I got "x" number of sweets. I can divide my sweets equally by 1,2,3,4,5,6,7 all the way to 101 friends such that I have no left overs. Find the least number of sweets I could possibly have at first?

Note: This logic is a troll, so I thought it is fun. Wanna see how people answer. Anyways, the answer is still a number. Not kidding and it is possible. No cutting of sweets happened.

  • 3
    $\begingroup$ Can I answer $-\infty$? $\endgroup$
    – Glen O
    Apr 29, 2015 at 15:10
  • 11
    $\begingroup$ It's a troll question, I figure a troll answer is appropriate. $\endgroup$
    – Glen O
    Apr 29, 2015 at 15:19
  • 2
    $\begingroup$ I know. But sometimes it's fun to intentionally reinterpret things. In mathematics, there's really nothing stopping $x$ being a negative number, unless you specify otherwise. The context does introduce a "natural" lower limit, but technically you can't divide zero sweets among 101 friends because there's nothing to divide - we overlook that nuance of words used in regular English because it's a mathematical question... but we could equally ignore other nuances. Mostly, though, I asked because it seemed like a fun idea to suggest it. I was hoping you'd laugh at it. $\endgroup$
    – Glen O
    Apr 29, 2015 at 15:26
  • 1
    $\begingroup$ I don't think you'll have friends after this is over :) $\endgroup$
    – Brian J
    Apr 29, 2015 at 20:18
  • 10
    $\begingroup$ For completeness' sake the LCM of 1..101 is 7041757898200960193617914702466542659236800 $\endgroup$
    – Snowbody
    Apr 29, 2015 at 21:46

8 Answers 8


Troll question?

The minimum number of sweets you can have is clearly 0.

  • 15
    $\begingroup$ What about -0? A troll answer for a troll question? :p $\endgroup$
    – Mark N
    Apr 29, 2015 at 15:22
  • $\begingroup$ Yeah, you got it.I wanted someone to give me a mathematical solution of finding the lowest common multiple of these and be like giving a large answer. $\endgroup$ Apr 29, 2015 at 15:23
  • 4
    $\begingroup$ @ministic2001 - then you shouldn't have listed it as a troll math question. Kind of a big red flag for "don't use the obvious answer you first think of". $\endgroup$
    – Glen O
    Apr 29, 2015 at 15:27
  • 1
    $\begingroup$ @ministic2001 But hang on, zero can also be divided by 102, so it can't be the answer... Unless that's your 'logical' weasel wording (i.e. you never said it can't also be split among 102, 103, 104... friends.) Right? $\endgroup$
    – grkvlt
    Apr 30, 2015 at 0:36
  • 1
    $\begingroup$ @grkvlt, there is no number divisible by 2, 3 and 17 that is not also divisible by 102, so that isn't a valid counter point (sorry). This is because 2, 3, and 17 are the prime factors of 102, and are all coprime. Yay math! $\endgroup$
    – Poik
    Apr 30, 2015 at 14:52

For the non-zero answer, in ruby, type:


Which yields:


Don't eat them all at once, I guess.


Just for fun, the answer is also:

$\int\limits_{-\infty}^{+\infty}(\frac{-(\sqrt{-0} \times sinh(f'(x)))^3}{π^e})dx$

  • 2
    $\begingroup$ BWAHAHA.Even though Im just 14 years old, I get this :P. $\endgroup$ Apr 29, 2015 at 15:39
  • 5
    $\begingroup$ SPOILER TO THE JOKE: The numerator has the sqrt(0) in it. $\endgroup$ Apr 29, 2015 at 17:36
  • 7
    $\begingroup$ Assuming $f$ is differentiable. $\endgroup$
    – wchargin
    Apr 30, 2015 at 1:00
  • 1
    $\begingroup$ Wouldn't this make the answer any constant value..? $\endgroup$ Apr 30, 2015 at 9:41
  • 2
    $\begingroup$ @No.7892142 Only if it was indefinite. You can look it up in wolfram if you want :) $\endgroup$
    – Mark N
    Apr 30, 2015 at 14:14

X = 1

You have one sweet for yourself, and your 1-101 friends are all imaginary friends for whom you conjure an appropriate amount of imaginary sweets.

  • $\begingroup$ Lol :) Nice little joke here.Sometimes math like my exam paper just doesnt make sense that we had to make jokes like these :) $\endgroup$ Apr 30, 2015 at 14:55
  • $\begingroup$ Maybe they could also just be 1-101 reflections of yourself in a mirror, so they are imaginary, but also real ;) $\endgroup$
    – Mark N
    Apr 30, 2015 at 14:57

Your sweets are powdered candy. You simply divide the amount you have into portions of equal weight, no cutting required. Probably something to the effect of a kilogram of candy for a satisfying portion when divided by 101.


The real answer is


In fact, you didn't specify that the solution is in $\mathbb{N}$, it could also be in $\mathbb{Z}$ ! The least number satisfying your problem is $-\infty$.
What does it mean that I have $-\infty$ sweets? It simply means that instead of having them, you have a debt of $-\infty$ sweets (exactly as $-20\$$ means a debt of $20$ dollars).

  • 14
    $\begingroup$ It means you have an infinite number of sweets, but they all want to eat you :) $\endgroup$
    – KnightOfNi
    Apr 29, 2015 at 21:20
  • 11
    $\begingroup$ Since when was $-\infty$ in $\mathbb Z$? $\endgroup$ Apr 30, 2015 at 2:16
  • 1
    $\begingroup$ @Chris There's no order in Z? So you mean that (-1)>(-2) is false? $\endgroup$
    – leoll2
    Apr 30, 2015 at 12:24
  • 4
    $\begingroup$ I mean an integer number which is always lower than any other integer. There is no such integer. $-\infty$ is not an integer just as $+\infty$ is not a natural number. $\endgroup$
    – Dennis
    Apr 30, 2015 at 14:23
  • 4
    $\begingroup$ The real answer is $- \infty$. Since when $- \infty$ is "real". :P $\endgroup$
    – Bhaskar
    May 1, 2015 at 7:39


In case you wanted a non-troll (less troll?) answer, the smallest number divisible by 1-101 that's also greater than zero would be:



I see how my previous answer was incorrect, I just missed a few steps. Not only do we need the product of every prime from 1 to 101, but we need the largest power of each prime that is itself less than 101. This is because every number less than 101 can be factored as a combination of prime numbers, but also, any factor of a number n is necessarily less than n.

So not only do we need...


We'll also need to include:
The largest power of 2 that is less than 101 (64 = 2^6)
The largest power of 3 that is less than 101 (81 = 3^4)
The largest power of 5 that is less than 101 (25 = 5^2)
And the largest power of 7 that is less than 101 (49 = 7^2).
The smallest power of 11, the next prime number, is 121, which is too large. Thus, our final product becomes:

= 7041757898200960193617914702466542659236800

Indeed, another answer using Ruby seems to have given the same result. This, above, is the mathematical proof of that.

  • 4
    $\begingroup$ 232862364358497360900063316880507363070 is obviously not divisible by 100, or any other non-squarefree number. $\endgroup$
    – Kyle G
    Apr 30, 2015 at 5:22
  • $\begingroup$ To expand on what @kgull said with a short counterexample: imagine you want a number divisible by 1-6. The product of all the primes between 1-6 would be 2*3*5 = 30. But 4∤15. $\endgroup$ Apr 30, 2015 at 10:23
  • 2
    $\begingroup$ Ah, indeed. My mistake. I shouldn't try large computations after three glasses of whiskey. $\endgroup$
    – Cubicon
    Apr 30, 2015 at 13:37
  • $\begingroup$ There we go. Think I did it right this time. $\endgroup$
    – Cubicon
    May 1, 2015 at 3:57

Perhaps I'm being too simplistic. If(as the question states)

I got "x" number of sweets

then I must have

something and thus not 0

My intuitive guess is

+ or - (n!) sweets (dont forget the units as my old maths teacher used to say) where n is the number of friends you have.

also the answer given by Mark N cannot be a full answer as

the solution to an integral adds a constant - there is no mention in this solution of how big (or little) the constant is.

  • $\begingroup$ Imagine this question:Solve "x" for the following equation 2x+6=6 where "x" becomes 0. Also, because to prevent obviousity, I have to put "number of sweets" rather than "no sweets"."I have 0 number of sweets", to me, make sense.Maybe it will be clearer if I said "I have "x" sweets. $\endgroup$ Apr 30, 2015 at 12:36
  • $\begingroup$ but like grkvit pointed out, 0 is also devisable by 102, 103, 104 ... so the aswer cannot be 0 $\endgroup$
    – Brian
    Apr 30, 2015 at 13:14
  • $\begingroup$ lol - I love the English language! I think to the last question the answer is 11 because if you were short of sweets then not everyone got one. If you had 29 sweets there would be 11 sweets left over (after everyone else got one sweet each). $\endgroup$
    – Brian
    Apr 30, 2015 at 14:36
  • $\begingroup$ The solution to an indefinite integral adds a constant. Mark N's answer as of 22 hours ago includes the limits negative infinity to infinity, making it a definite integral and thus an exact value. $\endgroup$
    – Poik
    Apr 30, 2015 at 14:38
  • $\begingroup$ @Brian, any number that satisfies the original question that isn't zero is also divisible by any number composed of two primes less than or equal to 101, so 101 and 2 which is 202. The answer didn't specify that it couldn't be divisible by anything else, and it would be impossible for it not to be. $\endgroup$
    – Poik
    Apr 30, 2015 at 14:41

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