If x is a positive integer with fewer than 3 digits, what is the

­

The total number of possible values of x is 99 from 1-99 ; (given x can be formed with either one or two digits).

In the equation x*(x+1),
For x,
From 1 to 99, total multiples of 4 is 24
total multiples of 5 is 19

For x+1,
For it to be divisible by 4, the x in the (x+1) needs to be 1 less than 4 (i.e 3,7,11,13 etc.)
From 1 to 99, we have 24 numbers as such for (x+1) to be divisible.

For it to be divisible by 5, the x in the (x+1) needs to be 1 less than 5 (i.e 4,9,14,19 etc.)
From 1 to 99, we have 19 numbers as such for (x+1) to be divisible.

So total probability stands = (24+19+24+19)/99 = 86/99. [which matches one of the answers here]



HOWEVER,
Taking it one step further to precision (which I believe could be the more appropriate answer) -

we need to deduct the double numbers from the total 86 possible outcomes. The double-counted numbers are the common multiples of 4 and 5 - as in multiples of 20 - 20,40,60,80.

So plausible outcome stands - (24+19+24+19-4)/99 = 82/99.

Well, if it were the exam day itself, I'd go with 86/99 being the closest solution.­

­There are other double counted numbers as well in your solution. I will explain only one of that rest you can find on your own after doing observation.
Like here, In this series the number 4, 24, 44.... are double counted with your initial count of 4s.

­If I understood you right, then my explanation would be following -

It infers two plausible events. One is when x is divisible by 4 or 5, and another is when (x+1) is divisible by 4 or 5.

The double count you're stating is not under one uniform event. I stated 2 different events - one for divisibility in the event of x and another for (x+1).­


Also the double count is between multiples of 4 and so of 5. So I'm not really sure how there could be more than the 4 double counts stated.­

Yes, I agree with this. I too ended up getting 82/99 as the answer.

Posted from my mobile device

Multiple of 4 : Every 2 out of 4 till 96 = 48 + 1 (from 99 * 100) = 49

Multiple of 5 : Every 2 out of 5 till 95 = 38 + 1 (from 99 * 100) = 39

Duplicate Counting : Every 4 out of 20 till 100 = 20 - 1 (Since we are not counting 100*101) = 19

Probability = (49 + 39 - 19) / (numbers from 1 to 99) = 69 / 99 (= 23 / 33)

­I did not understand the x+1 part...can you please help

Could you please explain the duplicate counting here?

If we visualise the trend here, x=4 and x=15 will suit our duplicate trend. Then each multiples of 20 will be counted twice. Once as x=19 and x+1=20; Second as x=20 and x+1=21.

So, till x=20, we have 4 pairs - (4,5); (15,16); (19,20); (20,21).

Distribution for the rest of the numbers will be symmetric.
For example, for the next 20 numbers, i.e., till x=40 we have - [(24,25); (35,36); (39,40); (40,41)]

Thus, we get 4 duplicates in every 20 values of x. Extrapolatiing the data, we get 20 values in first 100 values of x, i.e., x=1 to x=100. But x can get only as high as 99, so we need to eliminate 1 duplicate (where x=100).

So, the total duplicates are 20 - 1 = 19.

­i got 67 as total possible outcomes . can u please explain more deeply.

­Here the list of all the numbers that will satisfy the given condition.­

3 4 5 7 8 9 10
11 12 14 15 16 19 20
23 24 25 27 28 29 30
31 32 34 35 36 39 40
43 44 45 47 48 49 50
51 52 54 55 56 59 60
63 64 65 67 68 69 70
71 72 74 75 76 79 80
83 84 85 87 88 89 90
91 92 94 95 96 99

­The answer should be 69/99 or 23/33.

Your method has a couple of issues I think. For the 24 possibilities of (x+1)/4, you end at (95+1)/4 and the 19 possibilities of (94+1). You also have to have (99+1)/4 for both 4s and 5s so your total, without removing duplicates, should be 88.

Then you remove the duplicates multiples of 20 so we are at 84.

Then you remove the duplicates between (x+1)/4 and 5: 15, 35, 55, 75, and 95. Then you do the same for (x+1)/5 and 4: 4, 24, 44, 64, and 84. This brings us to 74.

Next, we remove the duplicates between (x+1)/4 and (x+1)/5 so we get rid of 19, 39, 59, 79, and 99. This eventually gets us to the right answer, 69 (confirmed by doing the problem in Python).

Feels crazy hard for a medium-difficulty GMAT problem. I solved it doing possible numbers 1-10, 11-20, 21-30, etc., and saw that there are 7 possible numbers in each range. Then for the last range, you do 90-99, and that one is 6 so you just do 7*10 - 1.­

­i got the answer to be 81/99 since the pairs (4,5)(24,25)(44,45)(64,65)(84,85) occur twice. Am i correct or am i not understanding the question correctly

This question seems either unnecessarily hard or unusually easy.

This is how I solved it:

4 has 24 multiples between 0 and 99.
5 has 19 multiples between 0 and 99.

some duplicates exist: 20, 40, 60, 80

so we already have 39/99, we can eliminate everything besides D and E (and I'm pretty sure we can eliminate D as well since I'm pretty sure the denominator can't be 100 in this case)

Anyway if we keep going - we notice that if this part (x + 1) is a multiple of 4 or five it should count towards are some as well. So numbers like 3, 7, 11, 15, 17, etc give us a multiple of 4 as well, and there are bunch of these cases from 0 to 99, way more than the 10 we need to eliminate D.

Therefore the answer is E.

This feels like a 600 rated question, but would be a 700+ question if the options weren't so obviously wrong. If you replaced some of the options with numbers like 75/99 and 82/99 it would be much harder, so I really doubt you would see a question like this on the actual GMAT.

Bunuel can you explain the answer please?

Can we get an official explanation?