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#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:

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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

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