Multiplying prices with one in cent is the same as them all added
up vote
0
down vote
favorite
Is it possible for all prices out of a list of prices (in Euro/Dollar) multiplied together, with one of them in cent however, to amount to the same total of 20.18€ as all of them added together? (multiplying them and having one price in cent cancel out)
Example:
1€+2.50€+0.5€= 4€
1€*2.50€*50ct= 125€
I can't figure out how to compile such a list where no rounding is necessary.
I tried to calculate the prices presuming there are only two articles:
a+b = 20.18
a*100b = 20.18
However, solved the solution is not possible with actual money (as in the amount is not even but irrational).
Any ideas?
Does it matter that I presumed there are only two articles?
Thanks in advance.
algebra-precalculus
edited Nov 15 at 13:47
xbh
4,9341421
asked Nov 15 at 13:44
Kedolind
104
add a comment |
up vote
0
down vote
favorite
Is it possible for all prices out of a list of prices (in Euro/Dollar) multiplied together, with one of them in cent however, to amount to the same total of 20.18€ as all of them added together? (multiplying them and having one price in cent cancel out)
Example:
1€+2.50€+0.5€= 4€
1€*2.50€*50ct= 125€
I can't figure out how to compile such a list where no rounding is necessary.
I tried to calculate the prices presuming there are only two articles:
a+b = 20.18
a*100b = 20.18
However, solved the solution is not possible with actual money (as in the amount is not even but irrational).
Any ideas?
Does it matter that I presumed there are only two articles?
Thanks in advance.
algebra-precalculus
edited Nov 15 at 13:47
xbh
4,9341421
asked Nov 15 at 13:44
Kedolind
104
1
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
1
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it possible for all prices out of a list of prices (in Euro/Dollar) multiplied together, with one of them in cent however, to amount to the same total of 20.18€ as all of them added together? (multiplying them and having one price in cent cancel out)
Example:
1€+2.50€+0.5€= 4€
1€*2.50€*50ct= 125€
I can't figure out how to compile such a list where no rounding is necessary.
I tried to calculate the prices presuming there are only two articles:
a+b = 20.18
a*100b = 20.18
However, solved the solution is not possible with actual money (as in the amount is not even but irrational).
Any ideas?
Does it matter that I presumed there are only two articles?
Thanks in advance.
algebra-precalculus
edited Nov 15 at 13:47
xbh
4,9341421
asked Nov 15 at 13:44
Kedolind
104
Is it possible for all prices out of a list of prices (in Euro/Dollar) multiplied together, with one of them in cent however, to amount to the same total of 20.18€ as all of them added together? (multiplying them and having one price in cent cancel out)
Example:
1€+2.50€+0.5€= 4€
1€*2.50€*50ct= 125€
I can't figure out how to compile such a list where no rounding is necessary.
I tried to calculate the prices presuming there are only two articles:
a+b = 20.18
a*100b = 20.18
However, solved the solution is not possible with actual money (as in the amount is not even but irrational).
Any ideas?
Does it matter that I presumed there are only two articles?
Thanks in advance.
algebra-precalculus
algebra-precalculus
edited Nov 15 at 13:47
xbh
4,9341421
asked Nov 15 at 13:44
Kedolind
104
edited Nov 15 at 13:47
xbh
4,9341421
asked Nov 15 at 13:44
Kedolind
104
edited Nov 15 at 13:47
xbh
4,9341421
edited Nov 15 at 13:47
xbh
4,9341421
edited Nov 15 at 13:47
xbh
4,9341421
4,9341421
asked Nov 15 at 13:44
Kedolind
104
asked Nov 15 at 13:44
Kedolind
104
asked Nov 15 at 13:44
Kedolind
104
104
1
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
1
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04
add a comment |
1
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
1
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04
1
1
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
1
1
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
First of all, let me say: I really, really don't like fractions, especially not when their denominator has to divide $100$. So, let's rewrite the equations to something with nice, whole numbers:
begin{align}
frac{x_1}{100}+frac{x_2}{100}+cdots+frac{x_k}{100}&=20.18\
100cdotfrac{x_1}{100}cdotfrac{x_2}{100}cdotcdotscdotfrac{x_k}{100}&=20.18\
end{align}
Where all $x_i$s are positive integers. Now we can rewrite this to
begin{align}
x_1+x_2+cdots+x_k&=2018\
x_1x_2cdots x_k&=2018cdot100^{k-2}\
end{align}
So, the equation with the product is actually fairly useful now. We need the product of all $x_i$s to be $2018cdot 100^{k-2}$, which factorizes as $2^{2k-3}5^{2k-4}cdot1009$. That means at least one of the numbers must have a factor $1009$. If it is not $1009$ itself, it is at least $2cdot1009=2018$; however, with the sum of all $x_i$s being $2018$, and all $x_i$s being positive, this is never going to work. So, one of the numbers we need is $1009$; let's say $x_k=1009$. The equations becomes
begin{align}
x_1+x_2+cdots+x_{k-1}&=1009\
x_1x_2cdots x_{k-1}&=2^{2k-3}5^{2k-4}\
end{align}
To get a feel for the problem we've turned this into, let's set some values for $k$. We need $k$ to be at least $2$, so let's start there.
Case 1: $k=2$. Now, $x_1=1009$ and $x_1=2$. That's very impossible.
Case 2: $k=3$. We find $x_1+x_2=1009$ and $x_1x_2=2^35^2$. Again, impossible. Whatever $x_i$s we choose, if the product is $2^35^2=200$, then the sum is never going to be as much as $1009$.
Case 3: $k=4$. This time, $x_1+x_2+x_3=1009$ and $x_1x_2x_3=2^55^4$. This might be possible; so, let's try something. We want the sum to get close to $1009$, so let's pick $x_3=1000$. Then we need $x_1+x_2=9$ and $x_1x_2=2^2cdot5$. Yes. YES. You see it? $x_1=4$, and $x_2=5$. We've found a solution!
So, we've found $(x_1,x_2,x_3,x_4)=(4,5,1000,1009)$. This translate to the following prices:
- €0.04 (4ct)
- €0.05
- €10,-
- €10.09
And indeed, their sum is €20.18, and their product, one being in cents, is (4ct)(€0.05)(€10)*(€10.09) which also is €20.18.
(Disclaimer: I don't actually dislike fractions, I just like integers more :D)
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
First of all, let me say: I really, really don't like fractions, especially not when their denominator has to divide $100$. So, let's rewrite the equations to something with nice, whole numbers:
begin{align}
frac{x_1}{100}+frac{x_2}{100}+cdots+frac{x_k}{100}&=20.18\
100cdotfrac{x_1}{100}cdotfrac{x_2}{100}cdotcdotscdotfrac{x_k}{100}&=20.18\
end{align}
Where all $x_i$s are positive integers. Now we can rewrite this to
begin{align}
x_1+x_2+cdots+x_k&=2018\
x_1x_2cdots x_k&=2018cdot100^{k-2}\
end{align}
So, the equation with the product is actually fairly useful now. We need the product of all $x_i$s to be $2018cdot 100^{k-2}$, which factorizes as $2^{2k-3}5^{2k-4}cdot1009$. That means at least one of the numbers must have a factor $1009$. If it is not $1009$ itself, it is at least $2cdot1009=2018$; however, with the sum of all $x_i$s being $2018$, and all $x_i$s being positive, this is never going to work. So, one of the numbers we need is $1009$; let's say $x_k=1009$. The equations becomes
begin{align}
x_1+x_2+cdots+x_{k-1}&=1009\
x_1x_2cdots x_{k-1}&=2^{2k-3}5^{2k-4}\
end{align}
To get a feel for the problem we've turned this into, let's set some values for $k$. We need $k$ to be at least $2$, so let's start there.
Case 1: $k=2$. Now, $x_1=1009$ and $x_1=2$. That's very impossible.
Case 2: $k=3$. We find $x_1+x_2=1009$ and $x_1x_2=2^35^2$. Again, impossible. Whatever $x_i$s we choose, if the product is $2^35^2=200$, then the sum is never going to be as much as $1009$.
Case 3: $k=4$. This time, $x_1+x_2+x_3=1009$ and $x_1x_2x_3=2^55^4$. This might be possible; so, let's try something. We want the sum to get close to $1009$, so let's pick $x_3=1000$. Then we need $x_1+x_2=9$ and $x_1x_2=2^2cdot5$. Yes. YES. You see it? $x_1=4$, and $x_2=5$. We've found a solution!
So, we've found $(x_1,x_2,x_3,x_4)=(4,5,1000,1009)$. This translate to the following prices:
- €0.04 (4ct)
- €0.05
- €10,-
- €10.09
And indeed, their sum is €20.18, and their product, one being in cents, is (4ct)(€0.05)(€10)*(€10.09) which also is €20.18.
(Disclaimer: I don't actually dislike fractions, I just like integers more :D)
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
add a comment |
up vote
3
down vote
accepted
First of all, let me say: I really, really don't like fractions, especially not when their denominator has to divide $100$. So, let's rewrite the equations to something with nice, whole numbers:
begin{align}
frac{x_1}{100}+frac{x_2}{100}+cdots+frac{x_k}{100}&=20.18\
100cdotfrac{x_1}{100}cdotfrac{x_2}{100}cdotcdotscdotfrac{x_k}{100}&=20.18\
end{align}
Where all $x_i$s are positive integers. Now we can rewrite this to
begin{align}
x_1+x_2+cdots+x_k&=2018\
x_1x_2cdots x_k&=2018cdot100^{k-2}\
end{align}
So, the equation with the product is actually fairly useful now. We need the product of all $x_i$s to be $2018cdot 100^{k-2}$, which factorizes as $2^{2k-3}5^{2k-4}cdot1009$. That means at least one of the numbers must have a factor $1009$. If it is not $1009$ itself, it is at least $2cdot1009=2018$; however, with the sum of all $x_i$s being $2018$, and all $x_i$s being positive, this is never going to work. So, one of the numbers we need is $1009$; let's say $x_k=1009$. The equations becomes
begin{align}
x_1+x_2+cdots+x_{k-1}&=1009\
x_1x_2cdots x_{k-1}&=2^{2k-3}5^{2k-4}\
end{align}
To get a feel for the problem we've turned this into, let's set some values for $k$. We need $k$ to be at least $2$, so let's start there.
Case 1: $k=2$. Now, $x_1=1009$ and $x_1=2$. That's very impossible.
Case 2: $k=3$. We find $x_1+x_2=1009$ and $x_1x_2=2^35^2$. Again, impossible. Whatever $x_i$s we choose, if the product is $2^35^2=200$, then the sum is never going to be as much as $1009$.
Case 3: $k=4$. This time, $x_1+x_2+x_3=1009$ and $x_1x_2x_3=2^55^4$. This might be possible; so, let's try something. We want the sum to get close to $1009$, so let's pick $x_3=1000$. Then we need $x_1+x_2=9$ and $x_1x_2=2^2cdot5$. Yes. YES. You see it? $x_1=4$, and $x_2=5$. We've found a solution!
So, we've found $(x_1,x_2,x_3,x_4)=(4,5,1000,1009)$. This translate to the following prices:
- €0.04 (4ct)
- €0.05
- €10,-
- €10.09
And indeed, their sum is €20.18, and their product, one being in cents, is (4ct)(€0.05)(€10)*(€10.09) which also is €20.18.
(Disclaimer: I don't actually dislike fractions, I just like integers more :D)
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
First of all, let me say: I really, really don't like fractions, especially not when their denominator has to divide $100$. So, let's rewrite the equations to something with nice, whole numbers:
begin{align}
frac{x_1}{100}+frac{x_2}{100}+cdots+frac{x_k}{100}&=20.18\
100cdotfrac{x_1}{100}cdotfrac{x_2}{100}cdotcdotscdotfrac{x_k}{100}&=20.18\
end{align}
Where all $x_i$s are positive integers. Now we can rewrite this to
begin{align}
x_1+x_2+cdots+x_k&=2018\
x_1x_2cdots x_k&=2018cdot100^{k-2}\
end{align}
So, the equation with the product is actually fairly useful now. We need the product of all $x_i$s to be $2018cdot 100^{k-2}$, which factorizes as $2^{2k-3}5^{2k-4}cdot1009$. That means at least one of the numbers must have a factor $1009$. If it is not $1009$ itself, it is at least $2cdot1009=2018$; however, with the sum of all $x_i$s being $2018$, and all $x_i$s being positive, this is never going to work. So, one of the numbers we need is $1009$; let's say $x_k=1009$. The equations becomes
begin{align}
x_1+x_2+cdots+x_{k-1}&=1009\
x_1x_2cdots x_{k-1}&=2^{2k-3}5^{2k-4}\
end{align}
To get a feel for the problem we've turned this into, let's set some values for $k$. We need $k$ to be at least $2$, so let's start there.
Case 1: $k=2$. Now, $x_1=1009$ and $x_1=2$. That's very impossible.
Case 2: $k=3$. We find $x_1+x_2=1009$ and $x_1x_2=2^35^2$. Again, impossible. Whatever $x_i$s we choose, if the product is $2^35^2=200$, then the sum is never going to be as much as $1009$.
Case 3: $k=4$. This time, $x_1+x_2+x_3=1009$ and $x_1x_2x_3=2^55^4$. This might be possible; so, let's try something. We want the sum to get close to $1009$, so let's pick $x_3=1000$. Then we need $x_1+x_2=9$ and $x_1x_2=2^2cdot5$. Yes. YES. You see it? $x_1=4$, and $x_2=5$. We've found a solution!
So, we've found $(x_1,x_2,x_3,x_4)=(4,5,1000,1009)$. This translate to the following prices:
- €0.04 (4ct)
- €0.05
- €10,-
- €10.09
And indeed, their sum is €20.18, and their product, one being in cents, is (4ct)(€0.05)(€10)*(€10.09) which also is €20.18.
(Disclaimer: I don't actually dislike fractions, I just like integers more :D)
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
First of all, let me say: I really, really don't like fractions, especially not when their denominator has to divide $100$. So, let's rewrite the equations to something with nice, whole numbers:
begin{align}
frac{x_1}{100}+frac{x_2}{100}+cdots+frac{x_k}{100}&=20.18\
100cdotfrac{x_1}{100}cdotfrac{x_2}{100}cdotcdotscdotfrac{x_k}{100}&=20.18\
end{align}
Where all $x_i$s are positive integers. Now we can rewrite this to
begin{align}
x_1+x_2+cdots+x_k&=2018\
x_1x_2cdots x_k&=2018cdot100^{k-2}\
end{align}
So, the equation with the product is actually fairly useful now. We need the product of all $x_i$s to be $2018cdot 100^{k-2}$, which factorizes as $2^{2k-3}5^{2k-4}cdot1009$. That means at least one of the numbers must have a factor $1009$. If it is not $1009$ itself, it is at least $2cdot1009=2018$; however, with the sum of all $x_i$s being $2018$, and all $x_i$s being positive, this is never going to work. So, one of the numbers we need is $1009$; let's say $x_k=1009$. The equations becomes
begin{align}
x_1+x_2+cdots+x_{k-1}&=1009\
x_1x_2cdots x_{k-1}&=2^{2k-3}5^{2k-4}\
end{align}
To get a feel for the problem we've turned this into, let's set some values for $k$. We need $k$ to be at least $2$, so let's start there.
Case 1: $k=2$. Now, $x_1=1009$ and $x_1=2$. That's very impossible.
Case 2: $k=3$. We find $x_1+x_2=1009$ and $x_1x_2=2^35^2$. Again, impossible. Whatever $x_i$s we choose, if the product is $2^35^2=200$, then the sum is never going to be as much as $1009$.
Case 3: $k=4$. This time, $x_1+x_2+x_3=1009$ and $x_1x_2x_3=2^55^4$. This might be possible; so, let's try something. We want the sum to get close to $1009$, so let's pick $x_3=1000$. Then we need $x_1+x_2=9$ and $x_1x_2=2^2cdot5$. Yes. YES. You see it? $x_1=4$, and $x_2=5$. We've found a solution!
So, we've found $(x_1,x_2,x_3,x_4)=(4,5,1000,1009)$. This translate to the following prices:
- €0.04 (4ct)
- €0.05
- €10,-
- €10.09
And indeed, their sum is €20.18, and their product, one being in cents, is (4ct)(€0.05)(€10)*(€10.09) which also is €20.18.
(Disclaimer: I don't actually dislike fractions, I just like integers more :D)
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
answered Nov 15 at 14:59
vrugtehagel
10.7k1549
10.7k1549
add a comment |
add a comment |
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1
Well the solution to $x^2=8$ is also irrational, but the solution to $x^3=8$ isn't. The fact that the solution is irrational with two variables doesn't necessarily say anything about the solution in three variables.
– vrugtehagel
Nov 15 at 13:48
1
So essentially the problem is finding numbers $x_0, x_1, x_2$ etc so that $$ x_0 + x_1 + ldots + x_n = 100 x_1 times x_2 times ldots times x_n $$ Indeed a good idea would be to start with two numbers and then work your way up. However, it's clear that you can only use relatively small numbers because multiplying by a large number has a much larger effect than adding a large number.
– Matti P.
Nov 15 at 13:50
Actually, you can build a list by starting with any (two or more) numbers whose product and sum is a whole number, and then adding $1$'s to the list until both sides of the equation are equal. Adding 1 increases the left-hand side but does nothing to the right-hand side.
– Matti P.
Nov 15 at 13:54
Fun fact: it's never going to work with more than $21$ numbers.
– vrugtehagel
Nov 15 at 14:04