The ceiling is breaking

Algebra Level 5

⌈ n 2 ⌉ + ⌈ n 3 ⌉ + ⌈ n 6 ⌉ = n + 1 \large \left \lceil \dfrac n2 \right \rceil + \left \lceil \dfrac n3 \right \rceil + \left \lceil \dfrac n6 \right \rceil = n+1

How many positive integers 1 ≤ n ≤ 100 1\leq n \leq100 satisfy the equation above?

Notation : ⌈ ⋅ ⌉ \lceil \cdot \rceil denotes the ceiling function .

Inspiration .


The answer is 67.

Harsh Shrivastava by Harsh Shrivastava , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Aakash Khandelwal
May 22, 2016

Let's see how you can avoid breaking of your ceiling😉

We know every positive integer is either of these 6 forms

6 k , 6 k + 1 , 6 k + 2 , 6 k + 3 , 6 k + 4 , 6 k + 5 6k , 6k+1, 6k+2 , 6k+3, 6k+4 , 6k+5 .

∀ k ∈ W \forall k \in W

Now putting each of these one by one we find only last four are possible.

Thus total such values are 67 \boxed{67} .

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Yup, good standard approach +(0!).

Harsh Shrivastava - 5 years ago

There was a typo in 2nd line that every real no. exists in 5 forms.

No one pointed out.

I have amended now.

Aakash Khandelwal - 5 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...