Math Is Fun Forum

Vishalini

Member

Registered: 2014-02-20

Posts: 6

In the following series, find the 2048th number

In the following sequence, what is the 2048th number?? How to find it?

1 2 2 3 3 3 4 4 4 4 1 1 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 4 4 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 ............
Please explain the concept..

gAr

Member

Registered: 2011-01-09

Posts: 3,482

Re: In the following series, find the 2048th number

One way by programming:

In python:

(sum([i*[1]+2*i*[2]+3*i*[3]+4*i*[4] for i in range(1,21)],[]))[2047]

=4

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bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: In the following series, find the 2048th number

Hi;

It is a 4.

1) Each grouping contributes {n, 2n, 3n, 2^n+1} numbers.

The first 8 of those groupings contributes.

The last group the 9th one has

Since 1236 + 1078 > 2048 the 2048th number is in the 9th group.

2048 - 1236 is 812, so we are interested in the 812 member of the 9 the group. Using 1) we can tell how many of each are in group 9.

The first 9 members of it are 1

the next 18 members of it are 2

the next 27 members of it are 3

the rest are 4's. The answer is 4.

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.