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 1 hour ago
J. W. Tanner
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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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 1 hour ago
J. W. Tanner
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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 1 hour ago
J. W. Tanner
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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 1 hour ago
J. W. Tanner
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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 1 hour ago
J. W. Tanner
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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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edited 1 hour ago
J. W. Tanner
2,6271217
edited 1 hour ago
J. W. Tanner
2,6271217
edited 1 hour ago
J. W. Tanner
2,6271217
2,6271217
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
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 2 hours ago
a_man_with_no_namea_man_with_no_name
573
asked 2 hours ago
a_man_with_no_namea_man_with_no_name
573
573
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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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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 2 hours ago
Mark FischlerMark Fischler
32.7k12251
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
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 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
$endgroup$
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 2 hours ago
Mark FischlerMark Fischler
32.7k12251
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
1 hour 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 2 hours ago
Mark FischlerMark Fischler
32.7k12251
$endgroup$
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
1 hour 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 2 hours ago
Mark FischlerMark Fischler
32.7k12251
$endgroup$
HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?
answered 2 hours ago
Mark FischlerMark Fischler
32.7k12251
answered 2 hours ago
Mark FischlerMark Fischler
32.7k12251
answered 2 hours ago
Mark FischlerMark Fischler
32.7k12251
answered 2 hours ago
Mark FischlerMark Fischler
32.7k12251
32.7k12251
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
1 hour ago
add a comment |
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
The sum od its digits Has to be divisible by 9
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
1 hour ago
$begingroup$
How could i miss it!
$endgroup$
– a_man_with_no_name
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 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
$endgroup$
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 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
$endgroup$
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 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
$endgroup$
The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.
answered 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
answered 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
answered 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
answered 1 hour ago
Dietrich BurdeDietrich Burde
80k647103
80k647103
add a comment |
add a comment |
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