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Use 2, 0, 1 and 8 to make 199
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Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order so that the result equals $199$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(sqrt8*2)!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $199$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $199$, but will not mark them as correct. (This is particularly important because otherwise the method of @CarlSchildkraut in answer to this question could be used to get a solution with little effort).
Here are some examples to this problem:
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
many thanks to the authors of these questions for inspiring this question.
This question is based on a previous question where the target was 109. Many thanks to @Hugh who edited and improved that question and others who made useful comments and suggestions...
mathematics
asked 2 hours ago
tom
1,691323
add a comment |Â
up vote
5
down vote
favorite
Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order so that the result equals $199$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(sqrt8*2)!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $199$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $199$, but will not mark them as correct. (This is particularly important because otherwise the method of @CarlSchildkraut in answer to this question could be used to get a solution with little effort).
Here are some examples to this problem:
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
many thanks to the authors of these questions for inspiring this question.
This question is based on a previous question where the target was 109. Many thanks to @Hugh who edited and improved that question and others who made useful comments and suggestions...
mathematics
asked 2 hours ago
tom
1,691323
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order so that the result equals $199$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(sqrt8*2)!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $199$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $199$, but will not mark them as correct. (This is particularly important because otherwise the method of @CarlSchildkraut in answer to this question could be used to get a solution with little effort).
Here are some examples to this problem:
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
many thanks to the authors of these questions for inspiring this question.
This question is based on a previous question where the target was 109. Many thanks to @Hugh who edited and improved that question and others who made useful comments and suggestions...
mathematics
asked 2 hours ago
tom
1,691323
Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order so that the result equals $199$. You may use the operations $x + y$, $x - y$, $x times y$, $x div y$, $x!$, $sqrtx$, $sqrt[leftroot-2uproot2x]y$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(sqrt8*2)!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $199$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get $199$, but will not mark them as correct. (This is particularly important because otherwise the method of @CarlSchildkraut in answer to this question could be used to get a solution with little effort).
Here are some examples to this problem:
- Use 2 0 1 and 8 to make 67
- Make numbers 93 using the digits 2, 0, 1, 8
- Make numbers 1 - 30 using the digits 2, 0, 1, 8
many thanks to the authors of these questions for inspiring this question.
This question is based on a previous question where the target was 109. Many thanks to @Hugh who edited and improved that question and others who made useful comments and suggestions...
mathematics
mathematics
asked 2 hours ago
tom
1,691323
asked 2 hours ago
tom
1,691323
asked 2 hours ago
tom
1,691323
asked 2 hours ago
tom
1,691323
asked 2 hours ago
tom
1,691323
1,691323
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
How about:
$$sqrt8!-((2+1)!)!+0!=sqrt40320-720+1=sqrt39601=199$$
answered 2 hours ago
JonMark Perry
14.4k42870
well done - that was quick :-)
– tom
12 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
How about:
$$sqrt8!-((2+1)!)!+0!=sqrt40320-720+1=sqrt39601=199$$
answered 2 hours ago
JonMark Perry
14.4k42870
well done - that was quick :-)
– tom
12 mins ago
add a comment |Â
up vote
7
down vote
accepted
How about:
$$sqrt8!-((2+1)!)!+0!=sqrt40320-720+1=sqrt39601=199$$
answered 2 hours ago
JonMark Perry
14.4k42870
well done - that was quick :-)
– tom
12 mins ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
How about:
$$sqrt8!-((2+1)!)!+0!=sqrt40320-720+1=sqrt39601=199$$
answered 2 hours ago
JonMark Perry
14.4k42870
How about:
$$sqrt8!-((2+1)!)!+0!=sqrt40320-720+1=sqrt39601=199$$
answered 2 hours ago
JonMark Perry
14.4k42870
edited 1 hour ago
answered 2 hours ago
JonMark Perry
14.4k42870
answered 2 hours ago
JonMark Perry
14.4k42870
answered 2 hours ago
JonMark Perry
14.4k42870
14.4k42870
well done - that was quick :-)
– tom
12 mins ago
add a comment |Â
well done - that was quick :-)
– tom
12 mins ago
well done - that was quick :-)
– tom
12 mins ago
well done - that was quick :-)
– tom
12 mins ago
add a comment |Â
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