Using only 1s, make 29 with the minimum number of digits
$begingroup$
Operations permitted:
- Standard operations: +, −, ×, ÷
- Negation: −
- Exponentiation of two numbers: x^y
- Square root of a number: √
- Factorial: !
- Concatenation of the original digits: dd
mathematics calculation-puzzle formation-of-numbers
New contributor
$endgroup$
add a comment |
$begingroup$
Operations permitted:
- Standard operations: +, −, ×, ÷
- Negation: −
- Exponentiation of two numbers: x^y
- Square root of a number: √
- Factorial: !
- Concatenation of the original digits: dd
mathematics calculation-puzzle formation-of-numbers
New contributor
$endgroup$
add a comment |
$begingroup$
Operations permitted:
- Standard operations: +, −, ×, ÷
- Negation: −
- Exponentiation of two numbers: x^y
- Square root of a number: √
- Factorial: !
- Concatenation of the original digits: dd
mathematics calculation-puzzle formation-of-numbers
New contributor
$endgroup$
Operations permitted:
- Standard operations: +, −, ×, ÷
- Negation: −
- Exponentiation of two numbers: x^y
- Square root of a number: √
- Factorial: !
- Concatenation of the original digits: dd
mathematics calculation-puzzle formation-of-numbers
mathematics calculation-puzzle formation-of-numbers
New contributor
New contributor
New contributor
asked 46 mins ago
Allan CaoAllan Cao
1063
1063
New contributor
New contributor
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a 7 digits solution:
7 digits: (11-1)x(1+1+1)-1
$endgroup$
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
add a comment |
$begingroup$
Lowest I managed so far is 9 digits:
(1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1
11*(1 + 1 + 1) - (1 + 1 + 1 + 1)
Some other ways I came up with:
(1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)
11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)
$endgroup$
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: "559"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
});
}
});
Allan Cao is a new contributor. Be nice, and check out our Code of Conduct.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80050%2fusing-only-1s-make-29-with-the-minimum-number-of-digits%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a 7 digits solution:
7 digits: (11-1)x(1+1+1)-1
$endgroup$
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
add a comment |
$begingroup$
Here's a 7 digits solution:
7 digits: (11-1)x(1+1+1)-1
$endgroup$
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
add a comment |
$begingroup$
Here's a 7 digits solution:
7 digits: (11-1)x(1+1+1)-1
$endgroup$
Here's a 7 digits solution:
7 digits: (11-1)x(1+1+1)-1
answered 23 mins ago
Dr XorileDr Xorile
12.9k22569
12.9k22569
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
add a comment |
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
That is the minimum I achieved by referencing the Single Digit Representations of Natural Numbers paper. Hopefully 6 is possible.
$endgroup$
– Allan Cao
13 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
Do you mean that allowing concatenations should reduce it from 7 to 6? Or are the constraints the same in the paper you cite as in the question above?
$endgroup$
– Dr Xorile
10 mins ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
$begingroup$
The paper uses different rules.
$endgroup$
– Allan Cao
13 secs ago
add a comment |
$begingroup$
Lowest I managed so far is 9 digits:
(1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1
11*(1 + 1 + 1) - (1 + 1 + 1 + 1)
Some other ways I came up with:
(1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)
11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)
$endgroup$
add a comment |
$begingroup$
Lowest I managed so far is 9 digits:
(1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1
11*(1 + 1 + 1) - (1 + 1 + 1 + 1)
Some other ways I came up with:
(1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)
11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)
$endgroup$
add a comment |
$begingroup$
Lowest I managed so far is 9 digits:
(1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1
11*(1 + 1 + 1) - (1 + 1 + 1 + 1)
Some other ways I came up with:
(1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)
11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)
$endgroup$
Lowest I managed so far is 9 digits:
(1 + 1 + 1 + 1)! + 1 + 1 + 1 + 1 + 1
11*(1 + 1 + 1) - (1 + 1 + 1 + 1)
Some other ways I came up with:
(1 + 1)^(1 + 1 + 1 + 1 + 1) - 1 - 1 - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)!/(1 + 1 + 1 + 1) - 1 (10 digits)
(1 + 1 + 1 + 1 + 1)^(1 + 1) + 1 + 1 + 1 + 1 (11 digits)
11*(1 + 1) + 1 + 1 + 1 + 1 + 1 + 1 + 1 (11 digits)
edited 25 mins ago
answered 33 mins ago
simonzacksimonzack
267110
267110
add a comment |
add a comment |
Allan Cao is a new contributor. Be nice, and check out our Code of Conduct.
Allan Cao is a new contributor. Be nice, and check out our Code of Conduct.
Allan Cao is a new contributor. Be nice, and check out our Code of Conduct.
Allan Cao is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Puzzling Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f80050%2fusing-only-1s-make-29-with-the-minimum-number-of-digits%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown