Ggthjy

Early credit roll before the end of the film

Can an insurance company drop you after receiving a bill and refusing to pay?

Advice for a new journal editor

Why isn't there a non-conducting core wire for high-frequency coil applications

Why has the mole been redefined for 2019?

What is the wife of a henpecked husband called?

Does SQL Server 2017, including older versions, support 8k disk sector sizes?

Porting Linux to another platform requirements

How can my powered armor quickly replace its ceramic plates?

Word or phrase for showing great skill at something WITHOUT formal training in it

Can we use the stored gravitational potential energy of a building to produce power?

Why avoid shared user accounts?

How do you funnel food off a cutting board?

If I delete my router's history can my ISP still provide it to my parents?

How to count the characters of jar files by wc

Can I write a book of my D&D game?

Can I become debt free or should I file bankruptcy ? How to manage my debt and finances?

Why is mind meld hard for T'pol in Star Trek: Enterprise?

Avoiding morning and evening handshakes

How would an AI self awareness kill switch work?

Are there any modern advantages of a fire piston?

Could a phylactery of a lich be a mirror or does it have to be a box?

Why would the Pakistan airspace closure cancel flights not headed to Pakistan itself?

Table formatting top left corner caption

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

17

$begingroup$

$17!$ 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

share|cite|improve this question

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 | 

17

$begingroup$

$17!$ 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

share|cite|improve this question

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

share|cite|improve this question

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.

$endgroup$

share|cite|improve this question

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.

share|cite|improve this question

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

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

2 Answers
2

active

oldest

votes

27

$begingroup$

HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?

share|cite|improve this answer

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 | 

15

$begingroup$

The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.

share|cite|improve this answer

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 | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

a_man_with_no_name is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3130579%2ffind-some-digits-of-factorial-17%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

27

$begingroup$

HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?

share|cite|improve this answer

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 | 

27

$begingroup$

HINT $17!$ is divisible by $9$. What is an easy test for divisibility by 9, involving the digits of a number?

share|cite|improve this answer

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?

share|cite|improve this answer

answered 10 hours ago

Mark FischlerMark Fischler

32.9k12452

$endgroup$

share|cite|improve this answer

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

15

$begingroup$

The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.

share|cite|improve this answer

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 | 

15

$begingroup$

The alternating sum of digits must be divisible by $11$, i.e., $11mid 18-x$. It follows that $x=7$.

share|cite|improve this answer

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 $x=7$.

share|cite|improve this answer

answered 10 hours ago

Dietrich BurdeDietrich Burde

80.1k647104

$endgroup$

share|cite|improve this answer

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