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For all real numbers, [x] = the greatest integer less . . .

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For all real numbers, [x] = the greatest integer less . . .

by VJesus12 » Thu Jan 18, 2018 7:13 am
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + ... + [28/6]?

A.48
B.50
C.52
D.54
E.56

The OA is the option E.

I need some help here. Experts, how can I solve this PS question? Thanks.
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by EconomistGMATTutor » Thu Jan 18, 2018 7:33 am
Hello Vjesus12.

Let's take a look at your question.

[x] = the greatest integer less than or equal to x, it implies that

- [1/6] to [5/6], the argument is less than 1, hence 5*0=0.
- [6/6] to [11/6], the argument is less than 2 and greater than 1, hence 6*1=6.
- [12/6] to [17/6], the argument is less than 3 and greater than 2, hence 6*2=12.
- [18/6] to [23/6], the argument is less than 4 and greater than 3, hence 6*3=18.
- [24/6] to [28/6], the argument is less than 5 and greater than 4, hence 5*4=20.

Finally, we have to add 0+6+12+18+20=56.

This is why the correct answer is the option [spoiler]E=56[/spoiler].

I hope this answer can help you.

Feel free to ask me again if you have a doubt.

Regards.
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by DrMaths » Thu Jan 18, 2018 7:59 am
"Arguments", as they are called, are divisions of numbers with any remainders ignored.
Hence [5/2] = 2 (as the fractional part or remainder is omitted).

[1/6] + [2/6] + [3/6] + ... + [28/6]?

Considering the above sum, the argument ,An, of each term create...
(1) For numerators < 6, A1 = 6 *0 = 0
(2) For 6<= numerator < 12, A2 = 6 *1=6
(3) For 12<= numerator < 18, A3 = 6 *2=12
(4) For 18<= numerator < 24, A4=6 *3=18
(5) For 24<= numerator < 29, A5=5 *4=20

The Sum of A (1 to 5) = 56
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