How to find the unknown number while only LCM is given
$begingroup$
Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
I have searched many ways to solve but could not find a step wise method to teach the students, please help.
least-common-multiple
edited Mar 5 '18 at 17:48
rbird
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
$endgroup$
add a comment |
$begingroup$
Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
I have searched many ways to solve but could not find a step wise method to teach the students, please help.
least-common-multiple
edited Mar 5 '18 at 17:48
rbird
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
$endgroup$
add a comment |
$begingroup$
Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
I have searched many ways to solve but could not find a step wise method to teach the students, please help.
least-common-multiple
edited Mar 5 '18 at 17:48
rbird
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
$endgroup$
Find the smallest value of $n$ such that the lcm of $n$ and $15$ is $45$.
I have searched many ways to solve but could not find a step wise method to teach the students, please help.
least-common-multiple
least-common-multiple
edited Mar 5 '18 at 17:48
rbird
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
edited Mar 5 '18 at 17:48
rbird
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
edited Mar 5 '18 at 17:48
rbird
1,194114
edited Mar 5 '18 at 17:48
rbird
1,194114
edited Mar 5 '18 at 17:48
rbird
1,194114
1,194114
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
asked Mar 5 '18 at 17:41
Yunas KhanYunas Khan
11
11
add a comment |
add a comment |
4 Answers
4
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oldest
votes
$begingroup$
$15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
$endgroup$
add a comment |
$begingroup$
Hint: $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
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add a comment |
$begingroup$
I would present the following as a systematic approach
$$
eqalign{
& {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
& Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
$$
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
$$1,3,5,9,15,45.$$
Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
$endgroup$
add a comment |
$begingroup$
$15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
$endgroup$
add a comment |
$begingroup$
$15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
$endgroup$
$15=3times 5$ and $45=3^2times 5$. Your number $n=3^2=9$.
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
answered Mar 5 '18 at 17:43
rbirdrbird
1,194114
1,194114
add a comment |
add a comment |
$begingroup$
Hint: $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
$endgroup$
add a comment |
$begingroup$
Hint: $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
$endgroup$
add a comment |
$begingroup$
Hint: $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
$endgroup$
Hint: $,n mid 45 = 3^2 cdot 5,$ so $,n=3^a cdot 5^b,$ with $,0 le a le 2,$ and $,0 le b le 1,$. Find the smallest $,a,b,$ such that $,45 = operatorname{lcm}(15,n) = operatorname{lcm}(3 cdot 5, 3^a cdot 5^b) = 3^{max(1, a)}cdot 5^{max(1,b)},$.
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
answered Mar 5 '18 at 17:44
dxivdxiv
58k648102
58k648102
add a comment |
add a comment |
$begingroup$
I would present the following as a systematic approach
$$
eqalign{
& {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
& Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
$$
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
I would present the following as a systematic approach
$$
eqalign{
& {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
& Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
$$
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
I would present the following as a systematic approach
$$
eqalign{
& {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
& Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
$$
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
$endgroup$
I would present the following as a systematic approach
$$
eqalign{
& {rm lcm}(15,n) = 45 = {{15 cdot n} over {gcd (15,n)}}quad Rightarrow quad n = 3gcd (3 cdot 5,n)quad Rightarrow cr
& Rightarrow quad min (n) = 3min left( {3,5,15} right) = 9 cr}
$$
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
answered Mar 5 '18 at 18:06
G CabG Cab
20.4k31341
20.4k31341
add a comment |
add a comment |
$begingroup$
$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
$$1,3,5,9,15,45.$$
Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
$endgroup$
add a comment |
$begingroup$
$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
$$1,3,5,9,15,45.$$
Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
$endgroup$
add a comment |
$begingroup$
$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
$$1,3,5,9,15,45.$$
Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
$endgroup$
$n$ must be a divisor of $45$, however $n$ and $15$ must not be each other's divisor. Divisors of $45$:
$$1,3,5,9,15,45.$$
Note that $3,5,15$ are divisors of $15$. So $n$ can be $9$ or $45$. Therefore the smallest $n$ is $9$.
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
answered Mar 5 '18 at 18:52
farruhotafarruhota
21.3k2841
21.3k2841
add a comment |
add a comment |
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