6
$\begingroup$

Not sure if this is too hard, but it's a more or less simple math puzzle. Use the numbers

$1,2,3,4,5,6,7,8,9$

to get an answer of 1150. You can use the operators

$+,-,*,/,\hat{}$

You can use each number only once, each operator can be used as many times as you want. You can also round to the nearest 10. You also cannot put the numbers together(e.g. you can't do 1234 or such, each has to be a separate number). In addition, all numbers must be used.

Note: This is solvable, no lateral thinking needed. Pure calculations.

$\endgroup$
  • 3
    $\begingroup$ What is the operator $**$? $\endgroup$ – Rand al'Thor Oct 4 '16 at 0:45
  • $\begingroup$ can we use ( )? $\endgroup$ – gstats Oct 4 '16 at 1:03
  • $\begingroup$ @randal'thor it's the exponent $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:10
  • $\begingroup$ @gstats yeah that's fine $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:10
  • 1
    $\begingroup$ There are so many possible answers, how are you going to judge? $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 1:17

18 Answers 18

11
$\begingroup$

My answer is

$(((7+3)*4)+6)*(5**2)+(9-(1+8)) = 1150$

Details:

$(((7+3)*4)+6)*(5**2)+(9-(1+8))= ((10*4)+6)*25 + 0 = 46*25 = 1150$

$\endgroup$
  • $\begingroup$ Wow, that's impressive. I actually got an answer of around 1154, which then rounded to 1150. This definitely fits, though. $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:14
74
$\begingroup$

Here's an answer which

doesn't change the order of the digits, and doesn't even use any operations other than $\times$ and $+$:

$1\times (2+3)\times (4+5+(6+7)\times (8+9))$

$\endgroup$
  • 6
    $\begingroup$ The best as of now :) $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 8:11
22
$\begingroup$

I have a solution that doesn't change the order of the digits.

$1+2^3+4^5+6\times 7+8\times 9$

and round to the nearest multiple of 10.

$\endgroup$
  • 2
    $\begingroup$ That is...really impressive. mind explodes $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:58
  • $\begingroup$ Is order of operations followed, or do you go from left to right? $\endgroup$ – OldBunny2800 Oct 4 '16 at 2:58
  • $\begingroup$ @OldBunny2800 Normal operator precedence :-) $\endgroup$ – GOTO 0 Oct 4 '16 at 3:01
10
$\begingroup$

Bit late, just for fun

Prime factoring gives $1150=2\times5^2\times23$
So sub in for those four values $\frac{6}{3}\times5^\frac{4}{2}\times(9+8+7-1)=1150$

$\endgroup$
  • 1
    $\begingroup$ I always upvote posts when they give a correct answer, even if it's late. :P I'm actually out of daily votes right now, but I will upvote it in 7 hours $\endgroup$ – TrojanByAccident Oct 4 '16 at 16:07
  • $\begingroup$ This is cool. @Saiid I'll do the honours. $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 18:13
  • $\begingroup$ @AbhirathMahipal coolio. $\endgroup$ – TrojanByAccident Oct 4 '16 at 18:16
9
$\begingroup$

Answer:

$(6**4)-(2**7)-9-8-5+3+1$

This gives

$1296-128-17-1=1150$, exactly as required.

I found this by

more or less using a greedy algorithm: $6^4$ was the closest perfect power I found to $1150$, and subtracting $1150$ from it gives $146$, to which the closest perfect power I found was $2^7$; then after that it's a simple addition/subtraction game.

$\endgroup$
  • $\begingroup$ Nice answer. I didn't know the site accepted more than 1 answer. $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 1:21
  • $\begingroup$ Thumbs up for the exact answer. He forced me to search for the perfect equation. Likewise with you @gstats $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 1:26
  • $\begingroup$ I'll give a +1 here because it is cleaner than gstats's, however they did give a correct answer before you. Granted, this is partly my fault for not saying exactly what I was looking for the first time. $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:31
  • $\begingroup$ @Saiid Just edited to add some explanation of my method, how I actually found this solution. $\endgroup$ – Rand al'Thor Oct 4 '16 at 1:31
  • $\begingroup$ @randal'thor yup, I saw it. It was interesting. $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:32
6
$\begingroup$

$(1 \times 2^3 \times 4 \times 5 - 6) \times 7 + 8 \times 9$
and
($-1 + 2 \times 3 \times 4) \times 5 \times 6 \times \frac{7 + 8}{9}$

also preserve the order of digits, and give exactly 1150

$\endgroup$
4
$\begingroup$

One answer is

$1234+56/7-89$.

This gives

$1153$, and you said rounding to the nearest 10 was allowed.

Another answer is

$6789/5-213+4$.

This gives

$1148.8$, which again is correct when rounding to the nearest 10.

$\endgroup$
  • $\begingroup$ ah i didn't think about a solution like that. i should have put they cannot be put together $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:11
  • $\begingroup$ @Saiid OK, added a better answer now. $\endgroup$ – Rand al'Thor Oct 4 '16 at 1:21
4
$\begingroup$

I came up with a solution based on breaking 1150 into

$900 + 250$, or $30 ^ 2 + 2 * 5 ^ 3$

From that it was pretty easy to use each digit once

$2 *5 ^ 3 + (6 * (9 - 4))^{8 - 7 + 1}$

$\endgroup$
  • 1
    $\begingroup$ nice! will upvote in 5 hours after my daily count resets $\endgroup$ – TrojanByAccident Oct 4 '16 at 18:50
4
$\begingroup$

It's now updated to meet the edited question :)
Solution

$2^7 * (1 + 8 + 4 + 5 - 9) -(6/3) = 1150$

@MackTuesday Thanks for pointing out the huge calculation mistake in my previous answer, I've gotten rid of it.

$\endgroup$
  • $\begingroup$ edited to include something I should have put before that makes this invalid $\endgroup$ – TrojanByAccident Oct 4 '16 at 1:22
  • $\begingroup$ Um... 8^4 = 4096. $\endgroup$ – MackTuesday Oct 4 '16 at 17:46
  • $\begingroup$ Damn. I missed it. I'll have to edit my answer then. No longer useful. $\endgroup$ – Abhirath Mahipal Oct 4 '16 at 18:19
4
$\begingroup$

Here are 100 answers calculated using this little Haskell program that I wrote:

$(9 - (6 - 8) \cdot 7) \cdot ((1 + 2 + 3 + 4) \cdot 5)$
$((8 - 6) \cdot 7 + 9) \cdot ((1 + 2 + 3 + 4) \cdot 5)$
$(1 + 2 + 3 + 4 + 6) \cdot 8 \cdot 9 + 5 - 7$
$(1 + 2 + 3 + 4 + 6) \cdot 8 \cdot 9 - 7 + 5$
$5 - (7 - (1 + 2 + 3 + 4 + 6) \cdot 8 \cdot 9)$
$5 - 7 + (1 + 2 + 3 + 4 + 6) \cdot 8 \cdot 9$
$(1 + 2 + 3 + 4 + 6) \cdot 8 \cdot 9 - (7 - 5)$
$9 \cdot ((1 + 2 + 3 + 4 + 6) \cdot 8) + (5 - 7)$
$9 \cdot ((1 + 2 + 3 + 4 + 6) \cdot 8) - (7 - 5)$
$(1 + 2 + 3 + 4 + 6) \cdot 9 \cdot 8 - 7 + 5$
$5 - (7 - (1 + 2 + 3 + 4 + 6) \cdot 9 \cdot 8)$
$(1 + 2 + 3 + 4 + 6) \cdot 9 \cdot 8 + 5 - 7$
$(1 + 2 + 3 + 4 + 6) \cdot 9 \cdot 8 - (7 - 5)$
$5 - 7 + (1 + 2 + 3 + 4 + 6) \cdot 9 \cdot 8$
$8 \cdot ((1 + 2 + 3 + 4 + 6) \cdot 9) + (5 - 7)$
$8 \cdot ((1 + 2 + 3 + 4 + 6) \cdot 9) - (7 - 5)$
$9 \cdot 8 \cdot (1 + 2 + 3 + 4 + 6) - (7 - 5)$
$9 \cdot (1 + 2 + 3 + 4 + 6) \cdot 8 - (7 - 5)$
$8 \cdot (1 + 2 + 3 + 4 + 6) \cdot 9 - (7 - 5)$
$9 \cdot 8 \cdot (1 + 2 + 3 + 4 + 6) + (5 - 7)$
$9 \cdot (1 + 2 + 3 + 4 + 6) \cdot 8 + (5 - 7)$
$8 \cdot (1 + 2 + 3 + 4 + 6) \cdot 9 + (5 - 7)$
$8 \cdot 9 \cdot (1 + 2 + 3 + 4 + 6) - 7 + 5$
$5 - (7 - 8 \cdot 9 \cdot (1 + 2 + 3 + 4 + 6))$
$8 \cdot 9 \cdot (1 + 2 + 3 + 4 + 6) + 5 - 7$
$8 \cdot 9 \cdot (1 + 2 + 3 + 4 + 6) - (7 - 5)$
$5 - 7 + 8 \cdot 9 \cdot (1 + 2 + 3 + 4 + 6)$
$(1 + 2 + 3 + 4 + 6) \cdot (8 \cdot 9) + (5 - 7)$
$(1 + 2 + 3 + 4 + 6) \cdot (8 \cdot 9) - (7 - 5)$
$(6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9) + (5 - 7)$
$(6 + (1 + 2 + 3 + 4)) \cdot 9 \cdot 8 + (5 - 7)$
$(6 + (1 + 2 + 3 + 4)) \cdot 8 \cdot 9 + (5 - 7)$
$9 \cdot 8 \cdot (6 + (1 + 2 + 3 + 4)) + (5 - 7)$
$(6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9) - (7 - 5)$
$(6 + (1 + 2 + 3 + 4)) \cdot 9 \cdot 8 - (7 - 5)$
$(6 + (1 + 2 + 3 + 4)) \cdot 8 \cdot 9 - (7 - 5)$
$9 \cdot 8 \cdot (6 + (1 + 2 + 3 + 4)) - (7 - 5)$
$(6 \cdot 9 - 8) \cdot (5 \cdot 7 - (1 + 2 + 3 + 4))$
$(8 - 6 \cdot 9) \cdot (1 + 2 + 3 + 4 - 5 \cdot 7)$
$(5 \cdot 7 - (1 + 2 + 3 + 4)) \cdot (9 \cdot 6 - 8)$
$(1 + 2 + 3 + 4 - 5 \cdot 7) \cdot (8 - 9 \cdot 6)$
$(8 - 9 \cdot 6) \cdot (1 + 2 + 3 + 4 - 5 \cdot 7)$
$(9 \cdot 6 - 8) \cdot (5 \cdot 7 - (1 + 2 + 3 + 4))$
$(9 - (6 - 8) \cdot 7) \cdot (1 + 2 + 3 + 4) \cdot 5$
$(9 - (6 - 8) \cdot 7) \cdot 5 \cdot (1 + 2 + 3 + 4)$
$(1 + 2 + 3 + 4) \cdot 5 \cdot (9 - (6 - 8) \cdot 7)$
$(9 - (6 - 8) \cdot 7) \cdot (5 \cdot (1 + 2 + 3 + 4))$
$(9 - 7 \cdot (6 - 8)) \cdot (5 \cdot (1 + 2 + 3 + 4))$
$((8 - 6) \cdot 7 + 9) \cdot (1 + 2 + 3 + 4) \cdot 5$
$((8 - 6) \cdot 7 + 9) \cdot 5 \cdot (1 + 2 + 3 + 4)$
$(1 + 2 + 3 + 4) \cdot 5 \cdot ((8 - 6) \cdot 7 + 9)$
$(9 + (8 - 6) \cdot 7) \cdot (5 \cdot (1 + 2 + 3 + 4))$
$(7 \cdot (8 - 6) + 9) \cdot (5 \cdot (1 + 2 + 3 + 4))$
$(7 \cdot 5 - (1 + 2 + 3 + 4)) \cdot (6 \cdot 9 - 8)$
$(1 + 2 + 3 + 4 - 7 \cdot 5) \cdot (8 - 6 \cdot 9)$
$(7 \cdot 8 + (6 \cdot 9 + 5)) \cdot (1 + 2 + 3 + 4)$
$(8 \cdot 7 + 5 + 6 \cdot 9) \cdot (1 + 2 + 3 + 4)$
$(8 \cdot 7 + 6 \cdot 9 + 5) \cdot (1 + 2 + 3 + 4)$
$(5 + 6 \cdot 9 + 8 \cdot 7) \cdot (1 + 2 + 3 + 4)$
$(6 \cdot 9 - 8) \cdot (7 \cdot 5 - (1 + 2 + 3 + 4))$
$(8 - 6 \cdot 9) \cdot (1 + 2 + 3 + 4 - 7 \cdot 5)$
$(6 \cdot 9 + (7 \cdot 8 + 5)) \cdot (1 + 2 + 3 + 4)$
$(9 \cdot 6 + 5 + 7 \cdot 8) \cdot (1 + 2 + 3 + 4)$
$(9 \cdot 6 + 7 \cdot 8 + 5) \cdot (1 + 2 + 3 + 4)$
$(5 + 7 \cdot 8 + 9 \cdot 6) \cdot (1 + 2 + 3 + 4)$
$(6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9) - 7 + 5$
$5 - (7 - (6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9))$
$(6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9) + 5 - 7$
$5 - 7 + (6 + (1 + 2 + 3 + 4)) \cdot (8 \cdot 9)$
$8 \cdot 9 \cdot (6 + (1 + 2 + 3 + 4)) + (5 - 7)$
$8 \cdot 9 \cdot (6 + (1 + 2 + 3 + 4)) - (7 - 5)$
$(8 - 6 \cdot 9) \cdot ((1 + 2 + 3 - 4 - 7) \cdot 5)$
$(8 - 9 \cdot 6) \cdot (1 + 2 + 3 - 4 - 7) \cdot 5$
$(8 - 9 \cdot 6) \cdot 5 \cdot (1 + 2 + 3 - 4 - 7)$
$(1 + 2 + 3 - 4 - 7) \cdot 5 \cdot (8 - 9 \cdot 6)$
$(8 - 9 \cdot 6) \cdot (5 \cdot (1 + 2 + 3 - 4 - 7))$
$(6 \cdot 9 - 8) \cdot ((7 - (1 + 2 + 3 - 4)) \cdot 5)$
$(9 \cdot 6 - 8) \cdot (7 - (1 + 2 + 3 - 4)) \cdot 5$
$(9 \cdot 6 - 8) \cdot 5 \cdot (7 - (1 + 2 + 3 - 4))$
$(7 - (1 + 2 + 3 - 4)) \cdot 5 \cdot (9 \cdot 6 - 8)$
$(9 \cdot 6 - 8) \cdot (5 \cdot (7 - (1 + 2 + 3 - 4)))$
$(5 \cdot 8 + 6) \cdot ((1 + 2 + 3 - 4) \cdot 9 + 7)$
$(6 \cdot 7 + 8) \cdot ((1 + 2 + 3 - 4) \cdot 9 + 5)$
$(7 \cdot 8 - 6) \cdot ((1 + 2 + 3 - 4) \cdot 9 + 5)$
$(5 + (1 + 2 + 3 - 4) \cdot 9) \cdot (8 \cdot 7 - 6)$
$(8 \cdot 7 - 6) \cdot (5 + (1 + 2 + 3 - 4) \cdot 9)$
$(7 + (1 + 2 + 3 - 4) \cdot 9) \cdot (8 \cdot 5 + 6)$
$(6 + 8 \cdot 5) \cdot (7 + (1 + 2 + 3 - 4) \cdot 9)$
$((1 + 2 + 3 - 4) \cdot 9 + 5) \cdot (7 \cdot 6 + 8)$
$(8 + 7 \cdot 6) \cdot (5 + (1 + 2 + 3 - 4) \cdot 9)$
$(9 / (1 + 2 + 3 - 4) \cdot 8 \cdot 7 + 6) \cdot 5$
$(9 / (1 + 2 + 3 - 4) \cdot 7 \cdot 8 + 6) \cdot 5$
$(8 \cdot 7 \cdot (9 / (1 + 2 + 3 - 4)) + 6) \cdot 5$
$((1 + 2 + 3 - 4) \cdot 9 + 7) \cdot (5 \cdot 8 + 6)$
$(5 \cdot 8 + 6) \cdot (9 \cdot (1 + 2 + 3 - 4) + 7)$
$(7 + 9 \cdot (1 + 2 + 3 - 4)) \cdot (6 + 5 \cdot 8)$
$(9 \cdot (1 + 2 + 3 - 4) + 5) \cdot (6 \cdot 7 + 8)$
$(8 + 6 \cdot 7) \cdot (9 \cdot (1 + 2 + 3 - 4) + 5)$
$(5 + 9 \cdot (1 + 2 + 3 - 4)) \cdot (8 + 6 \cdot 7)$
$(7 - (1 + 2 + 3 - 4)) \cdot ((6 \cdot 9 - 8) \cdot 5)$

Feel free to edit this answer if it feels too long, and to run the program changing take 100 with whatever number of answers you might like to find out.

$\endgroup$
  • $\begingroup$ Well...that's one way of doing it. What language is that btw? I can't read it $\endgroup$ – TrojanByAccident Oct 5 '16 at 16:30
  • $\begingroup$ The language is Haskell, it's a functional programming language so much closer to math than the average, and good for expressing this sort of problems. This puzzle is combinatorial so I thought it would be cool to find a solution using programming :) $\endgroup$ – hoheinzollern Oct 5 '16 at 18:55
  • $\begingroup$ Great one! Welcome to puzzling $\endgroup$ – GOTO 0 Oct 5 '16 at 20:03
  • 1
    $\begingroup$ I see why this question is on hold... $\endgroup$ – Buffer Over Read Oct 5 '16 at 22:34
  • 1
    $\begingroup$ Is it not fair to find the solution programmatically? I'm new to forum. I just saw the question and thought it was interesting to solve this way. The program is not completely trivial to formulate. $\endgroup$ – hoheinzollern Oct 6 '16 at 16:47
4
$\begingroup$

My solution:

$2 * (1 + 4) * 5 * (7 * 8 - 3 * 9 - 6) = 1150$

$\endgroup$
2
$\begingroup$

$(9+8+6)*5*(7+3)*(4-2-1)$

The trick I used was to express 1150 as

a product of factors, and then try and find individual factors. Here I saw $1150 = 23*5*10*1$.

$\endgroup$
  • $\begingroup$ interesting. lel $\endgroup$ – TrojanByAccident Oct 4 '16 at 14:26
2
$\begingroup$

Here's mine, uses every number and doesn't round.

$(((9^4)/3)-1)/2 + (7*8) + 6 - 5 = 1150$

$\endgroup$
2
$\begingroup$

$$1 + 2 + 3 \times 6 + 4 \times 5 \times 7 \times 8 + 9 = 1150$$

Closed in on it as follows:

$$\begin{align}1*2*3*4*5*6*7*8*9 &= 62880 \\1*2*3*4*5/6*7*8*9 &= 10080 \\1*2/3*4*5/6*7*8*9 &= 1120 \\1*2*4*5/6*7*8*3+9 &= 1129 \\1*4*5*7*8+9+6+2+3 &= 1140 \\1*4*5*7*8+9+6+2*3 &= 1141 \\1*4*5*7*8+9+6*2+3 &= 1144 \\1*4*5*7*8+9+6*3+2 &= 1149\end{align}$$

$\endgroup$
2
$\begingroup$

By now it's been done to death, but here's my variant. I tried to do it in reverse order of digits, but failed. Each grouping is in reverse order just to sort of succeed. :)

$9\cdot 5+ 8\cdot 7\cdot 6\cdot 3-4+2-1$

With parentheses for clarity.

$(9\cdot 5)+ (8\cdot 7\cdot 6\cdot 3)-(4-2+1)$

I did succeed at keeping it all in one line, so that's a plus.

$\endgroup$
  • $\begingroup$ It has been done to death. lol. I never thought a puzzle of mine would get this popular $\endgroup$ – TrojanByAccident Oct 5 '16 at 13:34
2
$\begingroup$

Here's my attempt

$(8\times7\times5\times4)+(9\times3)+(6-2-1)$

$=1120+27+3$
$=1150$

$\endgroup$
1
$\begingroup$

(9 + 1) * 5 * (((7 * ((3 / ((6 ^ 2 / 4) – 8) ))) +2)

$\endgroup$
1
$\begingroup$

I've got this for an answer:

(9*8*7*2)+(3*(6+5+4-1))

Can also be changed into:

((5+4)*8*7*2)+(3*(6+9-1))

It uses only + - and * .

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.