3n+1 problem

The ProblemConsider the following algorithm: 1. input n 2. print n 3. if n = 1 then STOP 4. if n is odd then  n = 3*n+1 5. else    n = n/2 6. GOTO 2 Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified,...

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jelly beans

Problem: !-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:"Arial Unicode MS"; panose-1:2 11 6 4 2 2 2 2 2 4; mso-font-charset:128; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-134238209 -371195905 63 0 4129279 0;} @font-face {font-family:"\@Arial Unicode MS"; panose-1:2...

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Bad king

Problema bad king has a cellar of 1000 bottles of delightful and very expensive wine. a neighbouring queen plots to kill the bad king and sends a servant to poison the wine. (un)fortunately the bad king's guards catch the servant after he has only poisoned one bottle. alas, the guards don't know which bottle but know that the poison is so strong that even if diluted 1,000,000 times it would still kill the king. furthermore, it takes one month to have an effect. the bad king decides he will get some of the prisoners in his vast dungeons to drink...

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Pirates

Problemsfive pirates have 100 gold coins. they have to divide up the loot. in order of seniority (suppose pirate 5 is most senior, pirate 1 is least senior), the most senior pirate proposes a distribution of the loot. they vote and if at least 50% accept the proposal, the loot is divided as proposed. otherwise the most senior pirate is executed, and they start over again with the next senior pirate. what solution does the most senior pirate propose? assume they are very intelligent and extremely greedy (and that they would prefer not to die).solution:...

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sum it up

!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:"Arial Unicode MS"; panose-1:2 11 6 4 2 2 2 2 2 4; mso-font-charset:128; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-134238209 -371195905 63 0 4129279 0;} @font-face {font-family:"\@Arial Unicode MS"; panose-1:2...

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Palindrome years

Problem: This year on October 2, 2001, the date in MMDDYYYY format will be a palindrome (same forwards as backwards). 10/02/2001 when was the last date that this occurred on? (see if you can do it in your head!) Solution: we know the year has to be less than 2001 since we already have the palindrome for 10/02. it can't be any year in 1900 because that would result in a day of 91. same for 1800 down to 1400. it could be a year in 1300 because that would be the 31st day. so whats the latest year in 1300 that would make a month? at first i thought...

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daughter's ages

!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:"Arial Unicode MS"; panose-1:2 11 6 4 2 2 2 2 2 4; mso-font-charset:128; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-134238209 -371195905 63 0 4129279 0;} @font-face {font-family:"\@Arial Unicode MS"; panose-1:2...

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bumblebee

!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:"Arial Unicode MS"; panose-1:2 11 6 4 2 2 2 2 2 4; mso-font-charset:128; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-134238209 -371195905 63 0 4129279 0;} @font-face {font-family:"\@Arial Unicode MS"; panose-1:2...

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Red marbles, blue marbles

Problem You have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.   Solution Chance! chance is easy if you know how...

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100 doors in a row

Problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door. question: what state are the doors in after the last pass? which are open which are closed? ...

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Reverse a string - word by word

!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:"Arial Unicode MS"; panose-1:2 11 6 4 2 2 2 2 2 4; mso-font-charset:128; mso-generic-font-family:swiss; mso-font-pitch:variable; mso-font-signature:-134238209 -371195905 63 0 4129279 0;} @font-face {font-family:"\@Arial Unicode MS"; panose-1:2...

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17 September Solutions

Answers1) b2) c3) b4) a5) d6) dSolutions:1)Number of ordered pairs =  1;because D=5, G=1Step 1.a + b + c + d = d + e + f + g = g + h + i = 17Means, ( a + b + c + d )+ (d +e + f + g )+ (g +h + i ) = 17 +17+17 = 17 x 3 = 51a + b + c + d + e + f + g + h + i +( d + g ) = 51Step 2.But 'a' to ' i ' takes values only from 1 to 9So a + b + c + d + e + f + g + h + i = sum of the numbers from 1 to 9 = 45Step 3.From step 1 and step 2 d + g = 51 - 45 = 6possible values of d & g are (1, 5), ( 5,1 ), (2,4) & (4, 2 )( 2, 4 ) & ( 4 , 2...

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Create a copy of a tree

Node *copy(mynode *root){ Node *temp; if(root==NULL)return(NULL); temp = (mynode *) malloc(sizeof(mynode)); temp->value = root->value; temp->left = copy(root->left); temp->right = copy(root->right); return(temp...

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