for certain , between thespying on millions of Americansand thesitting on decisive security system vulnerabilities for eld , the National Security Agency ( NSA ) has a fairly tough repute .
But have you heard about its brain - teasers ?
The US undercover agent agency utilize some of America ’s best and bright as ghost and code - breakers , so it ’s no surprise that its employee might have an interest in enigma and puzzles .
Every month , the NSA print on its site a mentality - teaser written by an employee that members of the public can try out their hand at .
One month it ’s a mathematics challenge make by an use enquiry mathematician ; the next it ’s a logic puzzler by a system applied scientist . They ’re all published in what the NSA call its " Puzzle Periodical . "
" news . It ’s the power to think abstractly . Challenge the unknown . reset the impossible . NSA employees work on some of the Earth ’s most demanding and exhilarating mellow - tech engineering challenges . Applying complex algorithm and expressing difficult cryptographic trouble in terminal figure of mathematics is part of the work NSA employee do every Clarence Day , " the NSA state on its internet site .
We ’ve flesh out up a six of the most interesting encephalon - teasers below . So take a read , and see if you’re able to out - smart the NSA ’s most fiendish riddlers !
1. Here’s a relatively easy one to start off with, from July 2016:
Submitted by Sean A. , NSA Applied Mathematician
On a rainy summertime twenty-four hours , brothers Dylan and Austin drop the day playing game and competing for prizes as their grandfather watches nearby . After winning two chess game matches , three consecutive hands of salamander and five rounds of ping - pong , Austin settle to challenge his brother , Dylan , to a net winner - take - all competition . Dylan remove the kitchen table and Austin grab an old coffee can of tail that their pop keep on the counter .
The game seems simple-minded as explained by Austin . The chum take turn of events place a quarter flatly on the top of the substantial kitchen board . Whoever is the first one to not find a infinite on his twist drop off . The loser has to give his chum tonight ’s afters . mighty before the biz begins , Austin arrogantly asks Dylan , “ Do you want to go first or second ? ”
Dylan turns to his grandfather for advice . The grandad knows that Dylan is tired of losing every biz to his brother . What does he whisper to Dylan ?
figure credit : MoneyBlogNewz / Flickr ( CC )
And here ’s the solution :
Dylan should go first . By doing this , Dylan can assure a win by playing to a deliberate strategy . On his first turn , he can rate a quarter right on the center of the table . Because the tabular array is symmetrical , whenever Austin places a after part on the tabular array , Dylan simply " mirrors " his brother ’s placement around the meat quarter when it is his number . For example , if Austin places a fourth part near a corner of the mesa , Dylan can place one on the opposite street corner . This scheme ensures that even when Austin finds an open space , so can Dylan . As a event , Dylan acquire victory , since Austin will die hard out of barren space first !
2. This one, from June 2016, requires a bit more math.
Submitted by Robert B. , NSA Applied Mathematician
After a brief muteness , one of the pirates tell , “ I deserve an supernumerary coin because I stretch the ship while the rest of you slept . ” Another pirate state , “ Well , I should have an extra coin because I did all the cookery . ” Eventually , a bash ensues over who should get the remaining three coin . The tavern custodian , get at by the chaos , kicks out a pirate who has erupt a tabular array and who is forced to riposte her coins . Then the tavern proprietor yells , “ Keep the peace or all of you must go ! ”
The pirates return to their seats and the captain , left with only 12 total pirates , go forward to diffuse the coins - “ one for you , ” “ one for you . ” Now , as the pile is almost depleted , she realizes that there are five extra coin . Immediately , the sea robber again debate over the five extra coins . The captain , fearing that they will be kicked out , take hold of the angriest buccaneer and ushers her out of the tavern with no recompense . With only 11 buccaneer provide , she sum up distribution . As the pile draw near depletion , she sees that there wo n’t be any extra coins . The captain breathe a suspiration of substitute . No arguments come about and everyone go to bed in serenity .
If there were less than 1,000 coins , how many did the pirates have to divvy up ?
Image credit : Disney
The answer is 341 .
There are actually infinite answer to the trouble , but only one number if the answer is under 1,000 . This puzzle is an example of modular arithmetic and theChinese Remainder Theorem .
The modest solvent under 1,000 for this problem is 341 coin , and the solvent is found by working back . To find it , we first note that with 11 literary pirate the coin separate equally ; hence , the number of coins is in the list :
11 , 22 , 33 , 44 , 55 , 66 , 77 , 88 , 99 , 110 , 121 , 132 , 143 …
What happen if we take these numbers and divide them among 12 pirates ? How many coin would be left over ? Well , we want 5 coins to be left over after divide by 12 . Hence , we abbreviate the list above to :
77 , 209 , 341 , 473 …
These numbers separate by 11 evenly and have 5 left over when divided by 12 . Now we take these remain numbers and part them by 13 until we find the number that gives 3 spare coins left over . Hence , 341 coin .
3. Here’s a logic puzzle from August 2015. You need to figure two things out.
Puzzle created by Roger B. , Cryptanalytic Mathematician , NSA
Nadine is having a party , and she has bid three friend – Aaron , Doug , and Maura . The three of them make the conform to affirmation on the days leading up to the company :
Two twenty-four hour period Before the Party : Aaron : Doug is fit to the party . Doug : Maura is not live to the party . Maura : Aaron will go to the party if , and only if , I do .
The Day Before the Party : Aaron : Maura will go to the company if , and only if , I do n’t go . Doug : An even issue out of the three of us are going to the party . Maura : Aaron is proceed to the party .
The Day of the Party : Aaron : It is not yet 2018.Doug : Aaron will go to the party if , and only if , I do . Maura : At least one of the three of us is not going to the company .
Nadine also knows that out of Aaron , Doug , and Maura : One of them never lies . A dissimilar one of them lie on days of the month that are divisible by 2 , but is otherwise truthful . The stay on one of them lies on 24-hour interval of the month that are divisible by 3 , but is otherwise true .
( 1 ) Can you work out out who is going to attend?(2 ) Can you figure out on what day , month , and twelvemonth the political party will be held , assuming it takes place in the future ?
look-alike credit : YouTube / blackboxberlin
adhere ? The answer are Doug and Aaron , and March 1 , 2016 . Here ’s how you get there …
( 1 ) Attendees solution :
The first step is to realize that the escort rules imply that nobody may lie in on two consecutive day .
If Maura ’s third statement is mistaken , then everyone is going to the party and Doug ’s first two assertion are lies , which is impossible . Therefore Maura ’s third argument is on-key , and at least one individual is not go to the party .
If Doug ’s 2nd statement is false , then ( from the above ) exactly one mortal is going to the party . This ca n’t pass without making Doug ’s first or third statement also imitation , which would have Doug dwell on two back-to-back days — an impossibility . Therefore Doug ’s second statement is true , and there are an even figure of people die to the political party .
If Doug is not going to the party , then Aaron ’s first statement is false , so his 2d affirmation must be true . This would result in an odd number of hoi polloi at the party , which we have shown is not the case . Therefore Doug is pass to the political party .
Since an even routine of the great unwashed are going to the political party , only one of Aaron or Maura is cash in one’s chips , making Maura ’s first argument imitation . Therefore her 2nd statement must be true , and Aaron is going to the party while Maura is not . The attendees are Doug and Aaron .
( 2 ) Date solution :
Having established the attendees as Doug and Aaron , now we recognise that the only prevarication are Maura ’s first statement and potentially Aaron ’s third affirmation .
If the three Clarence Day do not traverse a month boundary , then either the second day of the month or both the first and third engagement would be divisible by 2 , but nobody is available to dwell with that pattern . Therefore either the 2d or third day is the first of a month .
If the 24-hour interval before the party was the first of a month , then the twenty-four hour period of the party is the second of a month ; Aaron would have to be the one who lies on date divisible by 2 . Then , for Maura but not Aaron to lie on the first day , it would have to be divisible by 3 but not 2 . This is never reliable of the last mean solar day of a month . Therefore , the sidereal day before the party is not the first of a calendar month , so the day of the party itself must be the first day of a month .
This makes the day before the political party the last day of a calendar month , and since nobody lies on that day it must not be divisible by 2 .
This means two day before the party , the appointment is divisible by 2 , so it must not also be divisible by 3 or there would be two liar on that daylight . The only way this can happen two day before the end of a calendar month is when that 24-hour interval is February 28 of a leap year .
Since nobody lies on the first of a month , Aaron ’s third statement is true and it is not yet 2018 .
Finally , since the only leap year before 2018 is 2016 , we conclude that the party is being held onMarch 1 , 2016 .
4. This one (April 2016) is simpler — but not any easier.
Puzzle create by Andy F. , Applied Research Mathematician , NSA
Mel has four weight unit . He weighs them two at a clock time in all possible duad and finds that his span of weights total 6 , 8 , 10 , 12 , 14 , and 16 pound . How much do they each count individually ?
Note : There is not one unequalled answer to this problem , but there is a finite number of solutions .
Image credit : Doug Pensinger / Getty Images
And the solvent is …
There are incisively two potential answer : Mel ’s weights can be 1 , 5 , 7 , and 9 pounds , or they can be 2 , 4 , 6 , and 10 pounds . No other combination are potential .
account
Let the weighting be a , b , c , and d , sorted such that a < b < c < d. We can chain inequalities to get a + b < a + c < a + d , b + c < b + d < c + d. Thus , a + b = 6 , a + c = 8 , b + d = 14 , and blow + ergocalciferol = 16 . But we do n’t bed if a + d = 10 and b + c = 12 or the other way around . This is how we get two solutions . If a + d = 10 , we get 1 , 5 , 7 , and 9 ; if b + c = 10 , we get 2 , 4 , 6 , and 10 .
More on the Problem
Where this problem really get eldritch is that the identification number of solution depends on the number of system of weights . For instance , if Mel has three weights and bed the weight of all possible distich , then there is only one possible solution for the individual weights . The same is true if he has five free weight .
But now suppose that Mel has eight free weight , and the sums of pairs are 8 , 10 , 12 , 14 , 16 , 16 , 18 , 18 , 20 , 20 , 22 , 22 , 24 , 24 , 24 , 24 , 26 , 26 , 28 , 28 , 30 , 30 , 32 , 32 , 34 , 36 , 38 , and 40 . Now what are the individual weights ?
This time , there are three root :
5. October 2015 was another logic puzzler, with a school theme. Take a look:
Puzzle created by Ben E. , Applied Research Mathematician , NSA
Kurt , a math professor , has to leave for a group discussion . At the airport , he realizes he forget to happen a substitute for the course of study he was teaching today ! Before shut out his computer off for the flight , he sends an e-mail : " Can one of you cover my class today ? I ’ll broil a pie for whomever can do it . " He send the email to Julia , Michael , and Mary Ellen , his three closest acquaintance in the maths department , and board the aeroplane .
As Kurt is well - known for his delicious PIE , Julia , Michael , and Mary Ellen are each eager to fill in for him . Julia , as department chair , knows which class Kurt had to teach , but she does n’t know the metre or construction . Michael plays racquetball with Kurt so he knows what metre Kurt teaches , but not the course of instruction or building . Mary Ellen helped Kurt secure a special projector for his class , so she bed what build Kurt ’s class is in , but not the real class or the fourth dimension .
Julia , Michael , and Mary Ellen get together to figure out which category it is , and they all agree that the first person to figure out which social class it is gets to teach it ( and get Kurt ’s pie ) . Unfortunately the college ’s servers are down , so Julia bring a overlord list of all math classes teach that day . After get across off each of their own division , they are left with the following possibility :
After await the list over , Julia says , " Does anyone get laid which class it is ? " Michael and Mary Ellen right away respond , " Well , you do n’t . " Julia ask , " Do you ? " Michael and Mary Ellen both rock their heads . Julia then smiles and sound out , " I do now . I hope he bake me a chocolate peanut butter pie . "
Which social class does Kurt necessitate a second-stringer for ?
The solution is Calc 2 at 10 in North Hall . But why ?
1 ) Since Julia only bang the stratum name , the only means she could straight off make love is if it was Calc 3 . Since Michael and Mary Ellen both know that Julia does n’t know , that think of they know the class is n’t Calc 3 . Since Michael only know the time , that mean the class ca n’t be at noon . Because Mary Ellen only knows the building , that means the building ca n’t be West Hall . That go forth only the following possibilities :
2 ) Since Michael does n’t experience which class it is , that means the time ca n’t be 9 or 11 . Since Mary Ellen does n’t know either , the class ca n’t be in South Hall . That leaves only three possibilities :
3 ) At this point , Julia now says she knows the response . Since there are two Calc 1 classes , it must be that Julia live the class is Calc 2 . Thus , the grade is Calc 2 at 10 in North Hall .
6. Last one: Can you figure out Charlie’s birthday? (July 2015)
Puzzle create by Stephen C. , Applied Research Mathematician , NSA
After observing Albert and Bernard watch Cheryl ’s birthday , Charlie decides he want to play . He presents a list of 14 potential day of the month for his birthday to Albert , Bernard and Cheryl .
He then herald that he is perish to separate Albert the calendar month , Bernard the twenty-four hours , and Cheryl the year .
After he tell them , Albert say , " I do n’t know Charlie ’s birthday , but neither does Bernard . "
Bernard then say , " That is true , but Cheryl also does not know Charlie ’s birthday . "
Cheryl says , " Yes and Albert still has not calculate out Charlie ’s birthday . "
Bernard then replies , " Well , now I cognize his natal day . "
At this point , Albert says , " Yes , we all have it away it now . "
What is Charlie ’s birthday ?
Image credit rating : Will Clayton / Flickr ( CC )
The solvent is Apr 16 , 2000 , and the rationale is …
When Albert take that Bernard does not lie with Charlie ’s birthday , he is enunciate that he know that the right daytime occurs more than once in the leaning . In other parole , he is saying that Charlie ’s natal day is not Feb 19 , 2000 , and the only way of life he could know that is that he have intercourse that the calendar month is not February .
So by making this claim Albert has reduced the list for everyone to :
When Bernard say that it is on-key that he does not lie with Charlie ’s natal day , it tells everyone that even with the qualified tilt the correct day occurs more than once in the list . So everyone can thus eliminate May 17 , 2001 .
Furthermore , the claim that Cheryl also does not recognise Charlie ’s birthday is a claim that Bernard live that the yr occurs more than once on the remaining list .
This rule out Apr 14 , 1999 , and since Bernard could only rule this out by knowing that the daytime was not 14 , everyone can further reduce the list to :
When Cheryl says that Albert still has not cipher out Charlie ’s natal day , she is telling everyone that given the Modern list the month occurs more than once and thus decree out May 16 , 2001 . This secern everyone that Cheryl knows that the year is not 2001 and the list can be abbreviate to :
When Bernard then claims to know the date he is say that the day occurs only once among the 3 remain selection , thus telling everyone that Charlie ’s natal day is Apr 16 , 2000 .
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