Minimum value of unknown exponents for a perfect cube
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The following question showed up in an aptitude test I took.
If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.
I solved it as follows:
My solution
However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?
Here is the solution provided by the testing agency:
Provided solution
number-theory
asked Nov 15 at 18:51
Aamir Khan
33
|
show 3 more comments
up vote
0
down vote
favorite
The following question showed up in an aptitude test I took.
If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.
I solved it as follows:
My solution
However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?
Here is the solution provided by the testing agency:
Provided solution
number-theory
asked Nov 15 at 18:51
Aamir Khan
33
The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
1
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The following question showed up in an aptitude test I took.
If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.
I solved it as follows:
My solution
However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?
Here is the solution provided by the testing agency:
Provided solution
number-theory
asked Nov 15 at 18:51
Aamir Khan
33
The following question showed up in an aptitude test I took.
If $N = 2197^P × 144^2 × 2^R × 3^S$ is the perfect cube of a natural
number, where $P$, $R$ and $S$ are distinct positive integers, then find the
minimum value of $(P + R + S)$.
I solved it as follows:
My solution
However, the correct answer is 6 as per the answer key released. Could anyone please tell where am I going wrong?
Here is the solution provided by the testing agency:
Provided solution
number-theory
number-theory
asked Nov 15 at 18:51
Aamir Khan
33
asked Nov 15 at 18:51
Aamir Khan
33
edited Nov 15 at 19:15
asked Nov 15 at 18:51
Aamir Khan
33
asked Nov 15 at 18:51
Aamir Khan
33
asked Nov 15 at 18:51
Aamir Khan
33
33
The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
1
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01
|
show 3 more comments
The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
1
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01
The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
1
1
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01
|
show 3 more comments
1 Answer
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0
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$$2197 = 13^3$$
$$144 = 2^4 3^2$$
exponent of $13$ is $3P$
exponent of $2$ is $4 + R$
exponent of $3$ is $2+S$
So $P$ is $1,2,3,4,5...$
$R$ is $2,5,8,...$
$S$ is $1,4,7,...$
If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$
With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$
answered Nov 15 at 19:44
Will Jagy
101k597198
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
$$2197 = 13^3$$
$$144 = 2^4 3^2$$
exponent of $13$ is $3P$
exponent of $2$ is $4 + R$
exponent of $3$ is $2+S$
So $P$ is $1,2,3,4,5...$
$R$ is $2,5,8,...$
$S$ is $1,4,7,...$
If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$
With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$
answered Nov 15 at 19:44
Will Jagy
101k597198
add a comment |
up vote
0
down vote
$$2197 = 13^3$$
$$144 = 2^4 3^2$$
exponent of $13$ is $3P$
exponent of $2$ is $4 + R$
exponent of $3$ is $2+S$
So $P$ is $1,2,3,4,5...$
$R$ is $2,5,8,...$
$S$ is $1,4,7,...$
If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$
With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$
answered Nov 15 at 19:44
Will Jagy
101k597198
add a comment |
up vote
0
down vote
up vote
0
down vote
$$2197 = 13^3$$
$$144 = 2^4 3^2$$
exponent of $13$ is $3P$
exponent of $2$ is $4 + R$
exponent of $3$ is $2+S$
So $P$ is $1,2,3,4,5...$
$R$ is $2,5,8,...$
$S$ is $1,4,7,...$
If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$
With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$
answered Nov 15 at 19:44
Will Jagy
101k597198
$$2197 = 13^3$$
$$144 = 2^4 3^2$$
exponent of $13$ is $3P$
exponent of $2$ is $4 + R$
exponent of $3$ is $2+S$
So $P$ is $1,2,3,4,5...$
$R$ is $2,5,8,...$
$S$ is $1,4,7,...$
If $P$ is $1,2$ we need a second choice, best would be $S=4,$ but then $P+R geq 3$ regardless, so sum is at least $7$
With $P=3, R=2, S=1 $ we get sum down to $6,$ and we cannot do better since $R+S geq 3$
answered Nov 15 at 19:44
Will Jagy
101k597198
answered Nov 15 at 19:44
Will Jagy
101k597198
answered Nov 15 at 19:44
Will Jagy
101k597198
answered Nov 15 at 19:44
Will Jagy
101k597198
101k597198
add a comment |
add a comment |
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The answer you provide appears to be $4$ as well, no?
– lulu
Nov 15 at 18:53
@lulu- It was a mistake. The answer key shows 6 as the answer. I corrected it. Apologies.
– Aamir Khan
Nov 15 at 18:54
Oh, the problem is that $P,Q,R$ are required to be distinct.
– lulu
Nov 15 at 18:56
Maybe. Let me just also attach the image of their solution.
– Aamir Khan
Nov 15 at 18:57
1
Well, I'm pretty old and (obviously) I still make mistakes like that. It's a problem with puzzle questions...if the question arises naturally then it's a lot more natural to keep track of all the requirements.
– lulu
Nov 15 at 19:01