Question

For, *Studants*

Given below is a very interesting maths problem.
This was set out by a German. It is a little difficult and can help prevent Alzheimer disease.

2 + 2 + 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

*Use any of mathematical sign, wherever you need*

Answer 1

ANSWER:

Given:

• 2 + 2 + 2 = 6,
• 3 _ 3 _ 3 = 6,
• 4 _ 4 _ 4 = 6,
• 5 _ 5 _ 5 = 6,
• 6 _ 6 _ 6 = 6,
• 7 _ 7 _ 7 = 6,
• 8 _ 8 _ 8 = 6,
• 9 _ 9 _ 9 = 6

To Do:

• Fill the blanks with mathematical symbols to make the equations true.

Solution:

We are given that,

⟹ 2 + 2 + 2 = 6

Now, we need to find such mathematical operators such that the following equations hold true.

• 3 _ 3 _ 3 = 6
• 4 _ 4 _ 4 = 6
• 5 _ 5 _ 5 = 6
• 6 _ 6 _ 6 = 6
• 7 _ 7 _ 7 = 6
• 8 _ 8 _ 8 = 6
• 9 _ 9 _ 9 = 6

Now, we will take the each part one by one.

1) 3 _ 3 _ 3 = 6

For this part, there can be 2 answers.

1. We subtract one 3 from the product of other 3s,

⟹ 3 × 3 - 3

So,

⟹ 9 - 3

⟹ 6

2. We subtract two 3s and add the value to the factorial of the third 3,

⟹ 3 - 3 + 3!

So,

⟹ 0 + 6 (3! = 6)

⟹ 6

Hence,

ANSWERS:

• 3 × 3 - 3 = 6
• 3 - 3 + 3! = 6

2) 4 _ 4 _ 4 = 6

For this part, there can be 4 answers.

1. We take the square roots of each of the 4s, and add them,

⟹ √4 + √4 + √4

So,

⟹ 2 + 2 + 2

⟹ 6

2) We take square root of the last 4, and divide the second 4, with the last one. And adding it to the first one,

⟹ 4 + 4 ÷ √4

So,

⟹ 4 + 4 ÷ 2

⟹ 4 + 2

⟹ 6

3. We take the square roots of each of the 4s, and multiply the last two and add them,

⟹ √4 + √4 × √4

So,

⟹ 2 + 2 × 2

⟹ 2 + 4

⟹ 6

4. We take the cube roots of the sum of 2 of the 4s, and add them to the third 4,

⟹ ∛(4 + 4) + 4

So,

⟹ ∛(8) + 4

⟹ 2 + 4

⟹ 6

Hence,

ANSWERS:

• √4 + √4 + √4 = 6
• 4 + 4 ÷ √4 = 6
• √4 + √4 × √4 = 6
• ∛(4 + 4) + 4 = 6

3) 5 _ 5 _ 5 = 6

For this part, there can be 1 answer.

1. We add one 5 to the quotient obtained after dividing the other 5s, we will get,

⟹ 5 ÷ 5 + 5

So,

⟹ 1 + 5

⟹ 6

Hence,

ANSWER:

• 5 ÷ 5 + 5 = 6

4) 6 _ 6 _ 6 = 6

For this part, there can be 3 answers.

1. We subtract two 6s and add the value obtained to the third 6, we get,

⟹ 6 - 6 + 6

So,

⟹ 0 + 6

⟹ 6

2. We multiply two of the 6s, and divide them by the third,

⟹ 6 ÷ 6 × 6

So,

⟹ 1 × 6

⟹ 6

3. We divide the two 6s, and multiply with the third,

⟹ 6 × 6 ÷ 6

So,

⟹ 36 ÷ 6

⟹ 6

Hence,

ANSWERS:

• 6 - 6 + 6 = 6
• 6 ÷ 6 × 6 = 6
• 6 × 6 ÷ 6 = 6

5) 7 _ 7 _ 7 = 6

For this part, there can be 1 answer.

1. We divide the last two 7s and subtract the quotient from the first 7,

⟹ 7 - 7 ÷ 7

So,

⟹ 7 - 1

⟹ 6

Hence,

ANSWERS:

• 7 - 7 ÷ 7 = 6

6) 8 _ 8 _ 8 = 6

For this part, there can be 2 answers.

1. We take the cube roots of each of the 4s, and add them,

⟹ ∛8 + ∛8 + ∛8

So,

⟹ 2 + 2 + 2

⟹ 6

2. We take the cube roots of each of the 4s, and multiply the first two and add the third to them,

⟹ ∛8 × ∛8 + ∛8

So,

⟹ 2 × 2 + 2

⟹ 4 + 2⟹ 6

Hence,

ANSWERS:

• ∛8 + ∛8 + ∛8 = 6
• ∛8 × ∛8 + ∛8 = 6

7) 9 _ 9 _ 9 = 6

For this part, there can be 3 answers.

1. We take square roots of each 9, and subtract one √9 from the product of other √9s,

⟹ √9 × √9 - √9

So,

⟹ 9 - 3

⟹ 6

2. We take square roots of each 9, and subtract two √9s and add the value to the factorial of the third √9,

⟹ √9 - √9 + (√9)!

So,

⟹ 0 + 3!

⟹ 0 + 6 (3! = 6)

⟹ 6

Hence,

ANSWERS:

• √9 × √9 - √9 = 6
• √9 - √9 + (√9)! = 6

There can be more cases, but the complexity level will increase, so i included just the normal ones.