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This is a problem from Martin Gardner.

For each digit $1\leq d\leq 9$, make 100 using exactly five number of $i$s. Any operation is allowed, brackets as well.

For $i=1,2$ I have a solution:

$$100=111-11$$

$$100=(2*2*2+2)^2$$

For the rest, please help.

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    $\begingroup$ In case anyone missed it, the same problem with exactly 6 /i/s is trivial: (iii - ii) / i works for any i in any base > i. $\endgroup$
    – Rupert Morrish
    Commented Oct 8, 2018 at 22:12
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    $\begingroup$ Hello @Pet123 and welcome to Puzzling.SE. You said that this problem is from Martin Gardner and as such, it would be nice if you can give a link to the problem. Anyways, this is still a good math puzzle. Enjoy your time here :D $\endgroup$
    – Kevin L
    Commented Oct 9, 2018 at 2:56
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    $\begingroup$ Presumably, $d$ and $i$ refer to the same variable. $\endgroup$
    – Lawrence
    Commented Oct 9, 2018 at 4:37
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    $\begingroup$ if any operation is allowed, i would like to use the function called hundred, which takes exactly five arguments, and gives constant 100 as a result $\endgroup$
    – elias
    Commented Oct 9, 2018 at 11:35
  • $\begingroup$ I wonder how Positional notation allows for other approaches here? $\endgroup$
    – chux
    Commented Oct 9, 2018 at 15:45

17 Answers 17

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Seven

$100 = 77/.7 - 7/.7$

Six

$100 = 66/.6 - 6/.6$

Five (and the rest)

$100 = 55/.5 - 5/.5, \; nn/.n - n/.n$

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    $\begingroup$ Nicely done. Universal solution. $\endgroup$
    – Weather Vane
    Commented Oct 8, 2018 at 23:27
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    $\begingroup$ I wonder whether the use of a decimal point counts as an operation? $\endgroup$
    – Jack Aidley
    Commented Oct 9, 2018 at 8:55
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    $\begingroup$ This solution is very nice but the usage of decimal point implicitly uses a 0, because .7 is actually 0.7. $\endgroup$
    – rhsquared
    Commented Oct 9, 2018 at 12:39
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    $\begingroup$ @rhsquared Arguably, any integer $x$ is actually $x.0$.. ;) $\endgroup$
    – maxathousand
    Commented Oct 9, 2018 at 14:17
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    $\begingroup$ @chux: I find that an unconvincing argument, but since the Asker's specification is vague I shan't pursue the point further. your solution is certainly clever. $\endgroup$
    – Jack Aidley
    Commented Oct 9, 2018 at 15:40
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For 9:

$100 = 99 + (\frac{9}{9})^9$

For 8:

$100 = 88 + 8 + \sqrt{8+8}$

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For $6$

$\frac {6! - (6 - \frac {6}{6} ) !} { 6} = \frac {6! - 5!}{6} = \frac {720 - 120} {6} = \frac{600}{6} = 100$

Edit: added $7$

$7!! + \frac{7+7}{7} -7 = (7 \times 5 \times 3) + 2 - 7 = 100$

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  • $\begingroup$ Note: $1 ... 5$ and $8 ... 9$ already have uncontroversial solutions posted. $\endgroup$
    – Weather Vane
    Commented Oct 9, 2018 at 0:08
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Using generalized factorial(or multifactorial or k-torial) we obtain:

1:

100=11!!!!!!+11!!!!!!!+1

2:

100=22!!!!!!!!!!!!!!!!!!+(2+2)!/2

3:

100=((3!)!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+3!-(3!+3!)/(3!)

4:

100=(4!)!!!!!!!!!!!!!!!!!!!!+4+(4-4)/4

5:

100=(5!!)!!!!!!!!!+5+5+5-5

6:

100=(6!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+6-(6+6)6

7:

100=(7!!!)!!!!!!!!!!!!!!!!!!!!!!!!!+7!!!!!+(7+7)/7

8:

100=(8!!!!+8!!!!!!+(8+8)/8)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

9:

100=((9+9/9)!!!!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+9-9

So we got 100 using all the i's from 1 to 9.

Edit:

I found a formula that works for a any integer greater than 9:

$$100=\left(\left(\left(\frac{\left(\frac{(n+n){!}_{(n)}}{n+n}\right){\large!}_{(n-5)}}{n}\right){\huge!}_{(3)}\right){\huge!}_{(5)}\right){\huge!}_{(48)}$$ where $a!_{(b)}=a\overbrace{!!!\cdots!!!}^{b\mbox{ times}}$

We can write a formula for every number without using $.i$

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  • $\begingroup$ Hey @Holo. Welcome to Puzzling.SE and I must say, this is a nice first answer. However, you should check number 5 again. Anyways, happy puzzling :D $\endgroup$
    – Kevin L
    Commented Oct 9, 2018 at 4:04
  • $\begingroup$ @KevinL oops, forgot to add and subtract ;) $\endgroup$
    – Holo
    Commented Oct 9, 2018 at 4:06
  • $\begingroup$ Nicely done. :D $\endgroup$
    – Kevin L
    Commented Oct 9, 2018 at 4:07
  • $\begingroup$ $n$ appears 7 times in your last formula. Of course, multiple factorials kill all the fun from such puzzles, so they are usually non accepted. $\endgroup$
    – Evargalo
    Commented Oct 11, 2018 at 9:48
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    $\begingroup$ Oh ok. Two $n$ are only there to precise the number of exclam signs. Sorry! $\endgroup$
    – Evargalo
    Commented Oct 11, 2018 at 11:20
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One:

$$ 100=111-11 $$

Two:

$$ 100=(2*2*2+2)^2 $$

Three:

$$ \ 100 = \ 33 * 3 + \log_33 $$

Four:

$$ \ 100 = \ ( \frac{44-4} 4 ) ^ \sqrt4 $$

Five:

$$ \ 100 = 5 * 5 * 5 - 5 * 5 $$

Six:

$$ \ 100 = (\log_\sqrt66) ^ 6 + 6 * 6 $$

Seven:

$$ \ 100 = \ 7 * ( 7 + 7) + \log_\sqrt77 $$

Eight:

$$ \ 100 = 88 + 8 + \sqrt{8+8} $$

Nine:

$$ \ 100 = 99 + \log_\sqrt{9*9}9 $$

logarithms <3

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    $\begingroup$ Neat 7. I was wondering how to employ a base change. $\endgroup$
    – chux
    Commented Oct 10, 2018 at 11:53
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    $\begingroup$ The first whole solution I see. Especially providing the 7s without cheating (floor function or double exponentiation is cheating imho) is beautiful. $\endgroup$
    – Evargalo
    Commented Oct 10, 2018 at 13:35
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Partial (will update as I figure more out)

For 3:

33*3+(3/3)

For 4:

$4*(4+\frac{4}{4})^\sqrt{4}$

For 5:

(5*5*5)-(5*5)

For 7: (possibly cheating?)

ceiling(sqrt(7!)) + ceiling(sqrt(7!)) - (7*7) + 7. you did say any function was allowed.

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    $\begingroup$ 5 could also be (5+5+5+5)*5 $\endgroup$
    – PotatoLatte
    Commented Oct 8, 2018 at 21:30
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    $\begingroup$ Definitely cheating to use ceiling and floor, I would think. $\endgroup$
    – Rupert Morrish
    Commented Oct 8, 2018 at 22:10
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    $\begingroup$ I think ceiling and floor is much better than the generalized factorial with 48 exclamation points... $\endgroup$
    – Matthew G Dippel
    Commented Oct 9, 2018 at 18:55
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Four

$100 = (\frac{44 - 4}{4}) ^{\sqrt{4}}$

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Four:

4 + 4! + 4! + 4! + 4!

Five:

5 x 5 x 5 - 5 x 5 or 5 x 5 x (5 - 5/5)

Six:

TBD

Seven:

TBD

Eight:

TBD

Nine:

TBD

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To be improved... Remains the 7s

For 1 :

$100=111-11$

For 2:

$100=(2*2*2+2)^2$

For 3:

$100=33*3+3/3$

For 4:

$100=4!*4+4-4+4$

For 5:

$100=(5+5+5+5)*5$

For 6:

$100=\frac{6!-\frac{6!}{6}}{\sqrt{6}\sqrt{6}}$

For 7:

$100=(7+7)*(7+7^\frac{-7}{7})$ (one too many...)

For 8:

$100=88+8+\sqrt{8+8}$

For 9:

$100=99+\frac{\sqrt{9}*\sqrt{9}}{9}$

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    $\begingroup$ One extra 9 there ;) $\endgroup$
    – Jafe
    Commented Oct 9, 2018 at 13:29
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Number 5:

5! - (5 + 5 + 5 + 5) = 100
(5 + 5) ^ ((5+5)/5) = 100

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For some solutions other users were already faster, lets give them credit for it.

3

33*3+3/3

4

4+4*4!*4/4

5

(5+5+5+5)*5

6

TODO

7

It is my cheating. Lets declare a variable k
k=(7+7)/7,
7*7*k + k
(but I have used 7 5 times in total anyway)

8

TODO

9

99+(sqrt(9)*sqrt(9)/9)

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Using a cheaty approach, I have simply tried all permutations of 5 numbers being seperated by 4 operands:

  • none, e.g. concat the numbers
  • +
  • -
  • *
  • /
  • ^
  • (
  • )
  • ^0.5 (Square Root)

I've found solutions for three values of i yet, all of which have of course already been posted:

111-11 = 100
33*3+3/3 = 100
3*33+3/3 = 100
3/3+33*3 = 100
3/3+3*33 = 100
5*5*5-5*5 = 100

I'll improve my algorithm as I go to include more operands in more locations and edit my answer accordingly.

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0
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Seven (almost)

$100 = (7 + 7) * (7 + \frac{7^0}{7})$

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All numbers can be solved with the following:

100 = ii / .ii, 1 <= i <= 9

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    $\begingroup$ Yes, but the question wants exactly 5 is. $\endgroup$
    – Rupert Morrish
    Commented Oct 9, 2018 at 2:21
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For i=2,9

●((2)^2)^2+92 = 100

Some randoms,

●9^2 + 5^2 - 6 = 100

●2*(3^2)+ 82 =100

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    $\begingroup$ Hello @neh and welcome to PSE. Nice answer however, the question seems to only allow 1 number in each equation. Using both 2 and 9 in 1 equation, like yours, is not allowed $\endgroup$
    – Kevin L
    Commented Oct 10, 2018 at 9:54
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Nobody mentioned the following "solution" using five zeroes:

$0! : (0 \times 0) : (0 + 0) = 1 : 0 : 0 = 100$, where $:$ means concatenation (e.g. $2:8=28$). It works because all operations are explicitly allowed (I think, at least all standard, i.e. not specifically invented for this problem, ones are - and concatenation is obviously standard under this definition)

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  • $\begingroup$ The question said "For each digit $1 ≤ d ≤ 9$". $\endgroup$
    – u-ndefined
    Commented Oct 11, 2018 at 10:57
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Perhaps another UNIVERSAL solution

Antilog ( n+n/(n+n-n) ) = 100

Example

Antilog ( 2+2/(2+2-2)) = 100

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  • $\begingroup$ Shouldn't this be Antilog ( n+n/(n+n-n) )? The idea being that antilog in this case means 10^x, so if you can get x=2, you can make 100 every time. Or am I misunderstanding what antilog means? $\endgroup$
    – Darrel Hoffman
    Commented Oct 16, 2018 at 13:24

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