction.
A good website to visit on related problems is cut-the-knot.
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
$endgroup$
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
add a comment |
17
17
17
$begingroup$
It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+cdots+9=45$ is divisble by $9$, and so it can never be $100$.
To comment on the problem where we can use a decimal point - we have the same problem. Multiply both sides by a power of $10^k$ that makes the solution $x$ integers only - and $10^kxequiv 1^kxequiv xequiv 100equiv 1mod 9$. But still the solution multiplied by $10^k$ only adds numbers with digits $1,cdots,9$ and a number of $0$'s, so $10^kxequiv 0mod 9$, which is again a contradiction.
A good website to visit on related problems is cut-the-knot.
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
$endgroup$
It is impossible, since the sum will always be divisible by $9$. Note that for a number to be divisible by $9$, the sum of its digits must be divisible by $9$; thus, we know that the solution must be divisble by $9$ as $1+cdots+9=45$ is divisble by $9$, and so it can never be $100$.
To comment on the problem where we can use a decimal point - we have the same problem. Multiply both sides by a power of $10^k$ that makes the solution $x$ integers only - and $10^kxequiv 1^kxequiv xequiv 100equiv 1mod 9$. But still the solution multiplied by $10^k$ only adds numbers with digits $1,cdots,9$ and a number of $0$'s, so $10^kxequiv 0mod 9$, which is again a contradiction.
A good website to visit on related problems is cut-the-knot.
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
edited Feb 11 '16 at 21:52
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
answered Feb 11 '16 at 21:41
vrugtehagelvrugtehagel
10.7k1649
10.7k1649
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
add a comment |
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
1
1
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
The best that can be done is $1+2+3+4+5+60+7+8+9=99$.
$endgroup$
– Yves Daoust
Feb 11 '16 at 21:49
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
@YvesDaoust, yes, and there are are many more solutions if we want it to be $99$
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:50
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Oh, thanks for sorting out the decimal case (+1). Much appreciated!
$endgroup$
– lulu
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
Well, I've been trying to counterexample this for a while but I have failed so yes I suppose your reasoning is correct and concatenations of numbers or a decimal point cannot lead to a solution. So the point in the puzzle must be a clever way of using the numbers. I just don't know what it is :(
$endgroup$
– Themistoklis Haris
Feb 11 '16 at 21:51
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
$begingroup$
@ThemistoklisHaris, we must have another operation to get this to work.
$endgroup$
– vrugtehagel
Feb 11 '16 at 21:53
add a comment |
-1
$begingroup$
I think it is possible if we write $49 + 50+ 1/2 + 38/76$, which sums up to $100$.
edited Jan 4 at 11:00
YiFan
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
$endgroup$
add a comment |
-1
$begingroup$
I think it is possible if we write $49 + 50+ 1/2 + 38/76$, which sums up to $100$.
edited Jan 4 at 11:00
YiFan
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
$endgroup$
add a comment |
-1
-1
-1
$begingroup$
I think it is possible if we write $49 + 50+ 1/2 + 38/76$, which sums up to $100$.
edited Jan 4 at 11:00
YiFan
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
$endgroup$
I think it is possible if we write $49 + 50+ 1/2 + 38/76$, which sums up to $100$.
edited Jan 4 at 11:00
YiFan
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
edited Jan 4 at 11:00
YiFan
2,7641422
edited Jan 4 at 11:00
YiFan
2,7641422
edited Jan 4 at 11:00
YiFan
2,7641422
2,7641422
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
answered Jan 4 at 10:38
Salim ur RehmanSalim ur Rehman
1
1
add a comment |
add a comment |
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That can't be done, if you mean that you can only add $1,2,3,4,5,6,7,8,9$ and only use each once.
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– Bobson Dugnutt
Feb 11 '16 at 21:34
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@Lovsovs He probably meant with concatenation of the digits, so you can make $89$ with an $8$ and $9$.
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– Trevor Norton
Feb 11 '16 at 21:36
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Of course just adding is not effective. By combining, one could for example take the number 12 or 43 into account. Maybe there are other more clever ways of combining but I haven't thought of any yet.
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– Themistoklis Haris
Feb 11 '16 at 21:36
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@TrevorNorton Yeah, that makes more sense. OP, I'd suggest changing the word "combining" with "concatenating".
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– Bobson Dugnutt
Feb 11 '16 at 21:38
1
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I can't make a solution using decimal points and sums. Doesn't mean it's impossible. I can do it with backslashes (division)! $1 = frac {148}{296} + frac {35}{70}$. But that really seems to be far from what you wanted.
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– lulu
Feb 11 '16 at 21:50