# How do you get 23 using the numbers 1, 2, 3 and 5?

You can only use Addition, Subtraction, Multiplication, Division. This is for a math project for my daughter.

You can only use the numbers once and all numbers do not need to be used.

• I don''t believe there is and answer to the question as stated. Nov 17 '17 at 20:37
• Is it possible a 4 got lost somewhere? $23=5*3+4*2*1$ Nov 17 '17 at 21:50
• Can you use concatenation? In other words, can you make a two digit number by putting two of the digits together? Like $1$ and $5$ could make the number $15$? Nov 18 '17 at 12:54
• @SarahFritz Please tell us the average age of your daughter classroom (7 years old?, 17 years old?) and what was the teacher expecting for this math project? "impossible", "usage of concatenation", "usage of decimal point", "usage of exclamation point" (factorial), "usage of power notation", "usage of non-decimal base", etc.? Or was it just a typo from teacher?
– Cœur
Nov 19 '17 at 10:23
• It's worth bearing in mind that setting a child a maths question that cannot be solved - without first introducing the concept of insoluble problems - may reinforce any feeling in the child that they are unable to answer maths questions and are therefore bad at maths. Such questions should be handled carefully to ensure that the discovery that there is no solution is a positive outcome for the child. Nov 19 '17 at 11:03

As stated the problem is not possible. Here's an online solver to show that.

Lateral thinking options could fix it (like @Apep (reinterpretation of the list), @jlars62(decimal point (very clever)), or @hoffmale (factorials), or @sousben and @D Krueger (non-decimal)). Or allowing powers:

$5^2-3+1=23$

Or allowing concatenation.

• Powers are not allowed: -1 No solution: +1 Nov 17 '17 at 22:41
• @ElementsinSpace: powers were used as "lateral thinking" after explaining there is no solution.
– user10179
Nov 18 '17 at 12:46
• Per original question: You can only use each number once Nov 19 '17 at 14:50
• This is what I arrived at independently. Using each number once. 5 squared uses 5 and 2. Squaring a number is multiplication. How semantic is the question? +1 - like the end of the calculation!!
– Tim
Nov 20 '17 at 9:25
• You are, however, making an assumption (usually a good one) that the 23 stated in the title is base 10. puzzling.stackexchange.com/a/57085/37225 and puzzling.stackexchange.com/a/57082/37225 cleverly answer the question you "prove" is unanswerable. Every fact has assumptions underlying it.
– NH.
Nov 21 '17 at 18:32

One solution could be:

(5+3)*3-1

under the (possibly invalid) assumption that

the problem could be considered as using "1, two 3, and 5"

• Wins the "lateral thinking" award...
– smci
Nov 18 '17 at 1:34

I guess we are not allowed to repeat the numbers:

35 - 12 = 23

• This has to be the simplest possible answer. :)
– Sid
Nov 17 '17 at 18:59
• concatenation is not allowed as per the OP, if it was , surely just concatenate the 2 and 3=> 23 Nov 17 '17 at 19:09
• @JasonV, OP didn't say anything about that. And you can't concatenate the 2 and 3 because you must use all those four numbers. Nov 17 '17 at 19:12
• when OP said "you can only use Addition, Subtraction, Multiplication, Division" they did not include concatenation. Therefore, this is out of scope. Also, OP did not say you must use all numbers nor only once. Nov 17 '17 at 19:14
• @JasonV, I am sure you know that concatenating is not a part of mathematical operation. I think it is better wait and see what the OP thinks about these answers and we shouldn't talk on his behalf. Nov 17 '17 at 19:19

$$\frac{5}{.2} - 3 + 1 = 23$$

– Nic
Nov 18 '17 at 0:22
• Doesn't .2 mean dividing 2 by 10?
– DrD
Nov 19 '17 at 13:25
• @DEEM : Just because $0.2 = 2/10$ does not mean "$.2$" entails division ... any more than $4 = 8/2$ means that "$4$" entails division. Nov 19 '17 at 23:49
• Brilliant, but no. Had the question said "digits" this would work, but the question says "numbers" and the number .2 is not the same as the number 2. Nov 20 '17 at 6:07

23, not using 1 or 5.

didn't even need to use any mathematical functions

• Specifically "2" + "3" Nov 18 '17 at 12:45
• -1. You used an invalid operation (concatenation) that wasn't allowed in the post.
– NH.
Nov 21 '17 at 18:20

It's very straightforward:

(5*3+2)/1

Or, as pointed out by Cœur, since all numbers need not be used:5*3+2

Why this works:

Calculations are performed in base-7.

• You can remove the /1 as not all symbols are required.
– Cœur
Nov 19 '17 at 10:15

This is the answer, only using 2 of the 4 proposed numbers:

5 * 3 = 23

How come, you say?

we used base 6 calculations

Using concatenation:

25 - 3 + 1

Using numbers more than once, but interesting sequence.

1*2+2*3+3*5 = 23

• Given that the question is unclear as to if we can use each number more than once, this answer is plausible... Nov 17 '17 at 20:55
• fibonacci new world order confirmed? Nov 19 '17 at 2:27

Assuming concatenation is allowed then this is another answer:

13 + 2 * 5

If factorial is allowed:

(5 - 1)! - 3 + 2 = 23

or

5 * (3! - 1) - 2 = 23

or

(2 + 1) * 3! + 5 = 23

or

5! / 3! + 2 + 1

• What's faculty? Nov 18 '17 at 17:41
• I think hoffmale means factorial, or at least that's what I would call the ! operator
– Foon
Nov 18 '17 at 18:35
• @Mazura yeah, i meant factorial... bad/wrong translation from german Nov 18 '17 at 18:36
• "factorial" is "Fakultät" in German, which is basically the same word used for an university "faculty" Nov 18 '17 at 18:43
• factorial was the first solution that came to my mind! +1 Nov 20 '17 at 6:56

(3*2+1)*5 = 23 using HEX

• HEX does not sound like addition, subtraction, multiplication or division. Nov 21 '17 at 8:29
• @boboquack Why can't I use hexadecimal, where the base is 16?
– SJFJ
Nov 21 '17 at 8:40
• Base conversion is a way around literally any of these puzzles, and gets quite boring if done over and over again (a couple of samples). It also doesn't answer the intended question, even if it does answer the literal question. Lastly, I argue that base 16 requires $(3\cdot2+1)\cdot5=23_{16}$, which requires an extra 16. Nov 21 '17 at 8:53
• @boboquack Ok, new to the site so I wasn't aware that it was considered boring.
– SJFJ
Nov 21 '17 at 9:09
• @boboquack, boring or not, the question had better state that the number they want is $23_{10}$ if they are trying to rule out creative solutions.
– NH.
Nov 21 '17 at 18:35

Dr Xorile has determined that there is no solution to the problem as stated. So all that remains are out-of-the-box solutions. Some good approaches have already been presented, treating "+" as string concatenation, and changing bases among the best.

Here's what might be jokingly termed a statistician's approach:

You could attempt the question twice and take the average:

• 5(3+1)+2 = 22
• 5(3+2)-1 = 24

Average = $\frac{22+24}{2}$ = 23.

All conditions are fulfilled on each attempt. :D

• so fuzzy math ?
– NH.
Nov 21 '17 at 18:34
• @NH. All crisp. Stochastic, maybe. :) Nov 21 '17 at 22:44

If we can use a number twice, then:

(2+2)*5+3 = 23.
OR: (2+2)*3*2*1-5+(2*2)

• in that case, how about 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 23? Nov 20 '17 at 14:52
• @DanielVestøl My answer was before the OP edited the question
– Sid
Nov 20 '17 at 15:21

(5^2)-3+1

Squaring a number is the same as multiplying it by itself, so this counts in my book.

• I don't think that was the OP's intent. All other operations can be reduced to simple addition, and by breaking it down to this you are saying 5*5 which gives two 5s. If the OP says we can use the number twice, this works. Nov 17 '17 at 20:53
• Basically the same as @DrXorile's answer. Nov 17 '17 at 21:12

Perhaps,

Round of (51/2) - 3

That is

26 - 3 to fetch 23

Of course, this involves concatenation.

• directly "23" is simpler
– Cœur
Nov 19 '17 at 10:25