The number of positive integers not greater than 100, which are not di

Question

The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5, is:

A. 18
B. 26
C. 31
D. 42
E. 43

fskilnik · Accepted Answer

Hi, Bunuel. Beautiful problem!

https://i.ibb.co/72hDT58/01-Feb19-4z.gif

? = {\rm{Remainder}}

\left. \matrix{
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,2\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over 2}} \right\rfloor = 50 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,3\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over 3}} \right\rfloor = {{99} \over 3} = 33 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,5\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over 5}} \right\rfloor = 20 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,2\,\,{\rm{and}}\,\,3\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over 6}} \right\rfloor = {{96} \over 6} = 16 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,2\,\,{\rm{and}}\,\,5\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over {10}}} \right\rfloor = 10 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,3\,\,{\rm{and}}\,\,5\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over {15}}} \right\rfloor = {{90} \over {15}} = 6 \hfill \cr 
 \# \,\,{\rm{div}}\,\,{\rm{by}}\,\,2\,\,{\rm{and}}\,\,3\,\,{\rm{and}}\,\,5\,\,\,:\,\,\,\,\,\left\lfloor {{{100} \over {30}}} \right\rfloor = {{90} \over {30}} = 3\,\,\, \hfill \cr} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,A \cup B \cup C\,\,\mathop = \limits^{{\rm{simplifier}}} \,\,50 + 33 + 20 - \left( {7 + 13 + 3} \right) - 2 \cdot 3 = 74

? = 100 - 74 = 26

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Archit3110 · Answer

total integers which are divisible by 2 = 50 odd & 49 even
divisible by 3 = 33 ; 17 odd & 16 even
divisible by 5 = 20 ; 10 odd & 9 even

over lap of 3 & 5 at 15 * 45 and of all even integers

so total +ve integers <100 which are divisible by 2,3 or 5 = 49+17+8 ; 74
not divisible would be 100-74 ; 26
IMO B

exc4libur · Answer

#100: 100
#2: 100-2/2+1=100/2=50
#3: 96/3=33
#5: 100/5=20
#(2•3=6): 96/6=16
#(2•5=10): 100/10=10
#(3•5=15): 90/15=6
#(2•3•5=30): 90/30=3

three–overlapping sets  : Total=A+B+C- -2 +none
  100=(50+33+20)-(16+10+6-9)+2(3)+none
  100=(103)-(23)+(6)+none… none=100-74=26

three–overlapping sets  : Total=A+B+C- + +none
  100=(103)-(16+10+6)+3+none… 
  100=(103)-(32)+3+none… none=100-74=26

Answer (B).

rahul12988 · Answer

Could not understand ur method. 
Can u elaborate more ?

Gokha · Answer

@BUENUEL
Sol: number divisible by 2,3 or 5 ends with: 1,2,3,4,5,6,7,8,9 & 0. 
that leaves only 1 & 7 between 1-10.
in 11-20: 11,13,17,19 - all prime numbers.. rest are divisible by 2,3,5
this means all prime # between 1-99 and 1 should be desired one.

with this logic: 1+ 25 primes = 26 ans.
  Ans: B

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The number of positive integers not greater than 100, which are not di

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