Does it seem impossible to count?

Logic Level 5

2 2 2 = 2 \LARGE 2 \ \ \ \ \ 2 \ \ \ \ \ 2 \ \ = \ \ 2

Without touching the right side of the equation, can you find the number of ways you can make the equation true? You are only allowed to use fundamental operators like addition, subtraction, multiplication, division and parenthesis.

Details and assumptions :

  • You must use parenthesis.

  • You can use the parenthesis in in two ways, i.e. ( 2 + 2 ) 2 = 2 (2+2)-2=2 and 2 + ( 2 2 ) = 2 2+(2-2)=2 are considered distinct.

  • You can combine digits.

  • BODMAS is applied.

  • You are not allowed to use negative integers, i.e, ( 2 × 2 ) 2 (-2\times-2)-2 or 2 × 2 ( 2 ) -2\times \dfrac2{(-2)} is not allowed


The answer is 10.

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2 solutions

Sravanth C.
Jun 13, 2015

1. ( 2 + 2 ) 2 = 2 2. 2 + ( 2 2 ) = 2 3. ( 2 + 2 ) ÷ 2 = 2 4. ( 2 2 ) + 2 = 2 5. 2 ( 2 2 ) = 2 6. ( 2 ÷ 2 ) × 2 = 2 7. 2 ÷ ( 2 ÷ 2 ) = 2 8. ( 2 × 2 ) 2 = 2 9. ( 2 × 2 ) ÷ 2 = 2 10. 2 × ( 2 ÷ 2 ) = 2 1. \quad (2+2)-2=2 \\ 2. \quad 2+(2-2)=2\\ 3. \quad (2+2)÷2=2\\ 4. \quad (2-2)+2=2\\ 5. \quad 2-(2-2)=2\\ 6. \quad (2÷2)×2=2\\ 7. \quad 2÷(2÷2)=2\\ 8. \quad (2×2)-2=2\\ 9. \quad (2×2)÷2=2\\ 10. \quad 2×(2÷2)=2

Do you think these are the only possible ways? If not, submit the other possible ways.

Moderator note:

Why are these the only ways?

Why can't I use two pairs of parenthesis?

What about ( 2 2 ) 2 (-2*-2)-2 , 2 ( 2 / ( 2 ) ) -2*(2/(-2)) etc.? There are many many more like this, I gave up counting when I realised this. Counting the negatives also will take quite a long time, so I gave up and tried random answers.

Vishnu Bhagyanath - 5 years, 12 months ago

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Thanks sir, I've edited the question accordingly.

Sravanth C. - 5 years, 11 months ago

What 2 pairs of parentheses can we use challenge master?

Parag Zode - 6 years ago

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Yeah. I didn't understand that too.

Sravanth C. - 6 years ago

Good sol'n Sravanth! Done by this way.. Upvoted! ¨ \ddot\smile

Parag Zode - 6 years ago

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Thanks sir! I tried all permutations and found this.

Sravanth C. - 6 years ago

Proper way is to list + × ÷ + - \times \div for (2 2) 2 of 6 among 16, and + × ÷ + - \times \div for 2 (2 2) of 4 among another 16. 6 + 4 = 10. Considered distinct would be 10 or otherwise multiples of 10.

Answer: 10 \boxed{10}

Lu Chee Ket - 5 years, 4 months ago

Can you explain the sixth . It would be half .

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I don't think so. Using BODMAS, we must solve the brackets i.e, 2 2 × 2 = 1 × 2 = 2 \dfrac22 \times2 \\ =1\times2 \\=2

Sravanth C. - 6 years ago

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Yeah , sorry for it , I don't know what I was that time. Might have taken the seventh one in which the bracket is the first 2 . SORRY .

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@Karandeep Singh Ludhar No need to say sorry, mistakes happen. ¨ \huge\ddot\smile

Sravanth C. - 6 years ago
Sungil Cho
Jul 14, 2015

With two 2's we can make: (1) 2-2=0; (2) 2÷2=1; or (3) 2+2=2×2=4.

In case (1), 0+2=2+0=2-0=2 (1×3 ways).

In case (2), 1×2=2×1=2÷1=2 (1×3 ways).

In case (3), 4÷2=4-2=2 (2×2 ways).

In total, we have 10 ways.

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