Elitmus
Exam
Numerical Ability
Number System
How many Integers from 0 to 100 are not divisible by 2,3,5 & 7
Read Solution (Total 10)
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- there are 25 primes under 100,
2,3,5,7, will be removed and 1 will be added hence ans =25-4+1=22
or u can also do it by 100-AUBUCUB=100-(50+33+20+14)+(16+10+7+6+4+2)-(3+2+1)-0=22 - 7 years agoHelpfull: Yes(20) No(4)
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- 7 years agoHelpfull: Yes(17) No(3)
- total no not divisible by 2,3,5,7 =total no - total no divisible by 2,3,5,7
total no divisible by 2= last no - first no /gap +1
; no divisible by 2 =100-2/2 +1 =50
;no divisible by 3= 99-3/6+1= 17 (excluding no already divisible by 2)
;no divisible by 5 =5,25,55,65,85,95 =7 (excluding no already divisible by 2 &3)
; no divisible by 7= 7, 49,77,91 =4
total no divisible by 2,3,5&7=78
So total no not divisible =101-78=23
So 23 is the total no of nos between 0 &100 not divisible by 2,3,5&7. - 7 years agoHelpfull: Yes(6) No(1)
- Numbers divisible by 2 = 50
Numbers divisible by 3 = 33
Numbers divisible by 5 = 20
Prime numbers from 0 to 100=25
let's divide the problem into smaller domain= 0 to 10
every number is divisible from 1 to 10= 2,3,,4,5,6,7,8,9,10
answer is 21(25-4 (2,3,5,7)) - 7 years agoHelpfull: Yes(3) No(7)
- Div by 2=50
Div by 3=33
Div by 2&3=16
So div by only 3=33-16=17
Div by 5=20
Div by 5&2&3=13
So div by only 5=7
Div only 7, not by 2,3,5=3
Not div by 2,3,5,7=100-50-17-7-3=23
I think this is correct - 7 years agoHelpfull: Yes(3) No(1)
- 0-100 =101 num
divisible by 2 , ;0,2,4...100 => 0+(d-1)2=100 =>d=51
divisible by 3.; 3,9.15...99 => 3+(d-1)6=99 => d=17
divisible by 5 .; 5 ,25 ,35 ,55,65,85 ,95 => total 7 num
divisible by 7 .; 7 ,49 ,77 ,91 => total 4 num
so toal divisible by 2, 3 5 , 7 from 0 to 100 => 51+17+7+4=79
now , for we hv to find not divisible by them => 101-79 =22
so ,22ans - 7 years agoHelpfull: Yes(3) No(1)
- Every one submit different - different answer
For check correct answer compile this code:
#include
int main(){
int a;
int count=0;
printf("Numbers are:n");
for(a=1;a - 7 years agoHelpfull: Yes(1) No(1)
- we have 25 prime nos and every no is divisible by 1-9 except primes so we have 2 ,3 ,5 ,7 here but 1 is not divisible by those nos which is given in que so 25-- 4 (bcoz those 4 no are prime) = 21
1 is also not divsble so 21+1 =22
Ans is 22. - 6 years agoHelpfull: Yes(1) No(0)
- lets start with numbers divisible by 2 3 5 7 and then subtracting it.
divisible by 2 = 50 no.s (2x50=100)
divisible by 3 = 33 no.s
divisible by 5 = 20
divisible by 7 = 14
Now starting from above total numbers is 50+33+20+14=117
now 3x33 has 16 even numbers added in divislbe by 2 = 117-16
in divisible by 5, 10(10,20,30....) numbers are even so 117-16-10 and has 3 numbers divisible by 3 (15,45,75)
in divisible by 7, 7 even and 2 numbers added in multiples of 3(21,63)
so total = 117-16-10-3-7-2
= 79 numbers divisible
so 21 numbers not divisible - 7 years agoHelpfull: Yes(0) No(7)
- 21
and explanation is same as MOHAMMAD ANSAB - 7 years agoHelpfull: Yes(0) No(4)
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