Consecutive integers and boxes

Logic Level 2

$\large 1 \; \underbrace{\square \; 2 \; \square \; \cdots \; \square \; n \; \square}_{n \text{ number of }\square\text{'s}} \; (n+1) = n+ 2$

What is the minimum value of the positive integer $n>1$ such that we can fill in all the boxes above by using at least one of the four mathematical operators ( $+, -, \times , \div$ ) and the equation holds true?

Note : Order of operations (BODMAS) applied.

The answer is 4.

1 solution

Rajen Kapur
Jul 1, 2016

As $1 + 2 * 3 + 4 - 5 = 6$ , $n = 4$ .

Yup. For completeness, you should show that $n=2$ and $n=3$ yields no solution.

Pi Han Goh - 4 years, 11 months ago

Apart from what Pihan said , is there any way of proving also which sequences don't have a solution ?

A A - 4 years, 11 months ago

1 x 2 + 3 = 5. n=3.

David Hiskiyahu - 4 years, 6 months ago

And n=2 not 3 in your situation

Marc De Ladurantaye - 2 years, 3 months ago

You're wrong. You need to use BODMAS.

Pi Han Goh - 4 years, 6 months ago

Consecutive integers and boxes

The answer is 4.

1 solution

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