Counting number of friendly triplets
up vote
0
down vote
favorite
A triplet (a,b,c) is called friendly if product of any two is equal to third number . Find number of such ordered triplet. a,b,c are real numbers
permutations
asked Nov 14 at 16:22
maveric
59511
add a comment |
up vote
0
down vote
favorite
A triplet (a,b,c) is called friendly if product of any two is equal to third number . Find number of such ordered triplet. a,b,c are real numbers
permutations
asked Nov 14 at 16:22
maveric
59511
1
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A triplet (a,b,c) is called friendly if product of any two is equal to third number . Find number of such ordered triplet. a,b,c are real numbers
permutations
asked Nov 14 at 16:22
maveric
59511
A triplet (a,b,c) is called friendly if product of any two is equal to third number . Find number of such ordered triplet. a,b,c are real numbers
permutations
permutations
asked Nov 14 at 16:22
maveric
59511
asked Nov 14 at 16:22
maveric
59511
asked Nov 14 at 16:22
maveric
59511
asked Nov 14 at 16:22
maveric
59511
asked Nov 14 at 16:22
maveric
59511
59511
1
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27
add a comment |
1
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27
1
1
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get
$$abcdot bc cdot ca = abc$$
$$(abc)^2=abc$$
The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.
If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.
If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=pm 1$, $b=pm 1$ and $c=pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.
answered Nov 14 at 16:33
lisyarus
10.3k21433
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get
$$abcdot bc cdot ca = abc$$
$$(abc)^2=abc$$
The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.
If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.
If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=pm 1$, $b=pm 1$ and $c=pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.
answered Nov 14 at 16:33
lisyarus
10.3k21433
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
add a comment |
up vote
2
down vote
Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get
$$abcdot bc cdot ca = abc$$
$$(abc)^2=abc$$
The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.
If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.
If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=pm 1$, $b=pm 1$ and $c=pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.
answered Nov 14 at 16:33
lisyarus
10.3k21433
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
add a comment |
up vote
2
down vote
up vote
2
down vote
Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get
$$abcdot bc cdot ca = abc$$
$$(abc)^2=abc$$
The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.
If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.
If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=pm 1$, $b=pm 1$ and $c=pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.
answered Nov 14 at 16:33
lisyarus
10.3k21433
Assuming that you are looking for triples $(a,b,c)$ of real numbers such that $ab=c$ and $bc=a$ and $ca=b$, then multiply all three equalities to get
$$abcdot bc cdot ca = abc$$
$$(abc)^2=abc$$
The only two real solutions of $x^2=x$ are $x=0$ and $x=1$. So, either $abc=0$ or $abc=1$.
If $abc=0$, then one of $a$,$b$ or $c$ is zero. Without loss of generality, assume $a=0$. Then, $b=ca=0$ and $c=ab=0$.
If $abc=1$, substitute $bc=a$ to get $a^2=1$. Similarly, $b^2=1$ and $c^2=1$. This means $a=pm 1$, $b=pm 1$ and $c=pm 1$. This gives 8 possible triples $(a,b,c)$ which can be checked by hand.
answered Nov 14 at 16:33
lisyarus
10.3k21433
answered Nov 14 at 16:33
lisyarus
10.3k21433
answered Nov 14 at 16:33
lisyarus
10.3k21433
answered Nov 14 at 16:33
lisyarus
10.3k21433
10.3k21433
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
add a comment |
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
but few of which having (1,1,-1) will get removed.
– maveric
Nov 14 at 16:37
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
@maveric Yep, you should get $(1,1,1)$ and permutations of $(1,-1,-1)$.
– lisyarus
Nov 14 at 16:49
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
and (0,0,0) will also be present
– maveric
Nov 14 at 16:53
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
@maveric Yes, this is mentioned in the answer.
– lisyarus
Nov 14 at 17:04
add a comment |
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2998465%2fcounting-number-of-friendly-triplets%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
To be clear, you are saying that $(a,b,c)$ is considered friendly iff $ab=c$ and $ac=b$ and $bc=a$? In that interpretation, I expect you should find that there are only the trivial options: $(0,0,0)$ and $(1,1,1)$.
– JMoravitz
Nov 14 at 16:24
how can we sure of finiteness of solution. why cant we have more solutions. i arrived at two conditions abc=0 or abc =1. now for 1 , how do i show (1,1,1) is only possible.
– maveric
Nov 14 at 16:27