A sequence of numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... n occurs n

Question

A sequence of numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... n occurs n times for 1 ≤ n ≤ 25. For the first 300 numbers in the sequence, what is the least n that is greater than at least 25% of the first 300 numbers in the sequence?

A. 10
B. 11
C. 12
D. 13
E. 14

A. 10
B. 11
C. 12
D. 13
E. 14

purplex · Answer

To get the first 300 numbers, you need to go up to 24. 25% of 24 = 6. So least number = 10

jgagne18 · Answer

I think I did this the long way, so any feedback would be appreciated

25% of 300 would be 75, so we're looking for a number that is greater than all numbers less than the 75th number.

I finished the sequence up to 75 by counting the numbers each time the list expands so: 1, 3, 6, 10, 15 ... 78.

The list expanded 12 times meaning that the number 12 would be in the 67th place to the 78th place. We're looking for the number that's greater than AT LEAST the number in the 75th place. So, I chose D as 13 would be in the 79th place.

Again, this probably isn't the best way to do this so please offer feedback. For reference, I finished this in 2 mins and 17 seconds.

YannCJ · Answer

Please, can someone help me understand the solution?

gmat147 · Answer

for the first 25% of 300 is 75. 
the sequence follows such that the pace of nth integer will be be sum of n integers. 
eg: 4 is at 10th place sum first 4 integers = (4 * 5) /2 = 10

Now, similarly try out the options. 
11 = (11 * 12) /2 = 66 now this should be at least 75
12 = (12 * 13) /2 = 78, No also has values below 75
therefore, answer must be 13 which starts from 79th position. 
Ans is D

gchandana · Answer

25% of 300 is 75, so n should be in a way that it is greater than at least the first 75 numbers.

If we see the sequence, to get to 75 numbers, it is just the sum of the first n positive numbers, but we have to be careful here, as we cannot consider the last n numbers. Why?
Because if we say n is 10, we can't consider the last 10 numbers, as those will all be 10, which is not greater than 10.

So we could modify the formula for the sum of the first n positive numbers
 \frac{n(n+1) }{ 2 }  as

\frac{n(n+1) }{ 2 } - n >= 75

Which gives (n-1)(n) >= 150, so n has to be at least 13.

Hope it helps.

P.S. We could do a brute force approach, like take 10, sum up to 10(excluding 10), gives 45(so n=10, n is greater than only 45 numbers), then add 10, gives 55(n=11, n is greater than only 55 numbers), then add 11, gives 66(n=12, n is greater than only 66 numbers), then add 12, gives 78(n=13, n is greater than 78 numebrs).
But yeah, it depends on which one clicks with you and gives you a confident answer.

Wadree · Answer

Understanding da problem....IS the main problem here......lets get into it......
We can see......1 comes 1 time....2 comes 2 times....3 comes 3 times.....blah blah......
So...first 300 numbers in dis sequence will include one 1....two 2s.....three 3s.....so da 300th number wont be 300.....it will be something far less....but wont worry bout that.....it says we gotta take a least number that is greater than at least 25 % of these 300 numbers....so 25 % of 300 is 75 ..... So...time for common sense....which 75 numbers among these 300 are the smaller ones.....of course da first 75 ....... And da 75th number will be da greatest among these 75.....so we gotta take da LEAST NUMBER THAT IS GREATER than 75th number in sequence......
So.....sum of all numbers from 1 to 10 = 55..... And...1 to 11 = 66.....and 1 to 12 = 78......
So....from 67 to 78 number in sequence.....is 12 .....so....da 75 th number in da sequence is = 12

So...least number greater than 12 is......13.......

! nah id win!

A sequence of numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... n occurs n

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A sequence of numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 ....... n occurs n

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards