Eight light bulbs numbered 1 through 8 are arranged in a circle as sho

Question

https://gmatclub.com/forum/download/file.php?id=47607 Eight light bulbs numbered 1 through 8 are arranged in a circle as shown above. The bulbs are wired so that every third bulb, counting in a clockwise direction, flashes until all bulbs have flashed once. If the bulb numbered 1 flashes first, which numbered bulb will flash last? A. 2 B. 3 C. 4 D. 6 E. 7 PS56602.01 Quantitative Review 2020 NEW QUESTION 2019-04-26_1808.png

750300 · Accepted Answer

Ans: just deduct 3 places from First bulb, because flashing will stop before it reaches the first bulb.
( save time if lot of numbers are present)

MahmoudFawzy · Answer

The sequence of flashing would be:
1 -> 4 -> 7 -> 2 -> 5 -> 8 -> 3 -> 6

so  D

Archit3110 · Answer

logical question
draw fig and solve
we see bulb 6 flashes last
IMO D

MahmoudFawzy · Answer

I hope to help you with this.
In general, every third means that the difference is 3

on a number line, every third number is 3,6,9,..... --> if the initial integer is 0
but if the initial integer is 1, then the sequence is 4,7,10,...

if you assume a sequence of 1,3,6,9,.. ---> it will be an incorrect sequence as the difference between the second and the first is not equal the difference between the third and the second.
if you assume a sequence of 1,3,5,... ---> it will be a sequence of every second number (as the difference is 2)

in other words: if the bulb no. 1 is not the first to flash, then the first one to flash will be bulb no. 3 followed by 6.

GMATinsight · Answer

I would like to do it in manual way

Lighting happens like this

1, 4, 7, 2, 5, 8, 3, 6

i.e. Bulb 6 flashes last

Answer: Option D

energetics · Answer

"The bulbs are wired so that  every third bulb , counting in a clockwise direction, flashes until all bulbs have flashed once. If the bulb numbered 1 flashes first, which numbered bulb will flash last?"

Every third bulb means that the number 1 bulb would be counted as 0... and we would move 3 spaces to number 4 and repeat until none are left. In other words, Last-First + 1 ... would be 4th bulb - 1st bulb + 1 = 3 spaces moved. Repeat until finished. 
I drew a diagram and just crossed out the successive bulbs.

EgmatQuantExpert · Answer

Solution 
  Given  
In this question, we are given
 • Eight light bulbs numbered 1 through 8 are arranged in a circle, as shown in the diagram.
• The bulbs are wired so that every third bulb, counting in a clockwise direction, flashes until all bulbs have flashed once. 
• The bulb numbered 1 flashes first.

To Find  
We need to determine
 • The number of the bulb that will flash last.

Approach & Working  
As every third bulb is flashing, starting from bulb 1, we can determine the next bulbs as follows
 • First: Bulb 1
• Second: Bulb 4
• Third: Bulb 7
• Fourth: Bulb 2
• Fifth: Bulb 5
• Sixth: Bulb 8
• Seventh: Bulb 3
• Eighth: Bulb 6

As we can see, all the eight bulbs are flashing at least once, and the last one to flash is bulb 6.

Hence, the correct answer is option D.

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-WP-e1533292133520.png

ScottTargetTestPrep · Answer

So the bulbs will flash in the following order:

1, 4, 7, 2, 5, 8, 3 ,6

Answer: D

altairahmad · Answer

chetan2u   VeritasKarishma   Bunuel   generis

How is bulb numbered '3' not the 3rd bulb.

1st (Bulb 1), 2nd (Bulb 2), 3rd (Bulb 3).

I have seen MahmoudFawzy's explanation but e.g given a sequence 123456789 and asked to pick every 3rd number, I will pick 3,6 and 9. Will that be wrong ?

Is this something that one is tested on and is expected to be clear (provided that there is a 'standard') ? Shouldn't GMAT not give answer choices which work with both approaches (i.e in this case choice D and E)

Will appreciate some help.

chetan2u · Answer

Yes if you are asked to pick every third number in 123456789 will be 3,6,9. But if you are told that you pick 1 and then every third, the numbers would be 1,4,7.

KarishmaB · Answer

Every third member means "out of every 3, only 1". It doesn't mean that it must be a multiple of 3. 
Which one, depends on which you start with. If I pick 1, then I will skip 2 and 3 and the next third member will be 4.
If I pick 2 first, I will skip 3 and 4 and the next third member will be 5. If I skip 1 and 2 and pick 3 first, then I will skip 4 and 5 and pick 6th.

goaltop30mba · Answer

While it is important to understand what every third means, this question can simply be done basis pure logic.

Let us consider that the third bulb is the bulb numbered 3. This way, we will have 1–3–5–7–1... and then again the same sequence. BUT, the question says “every third bulb flashes *until all bulbs have flashed once*”. This part of the question makes the above obtained sequence or whatever that is incorrect because we will have only 1357 bulbs as flashing bulbs. Therefore the third bulb must be the bulb numbered 4. Now if we go in the required clockwise direction, we will se that every bulb gets to flash once.

Hope I am making sense.

Still it is important to understand what every third really means, and the experts, as always, have done a pretty good job, explaining the same.

Posted from my mobile device

MHIKER · Answer

Starting from bulb  1  Every third bulb flashes, What will be the last bulb when all bulbs will light up once?

First flashed bulb number  = 1 
Second flashed bulb number  = 4 
Third flashed bulb number  = 7 
Fourth flashed bulb number  = 2 
Fifth flashed bulb number  = 5 
Sixth flashed bulb number  = 8 
Seventh flashed bulb number  = 3 
Eighth flashed bulb number  = 6

Now we see all bulbs flashed once, so the last bulb is  6

The answer is D

Hovkial · Answer

OFFICIAL GMAT EXPLANATION

Arithmetic Properties of integers

The easiest way to do this problem might be by just counting every third bulb going clockwise around the circle starting at Bulb 1, which flashes, skipping 2 bulbs and getting to Bulb 4, which flashes, skipping 2 bulbs and getting to Bulb 7, which flashes, skipping 2 bulbs and getting to Bulb 2, which flashes, skipping 2 bulbs and getting to Bulb 5, which flashes, skipping 2 bulbs and getting to Bulb 8, which flashes, skipping 2 bulbs and getting to Bulb 3, which flashes, and finally skipping 2 bulbs and getting to Bulb 6, which flashes. Now, all 8 bulbs have flashed once and the last one to flash was Bulb 6.

sriharsha4444 · Answer

A more formulaic approach for example if the question were if there are 100 such bulbs, which bulb would be fleshing last when the first bulb to flash is 1?

hinges on the idea that 0 ≤ rem < Divisor.

Since there are 8 bulbs -> we can have (0 to 7) generated from remainder theory and add 1 to it.

So something (x % 8) + 1 -> would give a value between 1 to 8

Then we have a condition in the question, every third bulb -> so (3n % 8) + 1 would fit in.

To find the last bulb to light on. We need to find when will the bulb 1 again turn on. Since 3 and 8 doesn't factor out, 1 turns on when n = 8. 
So n=7 would be when the last bulb will turn on -> (3 * 7 % 8) + 1 -> (5) + 1 -> 6 is the last bulb.

You can see how this formula based approach can easily scale to say 1000 bulbs. Then the formula becomes (3n % 1000) + 1

1st bulb turns on again only when n=1000. So when n = 999 is when the last bulb turns on.

KarishmaB   Bunuel  is my approach correct ?

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Eight light bulbs numbered 1 through 8 are arranged in a circle as sho

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