Probability that a divisor of $10^99$ is a multiple of $10^96$
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What is the probability that a divisor of $10^99$ is a multiple of $10^96$? How to solve this type of question. I know probability but I'm weak in number theory.
probability elementary-number-theory
asked May 26 '13 at 6:04
iostream007
3,65131238
add a comment |Â
up vote
3
down vote
favorite
What is the probability that a divisor of $10^99$ is a multiple of $10^96$? How to solve this type of question. I know probability but I'm weak in number theory.
probability elementary-number-theory
asked May 26 '13 at 6:04
iostream007
3,65131238
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
What is the probability that a divisor of $10^99$ is a multiple of $10^96$? How to solve this type of question. I know probability but I'm weak in number theory.
probability elementary-number-theory
asked May 26 '13 at 6:04
iostream007
3,65131238
What is the probability that a divisor of $10^99$ is a multiple of $10^96$? How to solve this type of question. I know probability but I'm weak in number theory.
probability elementary-number-theory
asked May 26 '13 at 6:04
iostream007
3,65131238
asked May 26 '13 at 6:04
iostream007
3,65131238
asked May 26 '13 at 6:04
iostream007
3,65131238
asked May 26 '13 at 6:04
iostream007
3,65131238
3,65131238
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
6
down vote
accepted
HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0le k,mle n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^ellmid 2^k5^m$ if and only if $elllemink,m$.
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
add a comment |Â
up vote
12
down vote
As $10^99=2^99cdot 5^99,$
using divisor function 1,2, the number of divisors is $(1+99)(1+99)$
which are the product of $$1, 2^1,cdots,2^98,2^99text and 1, 5^1,cdots,5^98,5^99$$
To be multiple of $10^96$ we need to take $$ 2^96,2^97,2^98,2^99text and 5^96,5^97,5^98,5^99$$ i.e., there are $4cdot4=16$ of them
So, the required probability is $$frac4^2100^2=frac1625$$
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
add a comment |Â
up vote
5
down vote
$10^99=2^99 times 5^99$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^99$ has $10000$ distinct factors.
To be divisible by $10^96$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $frac 1610000$.
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
add a comment |Â
up vote
0
down vote
Since $10^99 = (2times5)^99 = 2^99 times 5^99$,
Every factor of $10^99$ is of the form $2^ptimes 5^q$
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100times 100 = 10000$ factors of $10^99$
That's the denominator of the desired probability.
Next we calculate the numerator of the probability.
They are the positive integers of the form $2^ptimes5^q$
which are multiples of $10^96$.
So p and q can can each be chosen as $96,97,98,99$
So there are $4times 4$ = 16 factors of $10^99$ which are multiples of $10^96$.
Therefore the desired probability is $16/10000=1/625$.
answered Aug 11 at 18:25
Chanchal
11
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0le k,mle n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^ellmid 2^k5^m$ if and only if $elllemink,m$.
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
add a comment |Â
up vote
6
down vote
accepted
HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0le k,mle n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^ellmid 2^k5^m$ if and only if $elllemink,m$.
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0le k,mle n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^ellmid 2^k5^m$ if and only if $elllemink,m$.
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
HINT: $10^n=2^n5^n$, so the divisors of $10^n$ are the numbers of the form $2^k5^m$, where $0le k,mle n$. Since there are $n+1$ choices for each of $k$ and $m$, $10^n$ has $(n+1)^2$ divisors. And $10^ellmid 2^k5^m$ if and only if $elllemink,m$.
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
answered May 26 '13 at 6:07
Brian M. Scott
448k39492880
448k39492880
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
add a comment |Â
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
$"10^n has (N+1)^2"$ divisors.can you explain this line?
– iostream007
Jun 14 '13 at 9:45
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
@iostream007: The Since clause of the sentence is the explanation. There are $n+1$ possible choices for $k$, namely, the $n+1$ integers from $0$ through $n$, and similarly there are $n+1$ possible choices for $m$. The choices are independent of each other: you can combine any of the $n+1$ possible values of $k$ with any of the $n+1$ possible values of $m$. Thus, there are $(n+1)cdot(n+1)=(n+1)^2$ possible combinations. Each of them yields a divisor of $10^n$, and by the fundamental theorem of arithmetic every divisor of $10^n$ must be of this form, so $10^n$ has $(n+1)^2$ divisors.
– Brian M. Scott
Jun 14 '13 at 9:48
add a comment |Â
up vote
12
down vote
As $10^99=2^99cdot 5^99,$
using divisor function 1,2, the number of divisors is $(1+99)(1+99)$
which are the product of $$1, 2^1,cdots,2^98,2^99text and 1, 5^1,cdots,5^98,5^99$$
To be multiple of $10^96$ we need to take $$ 2^96,2^97,2^98,2^99text and 5^96,5^97,5^98,5^99$$ i.e., there are $4cdot4=16$ of them
So, the required probability is $$frac4^2100^2=frac1625$$
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
add a comment |Â
up vote
12
down vote
As $10^99=2^99cdot 5^99,$
using divisor function 1,2, the number of divisors is $(1+99)(1+99)$
which are the product of $$1, 2^1,cdots,2^98,2^99text and 1, 5^1,cdots,5^98,5^99$$
To be multiple of $10^96$ we need to take $$ 2^96,2^97,2^98,2^99text and 5^96,5^97,5^98,5^99$$ i.e., there are $4cdot4=16$ of them
So, the required probability is $$frac4^2100^2=frac1625$$
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
add a comment |Â
up vote
12
down vote
up vote
12
down vote
As $10^99=2^99cdot 5^99,$
using divisor function 1,2, the number of divisors is $(1+99)(1+99)$
which are the product of $$1, 2^1,cdots,2^98,2^99text and 1, 5^1,cdots,5^98,5^99$$
To be multiple of $10^96$ we need to take $$ 2^96,2^97,2^98,2^99text and 5^96,5^97,5^98,5^99$$ i.e., there are $4cdot4=16$ of them
So, the required probability is $$frac4^2100^2=frac1625$$
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
As $10^99=2^99cdot 5^99,$
using divisor function 1,2, the number of divisors is $(1+99)(1+99)$
which are the product of $$1, 2^1,cdots,2^98,2^99text and 1, 5^1,cdots,5^98,5^99$$
To be multiple of $10^96$ we need to take $$ 2^96,2^97,2^98,2^99text and 5^96,5^97,5^98,5^99$$ i.e., there are $4cdot4=16$ of them
So, the required probability is $$frac4^2100^2=frac1625$$
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
edited May 26 '13 at 6:13
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
answered May 26 '13 at 6:08
lab bhattacharjee
215k14152264
215k14152264
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
add a comment |Â
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
The question says that the divisor of 10^99 should be a multiple of 10^96 which means that the divisor must be divisible by 10^96 ,so how can u say that for the number to be a multiple of 10^96 we need to take numbers from 2^96 , I just took an example , like 2^4 =16 and 10^4 =10000, Now here 10^4 is a multiple of 2^4 ,not the other way round , since 2^4 divides 10^4 ,so then how can 10^96 divide 2^96 ?
– radhika
Jan 31 '16 at 10:22
add a comment |Â
up vote
5
down vote
$10^99=2^99 times 5^99$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^99$ has $10000$ distinct factors.
To be divisible by $10^96$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $frac 1610000$.
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
add a comment |Â
up vote
5
down vote
$10^99=2^99 times 5^99$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^99$ has $10000$ distinct factors.
To be divisible by $10^96$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $frac 1610000$.
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$10^99=2^99 times 5^99$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^99$ has $10000$ distinct factors.
To be divisible by $10^96$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $frac 1610000$.
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
$10^99=2^99 times 5^99$
There are $100$ ways to choose the exponent for $2$ and $100$ ways to choose the exponent for $5$, so that $10^99$ has $10000$ distinct factors.
To be divisible by $10^96$ both exponents must be at least $96$, which leaves the four cases $96, 97, 98, 99$ for each exponent. This amounts to $4 times 4= 16$ cases.
Assuming the distribution is such that any factor is equally likely to be chosen, this gives $frac 1610000$.
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
answered May 26 '13 at 6:10
Mark Bennet
76.8k773171
76.8k773171
add a comment |Â
add a comment |Â
up vote
0
down vote
Since $10^99 = (2times5)^99 = 2^99 times 5^99$,
Every factor of $10^99$ is of the form $2^ptimes 5^q$
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100times 100 = 10000$ factors of $10^99$
That's the denominator of the desired probability.
Next we calculate the numerator of the probability.
They are the positive integers of the form $2^ptimes5^q$
which are multiples of $10^96$.
So p and q can can each be chosen as $96,97,98,99$
So there are $4times 4$ = 16 factors of $10^99$ which are multiples of $10^96$.
Therefore the desired probability is $16/10000=1/625$.
answered Aug 11 at 18:25
Chanchal
11
add a comment |Â
up vote
0
down vote
Since $10^99 = (2times5)^99 = 2^99 times 5^99$,
Every factor of $10^99$ is of the form $2^ptimes 5^q$
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100times 100 = 10000$ factors of $10^99$
That's the denominator of the desired probability.
Next we calculate the numerator of the probability.
They are the positive integers of the form $2^ptimes5^q$
which are multiples of $10^96$.
So p and q can can each be chosen as $96,97,98,99$
So there are $4times 4$ = 16 factors of $10^99$ which are multiples of $10^96$.
Therefore the desired probability is $16/10000=1/625$.
answered Aug 11 at 18:25
Chanchal
11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Since $10^99 = (2times5)^99 = 2^99 times 5^99$,
Every factor of $10^99$ is of the form $2^ptimes 5^q$
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100times 100 = 10000$ factors of $10^99$
That's the denominator of the desired probability.
Next we calculate the numerator of the probability.
They are the positive integers of the form $2^ptimes5^q$
which are multiples of $10^96$.
So p and q can can each be chosen as $96,97,98,99$
So there are $4times 4$ = 16 factors of $10^99$ which are multiples of $10^96$.
Therefore the desired probability is $16/10000=1/625$.
answered Aug 11 at 18:25
Chanchal
11
Since $10^99 = (2times5)^99 = 2^99 times 5^99$,
Every factor of $10^99$ is of the form $2^ptimes 5^q$
p and q can each be $0,1,2,...,99$ which is $100$ choices each.
So there are $100times 100 = 10000$ factors of $10^99$
That's the denominator of the desired probability.
Next we calculate the numerator of the probability.
They are the positive integers of the form $2^ptimes5^q$
which are multiples of $10^96$.
So p and q can can each be chosen as $96,97,98,99$
So there are $4times 4$ = 16 factors of $10^99$ which are multiples of $10^96$.
Therefore the desired probability is $16/10000=1/625$.
answered Aug 11 at 18:25
Chanchal
11
edited Aug 11 at 18:49
answered Aug 11 at 18:25
Chanchal
11
answered Aug 11 at 18:25
Chanchal
11
answered Aug 11 at 18:25
Chanchal
11
11
add a comment |Â
add a comment |Â
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