Welcome to second semester; cue the leaning tower of textbooks, bottomless coffee cup and permanent residence in the lab. It sounds like you could use a serious study or grading break already…

You’re in luck! We’ve got a brain break for you. A set of riddles for you to crack that flex your engineering knowledge. Here’s our challenge: How fast can you solve the below? 

Challenge accepted? Try your logic skills below and then check our answers at the bottom.

Riddle 1: Doug forgot his briefcase combination. Luckily, we bet you can figure it out faster than he can figure out how to break it open…

Riddle 2: Now, show-off what those math courses taught you. Let’s play with some numbers…

Riddle 3: Time to play with some more numbers… Let’s give this one a try.

Riddle 4: We think you can, we think you can… Don’t run out of steam!


Riddle 5: You’ve almost completed them all… can you top your time to solve the last riddle?


Did you conquer our challenges? Congratulations! You’re a true engineer. But, like any engineer, be sure to check your answers by scrolling to the bottom of the page. After you’ve confirmed your success, return to your studies or get back to grading those lab reports and try these again later… maybe you’ll be even faster the next time we serve up a challenge. Make sure to follow our blogs and check out all that our new website – www.university.ti.com – has to offer! 



#UniversiTI Riddle Answers:

*Of note to those who answered "8 ft" with the previous riddle wording: "If the top of the ladder slips down 4 ft, how many feet will the bottom slide out?" -- you are also correct! 

  • Looking to learn the reasoning behind each riddle answer? Look no further than these answer paths from E2E guru Jens-Michael Gross:

    Riddle 1


    This riddle asks us to find the values of five variables N1 to N5 (the five digits of the code).

    To solve it, we are given five independent equations:

    1: N5+N3 = 14

    2: N4=N2+1

    3: N1=2*N2-1

    4: N2+N3=10

    5: N1+N2+N3+N4+N5=30

    Equation 2+3 allow us to replace N4 and N1 in equation 5 by N2. Also, we can

    replace (N3+N5) in equation 5 by 14. This gives us N2:

    N1+N2+N3+N4+N5 = N1+N2+N4+14 = 2*N2-1+N2+N2+1+14 = 4*N2+14 = 30. Therefore N2=4.

    Knowing N2, we can calculate N4 (=5), N1 (=7), N3 (=6) and finally N5(=8)

    Riddle 2


    If only one symbol shall be added, the only possible places are between the digits of one of the two numbers or 'around' one of the numbers (e.g. SQR or SQRT).

    So either 63 must be turned into 25 or 26 into 64.

    Turning 63 into 25 won't work with any symbol one could add.

    But 26 can be turned to 64 by adding an '^' (2^6 = 64).

    Riddle 3


    The first step is to find the pattern that forms the given sequences of 4 numbers.

    The pattern will be the same for all four columns (and also for the four rows).

    There is no standardized way to find the pattern. Some imagination is required.

    The first two numbers of each sequence (X0 and X1) seem to be random. The third

    (X2) is the sum of X0 and X1. The fourth (X3) is the sum of X2 and X1.

    Putting this into a formula gives us two possible interpretations:

    1) Xn = X(n-1)+X(n-2) _for n>1 or

    2) Xn = X(n-1)+X1 (for n>1)

    Both patterns do match not only for each column but also for each row of the 4x4 number square.

    Both give the same result for X3 (X0+X1+X1) but will differ for X4

    Using formula 1, we'll get the following

    3  6  9 15   24

    8  2 10 12   22

    11  8 19 27   46

    19 10 29 39   68

    30 18 48 66  114

    Interestingly, the 5th row as well as the 5th column also give a sequence matching the formula.

    Using the second formula, we'll get

    3  6  9 15   21

    8  2 10 12   14

    11  8 19 27   35

    19 10 29 39   49

    27 12 39 51   63

    Here again, the 5th row and column match the formula.

    Riddle 4


    This one is a bit more tricky.

    It is assumed that both trains move with a constant speed and the tracks are parallel and have exactly the same length.

    It is also assumed, that they start at the same time. Since no absolute distance is given and only the relative speed

    of the two trains is requested, the calculation doesn't use (nor need) any units.

    Now how to solve this?

    We have two distances, X and Y, going from the 'meeting point' to the two end points.

    Train A needs 1hr for distance Y while train B needs 4 hours for distance X.

    So the speed of train A is (Y/1= Y) while the speed of train B is (X/4).

    We also know that train A needed the same time for distance X (at speed Y) as train B needed for distance Y (at speed X/4).

    So the time passed until meeting point is the same for both and gives us the relation of the distances X and Y:

    Distance X by speed (Y) equals distance Y by speed (X/4):

    X/Y = 4Y/X -> X*X/Y = 4Y -> X*X = 4Y*Y -> X = 2Y.

    After this, we can say that if the speed of train A is Y, then the speed of train B must be (2Y/4) = Y/2.

    This means that train A is twice as fast as train B.

    And they met two hrs after starting, when train A had already done 2/3 of the distance while train B only did 1/3.

    Riddle 5


    Wall, floor and ladder form a right-angled triangle. The ladder is the hypotenuse while wall and floor are the catheti.

    For right-angled triangles, the Pythagorean theorem applies, which tells us that the sum of the squares of the catheti is the square of the hypotenuse.

    For a ladder of 25ft length, the square is 625. And the square of the 'floor cathetus' is 49.

    This means the square of the 'wall cathetus' is (625-49)=576. Therefore the wall cathetus length (= height) is SQRT(576)=24ft.

    If the ladder slips down the wall by 4 ft (= 20ft), now the square of the wall cathetus is 400 and therefore the square of the floor cathetus is 225.

    This gives us a new length of the floor cathetus of SQRT(225) = 15ft. I slid 8ft from the original floor position.

    Interestingly, the area 'under the ladder' has increased from 84 to 150 square ft by this slide. It will continue to increase until both catheti are equal.

    This means, for the same diagonal (hypotenuses), the screen area of a 5:4 monitor is by a factor of 1.14 larger than the screen area of a 16:9 monitor.

    You'll need a 22" 16:9 monitor to get the same screen area as on a 19" 5:4 monitor. Pythagoras has told you!