# 100 using only 5 number of digits

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:

$$100=111-11$$

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

• 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. Commented Oct 8, 2018 at 22:12
• 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 Commented Oct 9, 2018 at 2:56
• Presumably, $d$ and $i$ refer to the same variable. Commented Oct 9, 2018 at 4:37
• 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 Commented Oct 9, 2018 at 11:35
• I wonder how Positional notation allows for other approaches here? Commented Oct 9, 2018 at 15:45

Seven

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

Six

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

Five (and the rest)

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

• Nicely done. Universal solution. Commented Oct 8, 2018 at 23:27
• I wonder whether the use of a decimal point counts as an operation? Commented Oct 9, 2018 at 8:55
• This solution is very nice but the usage of decimal point implicitly uses a 0, because .7 is actually 0.7. Commented Oct 9, 2018 at 12:39
• @rhsquared Arguably, any integer $x$ is actually $x.0$.. ;) Commented Oct 9, 2018 at 14:17
• @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. Commented Oct 9, 2018 at 15:40

For 9:

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

For 8:

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

For $$6$$

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

Edit: added $$7$$

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

• Note: $1 ... 5$ and $8 ... 9$ already have uncontroversial solutions posted. Commented Oct 9, 2018 at 0:08

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 $$.i$$

• 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 Commented Oct 9, 2018 at 4:04
• @KevinL oops, forgot to add and subtract ;)
– Holo
Commented Oct 9, 2018 at 4:06
• Nicely done. :D Commented Oct 9, 2018 at 4:07
• $n$ appears 7 times in your last formula. Of course, multiple factorials kill all the fun from such puzzles, so they are usually non accepted. Commented Oct 11, 2018 at 9:48
• Oh ok. Two $n$ are only there to precise the number of exclam signs. Sorry! Commented Oct 11, 2018 at 11:20

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

• Neat 7. I was wondering how to employ a base change. Commented Oct 10, 2018 at 11:53
• The first whole solution I see. Especially providing the 7s without cheating (floor function or double exponentiation is cheating imho) is beautiful. Commented Oct 10, 2018 at 13:35

### 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.

• 5 could also be (5+5+5+5)*5 Commented Oct 8, 2018 at 21:30
• Definitely cheating to use ceiling and floor, I would think. Commented Oct 8, 2018 at 22:10
• I think ceiling and floor is much better than the generalized factorial with 48 exclamation points... Commented Oct 9, 2018 at 18:55

Four

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

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

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:

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

• One extra 9 there ;)
– Jafe
Commented Oct 9, 2018 at 13:29

Number 5:

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

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)

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.

Seven (almost)

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

All numbers can be solved with the following:

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

• Yes, but the question wants exactly 5 is. Commented Oct 9, 2018 at 2:21

For i=2,9

●((2)^2)^2+92 = 100

Some randoms,

●9^2 + 5^2 - 6 = 100

●2*(3^2)+ 82 =100

• 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 Commented Oct 10, 2018 at 9:54

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)

• The question said "For each digit $1 ≤ d ≤ 9$". Commented Oct 11, 2018 at 10:57

Perhaps another UNIVERSAL solution

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

Example

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

• 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? Commented Oct 16, 2018 at 13:24