Whole Number to Exponents [closed]
$begingroup$
I need to find the exponents from $3240$. The answer sheet says $2^3cdot3^4cdot5$. But how do I get the exponents from my whole number?
algebra-precalculus elementary-number-theory
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
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closed as off-topic by José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon Dec 17 '18 at 21:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need to find the exponents from $3240$. The answer sheet says $2^3cdot3^4cdot5$. But how do I get the exponents from my whole number?
algebra-precalculus elementary-number-theory
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
$endgroup$
closed as off-topic by José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon Dec 17 '18 at 21:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
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– user334732
Dec 17 '18 at 19:47
add a comment |
$begingroup$
I need to find the exponents from $3240$. The answer sheet says $2^3cdot3^4cdot5$. But how do I get the exponents from my whole number?
algebra-precalculus elementary-number-theory
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
$endgroup$
I need to find the exponents from $3240$. The answer sheet says $2^3cdot3^4cdot5$. But how do I get the exponents from my whole number?
algebra-precalculus elementary-number-theory
algebra-precalculus elementary-number-theory
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
edited Dec 18 '18 at 3:09
AryanSonwatikar
28812
28812
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
asked Dec 17 '18 at 18:51
Oliver GustafsonOliver Gustafson
71
71
closed as off-topic by José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon Dec 17 '18 at 21:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon Dec 17 '18 at 21:22
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Jyrki Lahtonen, amWhy, Pierre-Guy Plamondon
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
$endgroup$
– user334732
Dec 17 '18 at 19:47
add a comment |
1
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
$endgroup$
– user334732
Dec 17 '18 at 19:47
1
1
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
$endgroup$
– user334732
Dec 17 '18 at 19:47
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
$endgroup$
– user334732
Dec 17 '18 at 19:47
add a comment |
2 Answers
2
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oldest
votes
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For the powers of $2$; keep dividing by $2$ until you can no longer: 3240-> 1620 -> 810 -> 405 .. that's it: 405 is not divisible by $2$. So, you were able to divide by $2$ three times, hence the exponent of $2$ is $3$.
Now continue dividing by $3$ ...
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
$endgroup$
add a comment |
$begingroup$
Straight away we can identify that $3240$ is divisible by $10$, or equivalently $2cdot 5$; so we can start with $3240=2cdot 5cdot 324$. Now, by using this divisibility test we can see that $324$ is divisible by $3$ precisely $108$ times. A further inspection sees that $108$ divides by $3$ again. So so far, we have $3240=2cdot 5cdot 3cdot 3cdot 36$. From here, we can swiftly conclude this by realising that $36=6^2=(2cdot 3)^2=2^2cdot 3^2$. Thus, putting everything together gives us $$3240=2cdot 5cdot 3cdot 3cdot 2^2cdot 3^2=2^3cdot3^4cdot 5,$$ as desired.
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the powers of $2$; keep dividing by $2$ until you can no longer: 3240-> 1620 -> 810 -> 405 .. that's it: 405 is not divisible by $2$. So, you were able to divide by $2$ three times, hence the exponent of $2$ is $3$.
Now continue dividing by $3$ ...
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
$endgroup$
add a comment |
$begingroup$
For the powers of $2$; keep dividing by $2$ until you can no longer: 3240-> 1620 -> 810 -> 405 .. that's it: 405 is not divisible by $2$. So, you were able to divide by $2$ three times, hence the exponent of $2$ is $3$.
Now continue dividing by $3$ ...
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
$endgroup$
add a comment |
$begingroup$
For the powers of $2$; keep dividing by $2$ until you can no longer: 3240-> 1620 -> 810 -> 405 .. that's it: 405 is not divisible by $2$. So, you were able to divide by $2$ three times, hence the exponent of $2$ is $3$.
Now continue dividing by $3$ ...
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
$endgroup$
For the powers of $2$; keep dividing by $2$ until you can no longer: 3240-> 1620 -> 810 -> 405 .. that's it: 405 is not divisible by $2$. So, you were able to divide by $2$ three times, hence the exponent of $2$ is $3$.
Now continue dividing by $3$ ...
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
answered Dec 17 '18 at 18:54
Bram28Bram28
60.7k44590
60.7k44590
add a comment |
add a comment |
$begingroup$
Straight away we can identify that $3240$ is divisible by $10$, or equivalently $2cdot 5$; so we can start with $3240=2cdot 5cdot 324$. Now, by using this divisibility test we can see that $324$ is divisible by $3$ precisely $108$ times. A further inspection sees that $108$ divides by $3$ again. So so far, we have $3240=2cdot 5cdot 3cdot 3cdot 36$. From here, we can swiftly conclude this by realising that $36=6^2=(2cdot 3)^2=2^2cdot 3^2$. Thus, putting everything together gives us $$3240=2cdot 5cdot 3cdot 3cdot 2^2cdot 3^2=2^3cdot3^4cdot 5,$$ as desired.
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
$endgroup$
add a comment |
$begingroup$
Straight away we can identify that $3240$ is divisible by $10$, or equivalently $2cdot 5$; so we can start with $3240=2cdot 5cdot 324$. Now, by using this divisibility test we can see that $324$ is divisible by $3$ precisely $108$ times. A further inspection sees that $108$ divides by $3$ again. So so far, we have $3240=2cdot 5cdot 3cdot 3cdot 36$. From here, we can swiftly conclude this by realising that $36=6^2=(2cdot 3)^2=2^2cdot 3^2$. Thus, putting everything together gives us $$3240=2cdot 5cdot 3cdot 3cdot 2^2cdot 3^2=2^3cdot3^4cdot 5,$$ as desired.
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
$endgroup$
add a comment |
$begingroup$
Straight away we can identify that $3240$ is divisible by $10$, or equivalently $2cdot 5$; so we can start with $3240=2cdot 5cdot 324$. Now, by using this divisibility test we can see that $324$ is divisible by $3$ precisely $108$ times. A further inspection sees that $108$ divides by $3$ again. So so far, we have $3240=2cdot 5cdot 3cdot 3cdot 36$. From here, we can swiftly conclude this by realising that $36=6^2=(2cdot 3)^2=2^2cdot 3^2$. Thus, putting everything together gives us $$3240=2cdot 5cdot 3cdot 3cdot 2^2cdot 3^2=2^3cdot3^4cdot 5,$$ as desired.
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
$endgroup$
Straight away we can identify that $3240$ is divisible by $10$, or equivalently $2cdot 5$; so we can start with $3240=2cdot 5cdot 324$. Now, by using this divisibility test we can see that $324$ is divisible by $3$ precisely $108$ times. A further inspection sees that $108$ divides by $3$ again. So so far, we have $3240=2cdot 5cdot 3cdot 3cdot 36$. From here, we can swiftly conclude this by realising that $36=6^2=(2cdot 3)^2=2^2cdot 3^2$. Thus, putting everything together gives us $$3240=2cdot 5cdot 3cdot 3cdot 2^2cdot 3^2=2^3cdot3^4cdot 5,$$ as desired.
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
answered Dec 17 '18 at 18:59
thesmallprintthesmallprint
2,6211618
2,6211618
add a comment |
add a comment |
1
$begingroup$
Hi Oliver, and welcome to MSE. It ends in a $0$ so is a multiple of both $5$ and $2$. The digits add to a multiple of $3$, so it's a multiple of $3$. Now divide by those known factors and see if ther are more easy factors to factor out. I would say "show what you've tried" but I suspect you didn't even get started.
$endgroup$
– user334732
Dec 17 '18 at 19:47