GATE2010 MN: GA-8

Question

Consider the set of integers $\{1,2,3,\ldots,5000\}.$ The number of integers that is divisible by neither $3$ nor $4$ is $:$

• A. $1668$
• B. $2084$
• C. $2500$
• D. $2916$

Comments

The set given in the question ends at 500, whereas answer considers it 5000. Please get the details in question corrected.

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@Veenit yes . i also stuck for some time. but it's understandable by seeing options 

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@Veenit yes . i also stuck for some time. but it's understandable by seeing options .

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@Veenit yes  i also stuck for some time. but it's understandable by seeing options . 

Answer 1

$S = \{1,2,3,\ldots,4999,5000\}$

Let 'A' be the set of integers divisible by $3 \implies A =\{3,6,9,12,15,\ldots,4998\}$

Let 'B' be the set of integers divisible by $4 \implies B= \{4,8,12,16,\ldots,5000\}$

Let 'C' be the set of integers divisible by both $3$ and $4 \implies C = \{12,24,36,\ldots,4992\}$

$|A| =  3+(n-1)3 = 4998  \implies n = 4998/3 = 1666$

$|B| = 4+(n-1)4 = 5000  \implies n = 5000/4 = 1250$

$|C| =12+(n-1)12 = 4992 \implies n = 4992/12 = 416$

Divisible by $3$ or $4:$  $\{3,4,6,8\dots 5000\}$ : $|D|   =  |A| + |B| -|C|$

$= 1666+1250-416$

$= 2500$

Number of integers that is divisible by neither $3$ nor $4$ 

$= |S| - |D|$

$= 5000-2500$

$= 2500$

So, C is the correct answer.

Comments

Got it now thanks for the explanation:)

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$\textbf{the set of integers that are divisible by 3 can be represented by the general term 3n.}$

$\text{For n=1,2,3......we get the set as {3,6,9,......}}$

$\text{for finding the number of the elements in the set}$

$3n\leq 5000 \Rightarrow n\leq \frac{5000}{3}= 1666.66$

$\therefore{the ~last~ n~ would ~be~1666~which ~is~the~count}$

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@ankitgupta.1729, I used this method to get the closest multiple of  $12$   less than $5000$.

So,  $\frac{5000}{12}=416.6667$   Then I subtracted  $416$  from this answer, then  multiplied it by  $12$ .  Here I got  $8$.

So, subtracted  $8$  from  $5000$  to get  $4992$  which is the last multiple of 12 less than  $5000$.

 I want to know if I can do this in similar other problems?