Mixing
How many positive divisors of 2000?
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$$2000=2^4 cdot 5^3$$
meaning there $4 cdot 3=12$ divisors. Plus $1$ and $2000$ are divisors, which give a total number of (distinct) divisors.
But the answer for some reason is 20.
combinatorics divisibility
edited 3 hours ago
amWhy
191k27223436
asked 3 hours ago
ProgC
1015
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up vote
1
down vote
favorite
$$2000=2^4 cdot 5^3$$
meaning there $4 cdot 3=12$ divisors. Plus $1$ and $2000$ are divisors, which give a total number of (distinct) divisors.
But the answer for some reason is 20.
combinatorics divisibility
edited 3 hours ago
amWhy
191k27223436
asked 3 hours ago
ProgC
1015
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$2000=2^4 cdot 5^3$$
meaning there $4 cdot 3=12$ divisors. Plus $1$ and $2000$ are divisors, which give a total number of (distinct) divisors.
But the answer for some reason is 20.
combinatorics divisibility
edited 3 hours ago
amWhy
191k27223436
asked 3 hours ago
ProgC
1015
$$2000=2^4 cdot 5^3$$
meaning there $4 cdot 3=12$ divisors. Plus $1$ and $2000$ are divisors, which give a total number of (distinct) divisors.
But the answer for some reason is 20.
combinatorics divisibility
combinatorics divisibility
edited 3 hours ago
amWhy
191k27223436
asked 3 hours ago
ProgC
1015
edited 3 hours ago
amWhy
191k27223436
asked 3 hours ago
ProgC
1015
edited 3 hours ago
amWhy
191k27223436
edited 3 hours ago
amWhy
191k27223436
edited 3 hours ago
amWhy
191k27223436
191k27223436
asked 3 hours ago
ProgC
1015
asked 3 hours ago
ProgC
1015
asked 3 hours ago
ProgC
1015
1015
add a comment |Â
add a comment |Â
3 Answers
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The factors are of the form of $2^xcdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0$.
answered 3 hours ago
Siong Thye Goh
87.1k1459108
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up vote
2
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You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
answered 3 hours ago
Yves Daoust
118k667214
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up vote
2
down vote
A divisor of $2^4,5^3$ has the prime decomposition $2^i,5^j$, where $;0le ile 4$ ($5$ possibilities) and $0le jle3$ ($4$ possibilities). Combining these possibilities, this make $5cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
answered 3 hours ago
Bernard
114k636106
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The factors are of the form of $2^xcdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0$.
answered 3 hours ago
Siong Thye Goh
87.1k1459108
add a comment |Â
up vote
2
down vote
The factors are of the form of $2^xcdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0$.
answered 3 hours ago
Siong Thye Goh
87.1k1459108
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The factors are of the form of $2^xcdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0$.
answered 3 hours ago
Siong Thye Goh
87.1k1459108
The factors are of the form of $2^xcdot 5^y$ where $x$ takes value from $0$ to $4$ and $y$ takes values from $0$ to $3$.
Hence there is a total of $(4+1)(3+1)=20$ factors.
Your method miss out number such as $5^y, y>0$.
answered 3 hours ago
Siong Thye Goh
87.1k1459108
answered 3 hours ago
Siong Thye Goh
87.1k1459108
answered 3 hours ago
Siong Thye Goh
87.1k1459108
answered 3 hours ago
Siong Thye Goh
87.1k1459108
87.1k1459108
add a comment |Â
add a comment |Â
up vote
2
down vote
You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
answered 3 hours ago
Yves Daoust
118k667214
add a comment |Â
up vote
2
down vote
You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
answered 3 hours ago
Yves Daoust
118k667214
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
answered 3 hours ago
Yves Daoust
118k667214
You take $2$ to any power between $0$ and $4$, and $3$ with any power between $0$ and $3$, hence $5times4$ combinations.
You forgot the options such that one exponent is zero, $2,4,8,16$ and $5,25,125$, and you counted $2^4cdot5^3=2000$ twice. Hence, $14-1+4+3=20$.
answered 3 hours ago
Yves Daoust
118k667214
answered 3 hours ago
Yves Daoust
118k667214
answered 3 hours ago
Yves Daoust
118k667214
answered 3 hours ago
Yves Daoust
118k667214
118k667214
add a comment |Â
add a comment |Â
up vote
2
down vote
A divisor of $2^4,5^3$ has the prime decomposition $2^i,5^j$, where $;0le ile 4$ ($5$ possibilities) and $0le jle3$ ($4$ possibilities). Combining these possibilities, this make $5cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
answered 3 hours ago
Bernard
114k636106
add a comment |Â
up vote
2
down vote
A divisor of $2^4,5^3$ has the prime decomposition $2^i,5^j$, where $;0le ile 4$ ($5$ possibilities) and $0le jle3$ ($4$ possibilities). Combining these possibilities, this make $5cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
answered 3 hours ago
Bernard
114k636106
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A divisor of $2^4,5^3$ has the prime decomposition $2^i,5^j$, where $;0le ile 4$ ($5$ possibilities) and $0le jle3$ ($4$ possibilities). Combining these possibilities, this make $5cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
answered 3 hours ago
Bernard
114k636106
A divisor of $2^4,5^3$ has the prime decomposition $2^i,5^j$, where $;0le ile 4$ ($5$ possibilities) and $0le jle3$ ($4$ possibilities). Combining these possibilities, this make $5cdot 4$ divisors in all.
More generally, given the prime decomposition of a number $n$, the number of its positive divisors is the product of the exponents plus $1$ of its prime factors.
answered 3 hours ago
Bernard
114k636106
answered 3 hours ago
Bernard
114k636106
answered 3 hours ago
Bernard
114k636106
answered 3 hours ago
Bernard
114k636106
114k636106
add a comment |Â
add a comment |Â
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