From 29 to 27, real quick!
⌊ x + 2 9 8 ⌋ + ⌊ x + 2 9 9 ⌋ + ⌊ x + 2 9 1 0 ⌋ + ⌊ x + 2 9 1 1 ⌋ + ⌊ x + 2 9 1 2 ⌋ = 2 5
Find the sum of all possible values of ⌊ 2 7 x ⌋ .
Notation: ⌊ ⋅ ⌋ denotes the floor function .
The answer is 3324.
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2 solutions
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~~~~~
There are two repeats. Is it because 29-27=2 ? No. it is only coincident.
T h e a b o v e w o u l d n o t c o m p i l e w i t h o u t t h i s ! ! !
I was not able to understand why "And we just need to get the sum of all the natural numbers in this range". Though it is true.
at this moment, let's forget the floor function 2 9 ⋅ 5 x + 5 0 = 2 9 ⋅ 2 5 = 7 2 5 ⇒ 1 4 5 x = 6 7 5 ⇒ x = 1 4 5 6 7 5 = 2 9 1 3 5 , but in this case, substituing in the equation above, we get 2 3 = 2 5 (Remember 2 9 ⋅ 5 = 1 4 5 ) so this solution for x is impossible, so the first step is x ≥ 2 9 1 3 7 (Remember 2 9 ⋅ 5 = 1 4 5 ). Now 2 9 ⋅ 6 = 1 7 4 ⇒ x < 2 9 1 6 2 . The rest is the same as Manuel Kahayon...
We can see that as long as 4 + 2 9 2 1 ≤ x < 5 + 2 9 1 7 , this equality is satisfied.
Now, we can see that 2 7 ( 4 + 2 9 2 1 ) ≤ 2 7 x < 2 7 ( 5 + 2 9 1 7 ) , and taking the floor function of this inequality gives us
⌊ 1 2 7 . 5 5 ⌋ ≤ ⌊ 2 7 x ⌋ ≤ ⌊ 1 5 0 . 8 2 ⌋
Giving us 1 2 7 ≤ ⌊ 2 7 x ⌋ ≤ 1 5 0
And we just need to get the sum of all the natural numbers in this range, i.e.
2 ( 1 5 0 − 1 2 7 + 1 ) ( 1 5 0 + 1 2 7 ) = 3 3 2 4