Mixing
The last digit of $n!$ for $n ge 5$ is always $0$. What are the options for the last non-zero digit of $n!,...
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I've found formulas online that use the greatest integer function, but they seem to answer my question for specific values of $n$. Is there an easier approach to find all values the last non-zero digit of a random $n$ can take? Is there another way to find these values (so not necessarily using the formulas with $leftlfloorcdotsrightrfloor$)?
modular-arithmetic
asked Jan 28 at 17:31
ZacharyZachary
1939
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add a comment |
$begingroup$
I've found formulas online that use the greatest integer function, but they seem to answer my question for specific values of $n$. Is there an easier approach to find all values the last non-zero digit of a random $n$ can take? Is there another way to find these values (so not necessarily using the formulas with $leftlfloorcdotsrightrfloor$)?
modular-arithmetic
asked Jan 28 at 17:31
ZacharyZachary
1939
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1
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
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– vadim123
Jan 28 at 17:33
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Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
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– gandalf61
Jan 28 at 17:39
1
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This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03
add a comment |
$begingroup$
I've found formulas online that use the greatest integer function, but they seem to answer my question for specific values of $n$. Is there an easier approach to find all values the last non-zero digit of a random $n$ can take? Is there another way to find these values (so not necessarily using the formulas with $leftlfloorcdotsrightrfloor$)?
modular-arithmetic
asked Jan 28 at 17:31
ZacharyZachary
1939
$endgroup$
I've found formulas online that use the greatest integer function, but they seem to answer my question for specific values of $n$. Is there an easier approach to find all values the last non-zero digit of a random $n$ can take? Is there another way to find these values (so not necessarily using the formulas with $leftlfloorcdotsrightrfloor$)?
modular-arithmetic
modular-arithmetic
asked Jan 28 at 17:31
ZacharyZachary
1939
asked Jan 28 at 17:31
ZacharyZachary
1939
asked Jan 28 at 17:31
ZacharyZachary
1939
asked Jan 28 at 17:31
ZacharyZachary
1939
asked Jan 28 at 17:31
ZacharyZachary
1939
1939
1
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
$endgroup$
– vadim123
Jan 28 at 17:33
$begingroup$
Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
$endgroup$
– gandalf61
Jan 28 at 17:39
1
$begingroup$
This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03
add a comment |
1
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
$endgroup$
– vadim123
Jan 28 at 17:33
$begingroup$
Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
$endgroup$
– gandalf61
Jan 28 at 17:39
1
$begingroup$
This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03
1
1
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
$endgroup$
– vadim123
Jan 28 at 17:33
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
$endgroup$
– vadim123
Jan 28 at 17:33
$begingroup$
Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
$endgroup$
– gandalf61
Jan 28 at 17:39
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
$endgroup$
– gandalf61
Jan 28 at 17:39
1
1
$begingroup$
This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03
$begingroup$
This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03
add a comment |
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1
$begingroup$
geeksforgeeks.org/last-non-zero-digit-factorial
$endgroup$
– vadim123
Jan 28 at 17:33
$begingroup$
Note that you get $k$ additional zeros every $a5^k$ terms where $a in mathbb Z^+$.
$endgroup$
– Mohammad Zuhair Khan
Jan 28 at 17:37
$begingroup$
For $n>1$ the last non-zero digit of $n!$ has to be even since the exponent of $2$ in the prime expansion of $n!$ is greater than the exponent of $5$ (i.e. in ${1,2,dots,n}$ there are more even numbers than multiples of $5$).
$endgroup$
– gandalf61
Jan 28 at 17:39
1
$begingroup$
This is A008904.
$endgroup$
– lulu
Jan 28 at 18:03