Gif 2
⌊ x ⌋ + ⌊ 2 x ⌋ + ⌊ 3 2 x ⌋ + ⌊ 3 x ⌋ + ⌊ 4 x ⌋ + ⌊ 5 x ⌋
If 0 ≤ x < 3 , find the total number of distinct values of the expression above.
The answer is 30.
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2 solutions
i did the same but just calculated to 1/2 , it had 5 values By symmetry , the ans should be 30, by the you did u too use wolfram alpha ? @Aniket Sanghi
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and it had 30 solvers uptill now ! i was the 31st , co-incidences !
I didn't use wolfram alpha
Dammit I missed one range. I did similarly btw. Nice Sol.
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S i m i l a r l y , f ( 1 . 4 5 ) S i m i l a r l y , f ( 1 . 9 5 ) S i m i l a r l y , f ( 2 . 4 5 ) = 1 9 , = 2 6 , = 3 5 , f ( 1 . 5 ) = 2 2 . f ( 2 ) = 3 1 . f ( 2 . 5 ) = 3 7 . ⟹ 2 0 , 2 1 A R E m i s s i n g . ⟹ 2 7 , 2 8 , 2 9 , 3 0 A R E m i s s i n g . ⟹ 3 6 I S m i s s i n g .
A n d f ( 2 . 9 5 ) = 4 1 , f ( 3 ) = 4 7 . f ( 2 . 9 9 9 9 9 9 9 9 ) a l s o = 4 1 , ⟹ 4 1 I S t h e l a s t i n t e g e r . S o t o t a l 4 1 + 1 ( f o r 0 ) − 1 2 = 3 0 . N o t e t h a t , m i s s i n g n u m b e r s a r e a t i n t e r v a l s o f 0 . 5 . F o u r e a c h b e f o r e x = 1 a n d b e f o r e x = 2 . O n e e a c h b e f o r e x = 0 . 5 a n d b e f o r e x = 2 . 5 . T w o b e f o r e x = 1 . 5 .
Analysed for 0 to 1
First divide the region into diff ranges to get diff integers for each . Then make these ranges into one specific range series.
Got 10 diff ranges (0,1/5)(1/5,1/4)(1/4,1/3)(1/3,2/5)(2/5,1/2)(1/2,3/5)(3/5,2/3)(2/3,3/4)(3/4,4/5)(4/5,1)
For all these the equation yields different values
By symmetry for 1 to 2 & 2 to 3 also we would get 10 diff values
Hence total is 30 values.......