Sharp Brains: Brain Fitness and Cognitive Health News

Neuroplasticity, Brain Fitness and Cognitive Health News


#22 Brain Teaser: The Really, Really, Really Big Number

Here is new brain teas­er writ­ten by puz­zle mas­ter Wes Car­roll.

The Real­ly, Real­ly, Real­ly Big Num­ber

Dif­fi­cul­ty: HARD
Type: MATH (Numerical/Abstract)
Intim­i­da­tion Fac­tor: HIGH — but don’t be scared!

When you divide 12 by 5, the remain­der is 2; it’s what’s left over after you have removed all the 5s from the 12.  When you raise 4 to the fifth pow­er (that is, 45), you mul­ti­ply four by itself five times: 4x4x4x4x4, which equals 1,024.

What is the remain­der when you divide 100100 by 11?



This one is so sneaky.

First, con­sid­er 100 divid­ed by 11. The remain­der here is 1. Now con­sid­er the remain­der when 100x100 is divid­ed by 11. Don’t do it on your cal­cu­la­tor or on paper. Rather, con­sid­er that you have one hun­dred hun­dreds, and each of them has a remain­der of 1 when divid­ed by 11. So, go through each of your hun­dred hun­dreds and divide it by 11, leav­ing remain­der 1. Then col­lect up your remain­ders into a sin­gle hun­dred, and divide it by 11, leav­ing a remain­der of 1. This process can be extend­ed to divid­ing 100x100x100 by 11, and indeed, to divid­ing any pow­er of 100 by 11.



Next brain teas­er in Sharp­Brains’ top 25 series:

Leave a Reply...

Loading Facebook Comments ...

25 Responses

  1. Masa says:

    This was easy for me, but it still was quite delight­ful. I think that some peo­ple will think of this as impos­si­ble. *I’m so super * 🙂

    BTW, why do you post the solu­tion as doc­u­ments? That’s slow.

  2. Caroline says:

    Good for you! I think it sounds scary, but yes, it’s not that bad once you start work­ing on it!

    I post the answers as a sep­a­rate doc­u­ment to avoid the temp­ta­tion of just read­ing the answer before at least try­ing to solve it! I’m open to try­ing oth­er means of hid­ing the answer if you have ideas.

  3. lizzy says:

    I’m glad I came across this sight. Seems over time a per­son los­es the abil­i­ty to solve sim­ple Math. I’ve been accused of being stu­pid but now don’t feel as such. Thanks!

  4. Caroline says:

    Wel­come Lizzy! Keep com­ing back for prac­tice, and you’ll get bet­ter and faster at the teasers!

  5. Jonathan says:

    Very nice. I’m going to make the num­ber a lit­tle small­er (small enough to lure some poor souls into doing hand cal­cu­la­tions, but still way too big for a cal­cu­la­tor) and give it to my high school stu­dents.

    Btw, you may enjoy this trick (not real­ly a puz­zle) I share with my stu­dents.

    Nice site (found it through the Car­ni­val of Math­e­mat­ics)

  6. Caroline says:

    Let us know how it goes! I’m going to work on that trick!

  7. Dave Marain says:

    Won­der­ful prob­lem, Car­o­line!
    It’s very hard for me not to use Remain­der The­o­ry, bet­ter known as Con­gru­ence The­o­rems in Num­ber The­o­ry.
    10^2 is con­gru­ent to 1 mod­u­lo 11. Raise both sides to the 100th pow­er and the result fol­lows direct­ly. How­ev­er, your solu­tion is far more instruc­tive for stu­dents. Note that 10^1 leaves a remain­der of 10 when divid­ed by 11. Num­ber the­o­rists would write:
    10 is con­gru­ent to -1 mod 11.
    There­fore, 10^3 is also con­gru­ent to -1 mod 11. This trans­lates to 1001 being divis­i­ble by 11, which it is, since 1001 = 7x11x13. Thus every odd pow­er of 10 leaves a remain­der of 10 when divid­ed by 11. That’s why the prob­lem you posed used a base of 100 — it guar­an­teed an even expo­nent so the remain­der would be 1. Sor­ry for the tech­ni­cal stuff — num­ber the­o­ry has always been my pas­sion and my knowl­edge rep­re­sents an infin­i­tes­i­mal part of what the pros know…
    I hope you’ll vis­it my site — there are occa­sion­al­ly puz­zles like this, although my pri­ma­ry goal is to pro­vide enrich­ment les­son plans for math edu­ca­tors in grades 7–12. Good luck!
    Dave Marain

  8. Caroline says:

    Thanks for the analy­sis Dave- it’s great to hear oth­er solu­tions and the­o­ries. I’ll def­i­nite­ly check out your site and please keep com­ing back here! And PS- the cred­it for the prob­lem goes to Wes Car­roll who is writ­ing puz­zles for us. You can fol­low the links to him at the top of the post.

  9. Jyrki Leskelä says:

    There is much eas­i­er way to solve this. We are all aware that 99 is 9*11. We can eas­i­ly derive, that all num­bers 99…99 where the amount of nines is even, are a prod­uct of “some con­stant num­ber” and 9*11. 100^n has always even num­ber of zeros so there is an equiv­a­lent 99..99. No math need­ed to under­stand that the dif­fer­ence is 1.

  10. Rami Lehti says:

    Ok. The remain­der is 1.
    What is the solu­tion?

    The gener­ic solu­tion for all pos­i­tive expo­nents is:

    For this par­tic­u­lar prob­lem just sub­sti­tute n=100.

  11. Jonathan says:

    So, I final­ly used the puz­zle in class, though per­haps not exact­ly as you intend­ed. It was with 2 9th grade, pre-vaca­tion alge­bra class­es, and the results are here.

  12. Alvaro says:

    Hel­lo Jonathan, thanks for shar­ing that expe­ri­ence! We are very glad your stu­dents enjoyed the teas­er, very impres­sive 🙂

    Hap­py hol­i­days

  13. RC says:

    The num­ber 100^100 may be writ­ten as,

    (99 + 1)^(99 + 1). By the expo­nent law, a^m * a^n = a^(m + n),

    we can rewrite to this,

    (99 + 1)(99 + 1)^99

    and futher,

    (99 + 1)(99 + 1)(99 + 1)^98

    and so on, down the line. Let a = (9)(11). We get,

    (a^2 + 2a + 1)(a + 1)^98 =

    (a^3 + 3a^2 + 3a + 1)(a + 1)^97 =

    (a^4 + 4a^3 + 6a^2 + 4a + 1)(a + 1)^96 = .. = a^99 + [98 terms] + 1

    We find that every term is divis­i­ble by 11 (a is divis­i­ble) except the very last one, 1.

    The remain­der is 1.

    Here’s anoth­er quick one, which is learned from the above.

    Using only 9s, express 100^100.

    Well, it’s no trick at all. 9/9 = 1.

    Ans. (99 + 9/9)^(99 + 9/9)

    or (99 + 9/9)(99 + 9/9)^99

  14. Aaron says:

    I think it’s one, I don’t have any crazy solu­tion it’s just my gut instinct =P

  15. Anthuan says:

    My solu­tion was a bit dif­fer­ent.

    10^0 mod 11 = 1
    10^1 mod 11 = 10
    10^2 mod 11 = 1
    10^3 mod 11 = 10


    10^n mod 11 = 1, if n is even, and
    10^n mod 11 = 10, if n is odd!

  16. T-Mak1978 says:

    Rais­ing 100 to the 100 pow­er is real­ly not rel­e­vant. When you divide 100 by 11 the remain­der is 1. In this case, no mat­ter how many zeros will be in the “big num­ber”, you will still come out with a remain­der of 1.

  17. Luca says:

    I used an invari­ant prop­er­ty of mul­ti­ples of 11:

    n=k*11 in n the sum of odd posi­tion dig­its is equal to the sum of even posi­tion dig­its.

    Thus, 100^n=10^2n has 2n+1 dig­its, and the pre­vi­ous mul­ti­ple of 11 must have 2n dig­its.

    so it can only be a num­ber like 999…999, and the remain­der must be 1

  18. jumbo says:

    answer is one
    100 is of the form to 11k+1 .
    now (11k+1)*(11k+1)=11p+1
    hence you raise 100 to any pow­er it will always give remain­der 1 on being divid­ed by 11

  19. alexander says:

    childs play got the answer imme­di­ate­ly

  20. R V says:

    We can also prove this by the Method of Induc­tion:

    Prob­lem: (100)^n / 11 gives remain­der 1
    To prove that any pow­er of 100 gives Remain­der 1.

    (100)^1 / 11 gives remain­der 1 (after divid­ing to 99) .… 1
    (100)^2 / 11 gives remain­der 1 (after divid­ing to 9999)

    Thus (100)^n / 11 gives remain­der 1 (our Hypoth­e­sis from the above state­ments)

    Now, (100)^(n+1) = 100 * (100)^n
    Divid­ing the above by 11 we see that (100)^n already gives remain­der 1, and the oth­er 100 also gives remain­der 1 (as per state­ment .… 1).

    Thus, all pow­ers of 100 shall give remain­der 1 when divid­ed by 11. (Proved)

    I also liked the solu­tion giv­en by: Anthuan

  21. Kevin says:

    I did this the same way Dave did it, by instinct.

    10 mod 11 === -1 mod 11, so

    10^100 mod 11 === (-1)^100 mod 11 === 1 mod 11

Leave a Reply

Categories: Brain Teasers

Tags: , , , , ,

All Slidedecks & Recordings Available — click image below

Search for anything brain-related in our article archives

About SharpBrains

As seen in The New York Times, The Wall Street Journal, BBC News, CNN, Reuters, and more, SharpBrains is an independent market research firm and think tank tracking health and performance applications of brain science.