Find some digits of factorial 17Factorial number of digitsTwo questions on finding trailing digits in (large)...
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Find some digits of factorial 17
Factorial number of digitsTwo questions on finding trailing digits in (large) numbers and one on divisibilityFirst decimal digits of factorial $n$ divided by $x$Factorial related problemsLast digits of factorialCounting zeros in a factorial(terminal + zeros in between digits)Find last 5 significant digits of 2017!Number of digits in a factorial sum $1!+2!+cdots+100!$Number of digits of N factorial in base BLast two digits of odd products
$begingroup$
$17!$ is equal to $$35568x428096y00$$
Both x and y, are digits. Find x,y.
So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
If there`s a product of $(2*5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
elementary-number-theory factorial
edited 10 hours ago
J. W. Tanner
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
$17!$ is equal to $$35568x428096y00$$
Both x and y, are digits. Find x,y.
So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
If there`s a product of $(2*5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
elementary-number-theory factorial
edited 10 hours ago
J. W. Tanner
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
$17!$ is equal to $$35568x428096y00$$
Both x and y, are digits. Find x,y.
So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
If there`s a product of $(2*5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
elementary-number-theory factorial
edited 10 hours ago
J. W. Tanner
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$17!$ is equal to $$35568x428096y00$$
Both x and y, are digits. Find x,y.
So, $$17!=2^{15}*3^6*5^3*7^2*11*13*17=(2^3*5^3)*2^{12}*3^6*7^2*11*13*17$$
If there`s a product of $(2*5)^3$
Then this number has $3$ zeros at the end, so $y=0$
How do I find the $x$ now?
elementary-number-theory factorial
elementary-number-theory factorial
edited 10 hours ago
J. W. Tanner
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
J. W. Tanner
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 10 hours ago
J. W. Tanner
2,6121217
edited 10 hours ago
J. W. Tanner
2,6121217
edited 10 hours ago
J. W. Tanner
2,6121217
2,6121217
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
asked 11 hours ago
a_man_with_no_namea_man_with_no_name
1324
1324
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
a_man_with_no_name is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
add a comment |
$begingroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
$endgroup$
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
add a comment |
$begingroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
add a comment |
$begingroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
$endgroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
answered 10 hours ago
Mark FischlerMark Fischler
32.9k12452
32.9k12452
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
add a comment |
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
10 hours ago
add a comment |
$begingroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
$endgroup$
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
add a comment |
$begingroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
$endgroup$
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
add a comment |
$begingroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
$endgroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
answered 10 hours ago
Dietrich BurdeDietrich Burde
80.1k647104
80.1k647104
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
add a comment |
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
$begingroup$
If the answer was $0$ or $9$, the other method (divisibility by $9$) would not be enough. This method guarantees unambiguity by itself. On the other hand alternating sum requires little more attention and discipline.
$endgroup$
– Kamil Maciorowski
1 hour ago
add a comment |
a_man_with_no_name is a new contributor. Be nice, and check out our Code of Conduct.
a_man_with_no_name is a new contributor. Be nice, and check out our Code of Conduct.
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