Mixing
A Fairly Simple Equation
Clash Royale CLAN TAG#URR8PPP
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I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)
Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)
Equation: $(A+B)(C-D)(E/F)$
I'm fairly certain there is only one solution.
mathematics
asked 1 hour ago
Excited Raichu
314
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Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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up vote
4
down vote
favorite
I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)
Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)
Equation: $(A+B)(C-D)(E/F)$
I'm fairly certain there is only one solution.
mathematics
asked 1 hour ago
Excited Raichu
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)
Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)
Equation: $(A+B)(C-D)(E/F)$
I'm fairly certain there is only one solution.
mathematics
asked 1 hour ago
Excited Raichu
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm new, so I thought I'd start with a fairly simple puzzle.(Simple doesn't necessarily mean easy.)
Can you fit the numbers $2, 3, 4, 5, 6,$ and $10$ into this equation, each replacing one letter, so that the result is $93$ (rounded to the nearest whole number)
Equation: $(A+B)(C-D)(E/F)$
I'm fairly certain there is only one solution.
mathematics
mathematics
asked 1 hour ago
Excited Raichu
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Excited Raichu
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Excited Raichu
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Excited Raichu
314
asked 1 hour ago
Excited Raichu
314
314
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Excited Raichu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago
add a comment |Â
Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago
Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
I have the answer.
$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.
To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.
edited 10 mins ago
gabbo1092
82112
answered 20 mins ago
Sarthak Rout
313
add a comment |Â
up vote
1
down vote
Just to be that guy.
There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.
answered 4 mins ago
Robert S.
1736
add a comment |Â
up vote
1
down vote
Without doing any of the following, there are no solutions.
- Combining numbers.
- Reusing numbers.
- Rotating numbers.
- Rounding.
When allowing numbers to be combined, there are thousands of solutions given that:
There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.
However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.
When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.
Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:
$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$
answered 9 mins ago
PerpetualJ
2,263228
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
I have the answer.
$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.
To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.
edited 10 mins ago
gabbo1092
82112
answered 20 mins ago
Sarthak Rout
313
add a comment |Â
up vote
2
down vote
I have the answer.
$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.
To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.
edited 10 mins ago
gabbo1092
82112
answered 20 mins ago
Sarthak Rout
313
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I have the answer.
$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.
To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.
edited 10 mins ago
gabbo1092
82112
answered 20 mins ago
Sarthak Rout
313
I have the answer.
$(10+4)(6-2)(5/3)=93.333$ which when rounded off is 93.
To solve this I tried various ratios of E/F and divided it from 93 and tried to make the nearest integer from the given numbers.
edited 10 mins ago
gabbo1092
82112
answered 20 mins ago
Sarthak Rout
313
edited 10 mins ago
gabbo1092
82112
edited 10 mins ago
gabbo1092
82112
edited 10 mins ago
gabbo1092
82112
82112
answered 20 mins ago
Sarthak Rout
313
answered 20 mins ago
Sarthak Rout
313
answered 20 mins ago
Sarthak Rout
313
313
add a comment |Â
add a comment |Â
up vote
1
down vote
Just to be that guy.
There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.
answered 4 mins ago
Robert S.
1736
add a comment |Â
up vote
1
down vote
Just to be that guy.
There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.
answered 4 mins ago
Robert S.
1736
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just to be that guy.
There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.
answered 4 mins ago
Robert S.
1736
Just to be that guy.
There are technically two unique solutions:
$(10+4)(6−2)(5/3)=93.333$ rounds down to $93$. Found first by Sarthak Rout.
Switching the first two numbers $(4+10)(6−2)(5/3)=93.333$ which rounds down to $93$.
answered 4 mins ago
Robert S.
1736
answered 4 mins ago
Robert S.
1736
answered 4 mins ago
Robert S.
1736
answered 4 mins ago
Robert S.
1736
1736
add a comment |Â
add a comment |Â
up vote
1
down vote
Without doing any of the following, there are no solutions.
- Combining numbers.
- Reusing numbers.
- Rotating numbers.
- Rounding.
When allowing numbers to be combined, there are thousands of solutions given that:
There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.
However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.
When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.
Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:
$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$
answered 9 mins ago
PerpetualJ
2,263228
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
add a comment |Â
up vote
1
down vote
Without doing any of the following, there are no solutions.
- Combining numbers.
- Reusing numbers.
- Rotating numbers.
- Rounding.
When allowing numbers to be combined, there are thousands of solutions given that:
There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.
However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.
When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.
Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:
$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$
answered 9 mins ago
PerpetualJ
2,263228
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Without doing any of the following, there are no solutions.
- Combining numbers.
- Reusing numbers.
- Rotating numbers.
- Rounding.
When allowing numbers to be combined, there are thousands of solutions given that:
There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.
However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.
When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.
Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:
$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$
answered 9 mins ago
PerpetualJ
2,263228
Without doing any of the following, there are no solutions.
- Combining numbers.
- Reusing numbers.
- Rotating numbers.
- Rounding.
When allowing numbers to be combined, there are thousands of solutions given that:
There are 6 numbers, and this means there are 7 basic representations of combinations for each number, for example:
$2, 22, 23, 24, 25, 26, 210$.
However, if you go even further; you can create roughly 36 combinations for each number if you limit yourself to only a single repetition in a single combination.
When rotating numbers is allowed, this number of solutions increases substantially, especially if combination is also allowed. Throw in rounding on top of all of this and you've got yourself a solution celebration similar to the ball drop in Time Square.
Since rounding is allowed, but the rest are not; I can promise that there are at least 2 solutions:
$(4 + 10)(6 - 2)(5 / 3) = 93$
$(10 + 4)(6 - 2)(5 / 3) = 93$
answered 9 mins ago
PerpetualJ
2,263228
edited 20 secs ago
answered 9 mins ago
PerpetualJ
2,263228
answered 9 mins ago
PerpetualJ
2,263228
answered 9 mins ago
PerpetualJ
2,263228
2,263228
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
add a comment |Â
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
@RobertS. Sorry, I was still updating my answer.
– PerpetualJ
6 secs ago
add a comment |Â
Excited Raichu is a new contributor. Be nice, and check out our Code of Conduct.
Excited Raichu is a new contributor. Be nice, and check out our Code of Conduct.
Excited Raichu is a new contributor. Be nice, and check out our Code of Conduct.
Excited Raichu is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to Puzzling.SE! This is a great puzzle, are we allowed to reuse numbers to make additional numbers? Such as $(23 + 3)(45 - 5)(610 / 10)$? I just want to ensure we know the rules of this. Also, I would assume that the pairing of parenthesis is implying multiplication as with traditional math?
– PerpetualJ
21 mins ago
You can't combine or reuse numbers, and the pairs of parentheses is multiplication.
– Excited Raichu
12 mins ago