Mixing
The base in which $1065 = 13 cdot 54$ is true
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
How can one find the base $r$ in which $1065 = 13 cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.
$$r^3-5r^2-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.
Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
number-systems
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
asked Jul 21 at 12:22
Theorem
977413
add a comment |Â
up vote
6
down vote
favorite
How can one find the base $r$ in which $1065 = 13 cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.
$$r^3-5r^2-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.
Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
number-systems
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
asked Jul 21 at 12:22
Theorem
977413
1
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
How can one find the base $r$ in which $1065 = 13 cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.
$$r^3-5r^2-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.
Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
number-systems
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
asked Jul 21 at 12:22
Theorem
977413
How can one find the base $r$ in which $1065 = 13 cdot 54$ is a true statement? My attempt was constructing an equation including $r$ but because the left side has 4 digits, I got a polynomial of degree 3.
$$r^3-5r^2-13r-7=0$$
I could realize from the rational root theorem that $r=7$ is the solution, but I don't think that was what I expected to do.
Is there an alternative way to figure out $r$? Perhaps without "guessing" anything? Thanks in advance
number-systems
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
asked Jul 21 at 12:22
Theorem
977413
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
edited Jul 21 at 13:18
Jyrki Lahtonen
105k12161355
105k12161355
asked Jul 21 at 12:22
Theorem
977413
asked Jul 21 at 12:22
Theorem
977413
asked Jul 21 at 12:22
Theorem
977413
977413
1
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40
add a comment |Â
1
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40
1
1
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
11
down vote
accepted
Since $3cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law,
$$
13_rcdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12equiv12pmodr.
$$
On the other hand, in base $10$,
$$
1065_r=1cdot r^3+0cdot r^2+6r+5equiv 5pmodr.
$$
Therefore, for equality, it must be that $5equiv 12pmodr.$.
answered Jul 21 at 12:28
Michael Burr
25.6k13262
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
add a comment |Â
up vote
1
down vote
The base is clearly less than $11$, since $13cdot54lt20cdot54=T80lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
add a comment |Â
up vote
-1
down vote
Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5
$$
answered Jul 21 at 13:11
Cesareo
5,7352412
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Since $3cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law,
$$
13_rcdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12equiv12pmodr.
$$
On the other hand, in base $10$,
$$
1065_r=1cdot r^3+0cdot r^2+6r+5equiv 5pmodr.
$$
Therefore, for equality, it must be that $5equiv 12pmodr.$.
answered Jul 21 at 12:28
Michael Burr
25.6k13262
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
add a comment |Â
up vote
11
down vote
accepted
Since $3cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law,
$$
13_rcdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12equiv12pmodr.
$$
On the other hand, in base $10$,
$$
1065_r=1cdot r^3+0cdot r^2+6r+5equiv 5pmodr.
$$
Therefore, for equality, it must be that $5equiv 12pmodr.$.
answered Jul 21 at 12:28
Michael Burr
25.6k13262
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Since $3cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law,
$$
13_rcdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12equiv12pmodr.
$$
On the other hand, in base $10$,
$$
1065_r=1cdot r^3+0cdot r^2+6r+5equiv 5pmodr.
$$
Therefore, for equality, it must be that $5equiv 12pmodr.$.
answered Jul 21 at 12:28
Michael Burr
25.6k13262
Since $3cdot4-5$ ends in a zero, your base divides $12-5=7$. Therefore, your base is either $1$ or $7$. Base $1$ is rather silly and usually avoided (see the comments for a discussion for how to interpret base $1$).
To give some details, if $r$ is the base, then, in base $10$, $13_r=1cdot r+3$ and $54_r=5r+4$. Therefore, by the distributive law,
$$
13_rcdot54_r=(r+3)(5r+4)=5r^2+4r+15r+12equiv12pmodr.
$$
On the other hand, in base $10$,
$$
1065_r=1cdot r^3+0cdot r^2+6r+5equiv 5pmodr.
$$
Therefore, for equality, it must be that $5equiv 12pmodr.$.
answered Jul 21 at 12:28
Michael Burr
25.6k13262
edited Jul 21 at 14:21
answered Jul 21 at 12:28
Michael Burr
25.6k13262
answered Jul 21 at 12:28
Michael Burr
25.6k13262
answered Jul 21 at 12:28
Michael Burr
25.6k13262
25.6k13262
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
add a comment |Â
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
3
3
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
But the base is not $1$, because then it cannot have a $6$ as digit.
– wythagoras
Jul 21 at 12:31
1
1
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
Arguably (if only weakly so) it could be base $1$ - since all digit symbols have the same value, any variation is purely decorative, and the expression corresponds in decimal to $4=2cdot 2$
– Joffan
Jul 21 at 14:11
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
@Joffan I considered that, but base $1$ gets really silly and the interpretations are complicated. In base $1$, one could also interpret $1065$ as $1cdot 1^3+0cdot 1^2+6cdot 1+5=12$. Therefore, the entire statement becomes $4cdot 9=12$ (in base $10$), which is also silly.
– Michael Burr
Jul 21 at 14:21
add a comment |Â
up vote
1
down vote
The base is clearly less than $11$, since $13cdot54lt20cdot54=T80lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
add a comment |Â
up vote
1
down vote
The base is clearly less than $11$, since $13cdot54lt20cdot54=T80lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The base is clearly less than $11$, since $13cdot54lt20cdot54=T80lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
The base is clearly less than $11$, since $13cdot54lt20cdot54=T80lt1000$ (where $T$ stands for Ten) in any base greater than $10$. Because of the "$6$" in "$1065$," the base must presumably* be at least $7$. But the base can't be even, since $1065$ is odd in any even base, while $54$ is even. That leaves only $9$ and $7$ as possible bases. But $9$ doesn't work since $1065$ is not divisible by $3$ in base $9$, while $13$ is. So we're left with base $7$, which does indeed work.
*It's purely convention, based on the desire to have a unique expression for each number, that leads us to assume that a number written in base $r$ can only be written with symbols for digits up to $r-1$. (E.g., we could add the extra digit $T$ to base ten, but then we'd have duplications such as $1T=20$.) Note, the OP's approach, solving $r^3+6r+5=(r+3)(5r+4)$ for $r$, is agnostic with respect to this convention. If you reject the convention in this answer's approach, you merely need rule out the other odd possibilities less than $11$, $r=5$ and $r=3$. For $r=5$, the left hand side would be a multiple of $5$ but not the right; for $r=3$, the left hand side is not a multiple of $3$ but $13$ is.
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
answered Jul 21 at 15:26
Barry Cipra
56.5k652118
56.5k652118
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
add a comment |Â
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
I was going to sit down and work through this kind of approach, having done it in my head. It is easy to see that the base must be odd, and your observations at the end mean that it can't be divisible by $3$ or $5$. There are various ways of putting an upper limit on and yours is neater than my first thought (which had base $lt 12$). ++1
– Mark Bennet
Jul 21 at 16:25
add a comment |Â
up vote
-1
down vote
Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5
$$
answered Jul 21 at 13:11
Cesareo
5,7352412
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
add a comment |Â
up vote
-1
down vote
Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5
$$
answered Jul 21 at 13:11
Cesareo
5,7352412
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5
$$
answered Jul 21 at 13:11
Cesareo
5,7352412
Let $b$ be the base then
$$
(b+3)(5b+4)=b^3+6b+5Rightarrow 5b^2+15b+4b+12 = b^3+6b+5
$$
hence $b = 7$ then
$$
5b^2+(2b+1)b+4b+12=5b^2+2b^2+6b+5 =(5+2)b^2+6b+5 =b^3+6b+5
$$
answered Jul 21 at 13:11
Cesareo
5,7352412
edited Jul 21 at 13:30
answered Jul 21 at 13:11
Cesareo
5,7352412
answered Jul 21 at 13:11
Cesareo
5,7352412
answered Jul 21 at 13:11
Cesareo
5,7352412
5,7352412
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
add a comment |Â
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
@Theorem Of course I read the question. The solution involves polynomial and congruence representation. My answer is completely right.
– Cesareo
Jul 21 at 13:17
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
it is correct, however, if you rearrange your first equation, it results in the OP's equation: $b^3-5b^2-13b-7=0$, for which s/he wants an alternative...
– farruhota
Jul 21 at 13:49
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
@farruhota I saw that and I know also what steps to avoid. This is a trivial question and should be handled keeping in mind the congruence properties. There are many ways to obtain the right answer.
– Cesareo
Jul 21 at 14:01
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2858456%2fthe-base-in-which-1065-13-cdot-54-is-true%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
In what base does $3$ times $4$ have a units digit of $5$? There are not that many.
– Michael Burr
Jul 21 at 12:25
@MichaelBurr I thought of exactly that! However, this "move" is only valid in base 10. in other basis, the rightmost digit of a product isn't affected only by the rightmost digits of the numbers multiplied. Am I wrong?
– Theorem
Jul 21 at 12:28
Multiplication actually works the same way, it's only when you're changing bases do all the digits matter.
– Michael Burr
Jul 21 at 12:29
@MichaelBurr Interesting... so I claim that the rightmost digit of the result in base $r$ has to be $(3cdot 4) textmod r=5$?
– Theorem
Jul 21 at 12:32
If you think about it, that is the same as what you have done with the rational root theorem, but without the need to compute the whole polynomial.
– Mark Bennet
Jul 21 at 12:40