Latest TCS Aptitude Question SOLUTION: 122333444411222233333344444444 111222222333333333444444444444 .....what will be the 2024 number?

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122333444411222233333344444444 111222222333333333444444444444 .....what will be the 2024 number?

Asked In TCS PRIYANKA RANDIVE (10 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

ANKUR GUJRATI  (9 years ago)

in first slot there are total 10numbers
in second there are total 20
in third 30
in fourth 40
50
60
70 and goes on
....

so for 1900th term we will find the term lies in 19th slot
now remaining terms are 2024-1900=124

that means the answer lies between 19th slot and 20th slot
20th slot contains 200 digits among which there are:
twenty no. of 1's(since 1st slot contain one 1, 2nd two 1's, 15th fifteen 1's)
forty no. of 2's(since 1st slot contain two 2's, 2nd four 2's, 15th thirty 2's)
sixty no of 3's(since 1st slot contain three 3, 2nd six 3's, 15th fortyfive 3's)
and
eighty no of 4's(since 1st slot contain four 4's, 2nd eight 4's, 15th sixty 4's

since 124=20(1's)+40(2's)+60(3's)+4(4's)
hence 2024th term is 4

Is this solution Helpfull? Yes (23) | No   

SOUBHIK BISWAS  (10 years ago)

1*(1+2+3+4)+2*(1+2+3+4)+3*(1+2+3+4)+...........+19*(1+2+3+4)=1900...
20(1+2+3)=120...
120+1900=2020...
so,the 2024th term is 4

Is this solution Helpfull? Yes (11) | No (5)  

suhasini  (10 years ago)

soubhik please explain it clearly.. how that sum is 1900

Is this solution Helpfull? Yes (1) | No   

shubham agrawal  (9 years ago)

In first series there are 10 nos. Nd in 2nd series there are 20 nos in 3rd 30 so on...so 2020th no will be 4 nd after that new series will be start for 2030 so we can guess that 2024th no should be 4 so ans is 4.

Is this solution Helpfull? Yes (1) | No (1)  

Kethineni Vinodh  (9 years ago)

122333444411222233333344444444...
This question has some ambiguity as the next term is 111222222333333333444444444444 or 1111222222223333333333334444444444444444....
We solve this question by assuming this series as
1223334444,11222233333344444444,1111222222223333333333334444444444444444,....
i.e., in this series the number of digits in each segment are 10, 20, 40 , ...
So 10 + 20 + 30 + ..... ≤ 2888.
10 (1 + 2 + 4 + ...) ≤ 2888
Sum of the terms of the series 1 + 2 + 4 + ... = 2n−1
10 (2n−1) ≤ 2888
for n = 8
2550 ≤ 2888
Now remaining digits are 2888 - 2550 = 338
Now the 9th term has 256 "1's", 512 "2's"... ets
So after 256 "1's" we have 2 as 2888 digit.

Is this solution Helpfull? Yes (1) | No   

ARUMALLA DHEERAJ  (9 years ago)

take first 10 terms we get 1,2,3,4 as no of times repeated and second ten it is 2,4,6,8 and third 10 it is 3,6,9,12 and so on
add above no of times repeated value we get 10,20,30,............
therefore 10*(1+2+3+..........)=2024
using sum of n terms formula we get n as 20 (trail and error method) and multiply the result with ten of sum of n values we get 2100 ((n*(n+1)/2)*10). so 2100-2024 gives 76 and multiplies of 20 are 20,40,60,80 so 76 lies between 60-80 where 4 gets repeated so answer is 4.

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Sharvil Mayekar  (9 years ago)

Ans. 4 is 2024th term.
Explanation:
1st series of numbers is 1,2,2,3,3,3,4,4,4,4. total numbers=10
2nd series of numbers are 1,1,2,2,2,3,3,3,3,4,4,4,4,4. total numbers=14
therefore after every series the count increases by 4.
we have 10,14,18,22,...x
consider x=2024
we can check for 2024 whether it is divisible by any of the mid set of integers
in the given series.
checking 2024/14=not divisible.
2024/18=not divisible.
2024/22=divisible=92.

therefore we can consider it to be it in series with the last integer being "4"

Is this solution Helpfull? Yes | No (2)  

Prashant Saini  (8 years ago)

1st slot has 10 numbers, 2nd slot has 20 numbers, 3rd slot has 30 numbers and so on...
Now we have to find on which slot, 2024 falls. Because we can count the numbers then. I mean notice hear, 1st slot has only one 1, second slot has 2 ones, 3rd slot has 4 ones and so on. Once we find the slot, we can find number of 1's and thus number of 2's and 3.
now, to find on which slot 2024th number falls, let us do following thing
10+20+30+.......

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raviranjan  (8 years ago)

1223334444=10--------1st step
11222233333344444444=20---------------------2nd step
.
.
11111111...20th times.......................4444....80th times=2000.....................20th step
111111111..21th times222
hence 2 at 2024 position

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N Murali Krishna   (8 years ago)

1,2,3,4....10
2,4,6,8.....20
3,6,9,12.....30

Every 10 multiple ended with 4 so 2020 also will end with 4...after 1 series will start...so answer is 1

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