Xjyuk

Ambiguous sentences: How to tell when they need fixing?

Rearranging the formula

QGIS Linestring rendering curves between vertex

Does Google Maps take into account hills/inclines for route times?

Interpreting the word "randomly"

Did any of the founding fathers anticipate Lysander Spooner's criticism of the constitution?

Occasus nescius

Draw 3D Cubes around centre

How can an advanced civilization forget how to manufacture its technology?

School House Points (Python + SQLite)

How to check the quality of an audio sample?

Filtering fine silt/mud from water (not necessarily bacteria etc.)

What's the point of this scene involving Flash Thompson at the airport?

Why does the trade federation become so alarmed upon learning the ambassadors are Jedi Knights?

Mbed Cortex-m hardfault when sending data via TCP

nginx serves wrong domain site. It doenst shows default site if no configuration applies

Players of unusual orchestral instruments

Modeling, view and projection transformation using vector and point in homogenous form

If the derivative of a function is square of it then it is constant

How many matrices satisfy this equality?

How to convert 1k to 1000 and 1m to 1000000 in Excel

Are lithium batteries allowed in the International Space Station?

How is the idea of "X comes a distant third" commonly expressed in Russian?

As a DM, how to avoid unconscious metagaming when dealing with a high AC character?

Find the wrong number in the given series: 6, 12, 21, 36, 56, 81?

Guess the Function for Scatterplot of Number Theoretic FunctionOh no! Yet another number series… - Find the pattern in the two examples and provide an extensionWhen setting number-sequence puzzles, is there any accepted method to prevent arbitrary solutions?Hidden Number SequenceWhat is a RAP Number™?What is a Frightful Number™?I need help with this number sequenceWhat is a Well-toned Number™?Which number replaces the question mark in the circle?speciality of the binary sequence $underline100101100110100101101001$

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

$begingroup$

The following problem was given to me by one of my friend:

Find the wrong number in the given series:
$$6,, 12,, 21,, 36,, 56,, 81 $$
?

A) $21,$ B) $12,$ C) $36,$ D) $56$

and told me that its answer is B) $12$

but according to me answer should be D) $56$ ; as all the other numbers in the series is exactly divisible by $3$

So, is my answer right ? or there is more reasonable approach for option B) $12$

Any hints/suggestions or help please...

pattern number-sequence

share|improve this question

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

The following problem was given to me by one of my friend:

Find the wrong number in the given series:
$$6,, 12,, 21,, 36,, 56,, 81 $$
?

A) $21,$ B) $12,$ C) $36,$ D) $56$

and told me that its answer is B) $12$

but according to me answer should be D) $56$ ; as all the other numbers in the series is exactly divisible by $3$

So, is my answer right ? or there is more reasonable approach for option B) $12$

Any hints/suggestions or help please...

pattern number-sequence

share|improve this question

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

$endgroup$

add a comment | 

$begingroup$

The following problem was given to me by one of my friend:

Find the wrong number in the given series:
$$6,, 12,, 21,, 36,, 56,, 81 $$
?

A) $21,$ B) $12,$ C) $36,$ D) $56$

and told me that its answer is B) $12$

but according to me answer should be D) $56$ ; as all the other numbers in the series is exactly divisible by $3$

So, is my answer right ? or there is more reasonable approach for option B) $12$

Any hints/suggestions or help please...

pattern number-sequence

share|improve this question

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

$endgroup$

share|improve this question

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

share|improve this question

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

edited 13 hours ago

JonMark Perry

23.9k66 gold badges4646 silver badges106106 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

asked 13 hours ago

SureshSuresh

21577 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

The series should be:

$6, 11, 21, 36, 56, 81$

Because:

The differences will be $5, 10, 15, 20, 25$.

So the answer is:

$B)~12$

Note that:

This is given as a series, not a set, therefore it's wise to treat it as it should be.

share|improve this answer

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

My guess is:

A reasonable approach, following your solution is that all those numbers minus $1$:
$5$, $20$, $35$, $55$, $80$

are divisible by $5$.
$12 - 1 = 11$, isn't.

Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)

share|improve this answer

edited 12 hours ago

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to Occam's Razor?
  $endgroup$
  – M Oehm
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MOehm haha, good point, that's the simpliest I've found for now, but I'm still searching :) Editing my question to take it in account ;)
  $endgroup$
  – JKHA
  12 hours ago
  
  

  
  
  
  

add a comment | 

Your Answer

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
);



);

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%2fpuzzling.stackexchange.com%2fquestions%2f86130%2ffind-the-wrong-number-in-the-given-series-6-12-21-36-56-81%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

The series should be:

$6, 11, 21, 36, 56, 81$

Because:

The differences will be $5, 10, 15, 20, 25$.

So the answer is:

$B)~12$

Note that:

This is given as a series, not a set, therefore it's wise to treat it as it should be.

share|improve this answer

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

$endgroup$

add a comment | 

5

$begingroup$

The series should be:

$6, 11, 21, 36, 56, 81$

Because:

The differences will be $5, 10, 15, 20, 25$.

So the answer is:

$B)~12$

Note that:

This is given as a series, not a set, therefore it's wise to treat it as it should be.

share|improve this answer

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

$endgroup$

add a comment | 

$begingroup$

The series should be:

$6, 11, 21, 36, 56, 81$

Because:

The differences will be $5, 10, 15, 20, 25$.

So the answer is:

$B)~12$

Note that:

This is given as a series, not a set, therefore it's wise to treat it as it should be.

share|improve this answer

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

$endgroup$

share|improve this answer

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

answered 11 hours ago

athinathin

11.9k22 gold badges3939 silver badges9797 bronze badges

2

$begingroup$

My guess is:

A reasonable approach, following your solution is that all those numbers minus $1$:
$5$, $20$, $35$, $55$, $80$

are divisible by $5$.
$12 - 1 = 11$, isn't.

Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)

share|improve this answer

edited 12 hours ago

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to Occam's Razor?
  $endgroup$
  – M Oehm
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MOehm haha, good point, that's the simpliest I've found for now, but I'm still searching :) Editing my question to take it in account ;)
  $endgroup$
  – JKHA
  12 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

My guess is:

A reasonable approach, following your solution is that all those numbers minus $1$:
$5$, $20$, $35$, $55$, $80$

are divisible by $5$.
$12 - 1 = 11$, isn't.

Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)

share|improve this answer

edited 12 hours ago

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What happened to Occam's Razor?
  $endgroup$
  – M Oehm
  12 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MOehm haha, good point, that's the simpliest I've found for now, but I'm still searching :) Editing my question to take it in account ;)
  $endgroup$
  – JKHA
  12 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

My guess is:

A reasonable approach, following your solution is that all those numbers minus $1$:
$5$, $20$, $35$, $55$, $80$

are divisible by $5$.
$12 - 1 = 11$, isn't.

Nevertheless, I don't know where this puzzle comes from but that won't be surprising you are right and your friend may have made a mistake... Unless PSE community finds a simplier solution :)

share|improve this answer

edited 12 hours ago

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

$endgroup$

share|improve this answer

edited 12 hours ago

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges

answered 12 hours ago

JKHAJKHA

74533 silver badges1313 bronze badges