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Let [x] denote the greatest integer less than or equal to x. Find the value of [2/5]+[4/5]+[6/5]+...+[48/5]?
6 Answers
- ubiquitous_phiLv 71 decade agoFavourite answer
Greetings,
We must group the values by multiples of 5
2,4 < 5 so 2*0
6, 8, < 10 so 2*1
10, 12, 14 < 15 so 3*2
16, 18 < 20 so 2*3
20, 22, 24< 25 so 3*4
26, 28 < 30 so 2*5
30, 32, 34 < 35 so 3*6
36, 38 < 40 so 2*7
40, 42, 44 < 45 so 3*8
46, 48 < 50 so 2*9
2(1 + 3 + 5 + 7 + 9) + 3(2 + 4 + 6 + 8)
= 2*25 + 3*20
= 50 + 60
= 110
Regards
- 1 decade ago
If you're interested in just the answer, a program can quickly calculate this for you. If you're interested in how to obtain this answer, I suggest you review arithmethetic series.
--- Program Begin ---
0 + 0 + 1 + 1 + 2 + 2 + 2 + 3 + 3 + 4 + 4 + 4 + 5 + 5 + 6 + 6 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 9
sum = 110
--- Program End ---
Source(s): my programming skills: public class NumberTest { public static void main(String[] args) { int sum = 0; System.out.println("--- Program Begin ---"); for (int i=2; i<=48; i = i+2) { sum += f(i/5); System.out.print( f(i/5) + " + " ); } System.out.println("\n\nsum = " + sum + "\n\n" ); System.out.println("--- Program End ---"); } public static int f(double x) { int k = (int)Math.floor(x); return k; } } - Paschal HLv 61 decade ago
Notice that:
2/5 + 48/5 = 50/5 = 10
4/5 + 46/5 = 50/5 = 10
Get the idea?
I think there are 12 pairs like this, so 12*10 = 120
- cattbarfLv 71 decade ago
You can pair these up with greatest and least which are 2/5 + 48/5 = 10, and then the next pair
4/5 + 46/5 = 10, and keep going.
- vcs7578Lv 51 decade ago
apply the same concept that is applied to find sum of consecutive N natural numbers.
note [4/5]=0, [48/5]=9 etc. etc.
assume the sum is S. write it forward(as presented) and in reverse order.
S=[ 2/5]+[ 4/5]+[ 6/5]..............+[44/5]+[46/5]+[48/5]
S=[48/5]+[46/5]+[44/5]...........+[ 6/5] +[ 4/5] + [2/]
summing the two series
2S=(0+9)=(0+9)+(1+8) +(8+1)+(9+0)+(9+0)
There are 24 items of summation in the series.
2S=24*9
S=12*9=108
- ?Lv 45 years ago
often, [-x] + [x] = -a million if x isn't an integer and 0 if x is an integer. WLOG, we can assume that x is helpful. If x isn't an integer, then there exists an integer n such that n < x < n + a million wherein case, [x] = n. We be conscious that -n - a million < -x < -n. wherein case [-x] = -n - a million. hence [x] + [-x] = -a million whilst x is an integer, then [x] + [-x] = x + -x = 0.