2
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How do you form the number 1000 with the digits, 1,2,3 and 4?

The mathematical operations that can be are addition, subtraction, multiplication, division, parentheses, percentages and factorials.

Digits should not be repeated.

You can combine digits as well, example 12+34,etc.

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  • $\begingroup$ Can these digits be repeated, or should they be used only once each? $\endgroup$ – Kritixi Lithos Jun 13 '15 at 6:30
  • $\begingroup$ Why are you planning to cheat? $\endgroup$ – GohPiHan Jun 14 '15 at 5:11
  • $\begingroup$ Who said I am cheating and how am I cheating? $\endgroup$ – Swapnil Das Jun 14 '15 at 9:13
  • $\begingroup$ If you are referring to the problem, I failed to solve that long ago. And by the way, you asked the keep the numbers in order and I didn't. I wanted to learn other approaches if they are not in order. So keep in mind these conditions and then comment. $\endgroup$ – Swapnil Das Jun 14 '15 at 9:17
  • $\begingroup$ I wanted to know if there is any other approach if the no.s are not kept in order. BTW I failed in the problem long ago. So mind your comment. $\endgroup$ – Swapnil Das Jun 14 '15 at 9:22
6
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You can do it without repeating numbers, assuming you may arrange order of operations as you wish:

$\frac{4 \times 3 - 2 }{ 1\%}$

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  • $\begingroup$ Missed you by secs.. $\endgroup$ – Sharad Gautam Jun 13 '15 at 7:19
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Note: Puzzle poser, can exponentiation be used, as Charles29020 used in his attempt? I asked, because no one addressed that part of his solution, just the 1,000 versus the 10,000 part that I can see. The instructions do not mention exponentiation at all. I have other potential solutions that use exponentiation.

. .

$\dfrac{3*2 + 4}{1\%}$

$\dfrac{14 + 3!}{2\%}$

$\dfrac{4! - 1 - 3}{2\%}$

$\dfrac{(1 + 3)! - 4}{2\%}$

$\dfrac{(1 + 2)! + 4!}{3\%}$

$\dfrac{3! + 4!}{(1 + 2)\%}$

$\dfrac{4 + 3!}{(2 - 1)\%}$

$\dfrac{(3!/2)! + 4}{1\%}$

$\dfrac{(1 + 4)!}{(2*3!)\%}$

$\dfrac{2(3! - 1)!}{(4!)\%}$

$\dfrac{(3!)!}{4!*(1 + 2)\%}$

$\dfrac{((1 + 2)!)!}{(4!*3)\%}$

$\dfrac{4!/3 + 2}{1\%}$

$\dfrac{4(3! - 1)}{2\%}$

$\dfrac{2}{4\%*(3! - 1)\%}$

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-4
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$((3^2)+1) ^ 4$

In words, (3 squared) + 1 to the 4th power

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  • 2
    $\begingroup$ Isn't it 10000? $\endgroup$ – leoll2 Jun 13 '15 at 17:01
  • $\begingroup$ The expression submitted is correct. The first term (3^2) = 9. As 3 x 3 is typically referred to as "3 squared", using (3^2) meets the criteria of using the numbers 2 and 3. Adding the result of the previously mentioned expression to 1, the third of 4 digits required, you have a result of 10. When you multiply 10 to itself 4 times, your result is 1000. Each digit; 1,2,3,4 has been used and the calculation submitted yields a result yields a result of 1000, the desired result. I can write this out and submit a written copy of my logic if my explanation needs further clarification. $\endgroup$ – Charles29020 Jun 13 '15 at 20:44
  • $\begingroup$ I'm pretty sure that 10^4=10000, am I wrong? $\endgroup$ – leoll2 Jun 14 '15 at 7:02
  • $\begingroup$ You are not, 10^4 = 10,000 $\endgroup$ – FunkTheMonk Jun 15 '15 at 8:00

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