How many positive integers less than 100 are divisible by 5 and 2
Answers
Answer:
Hey there !
Solution:
Numbers divisible by 3, 5 and 7 is also divisible by the LCM of 3, 5 and 7.
=> LCM of 3,5,7 = 3 × 5 × 7 = 105
So if a number is divisible by 105, then all the numbers which are multiples of 105 are also divisible by 3, 5 and 7.
But since 105 > 100, there is no number less than 100 which is divisible by all 3,5 and 7.
So there are 0 positive integers which are less than 100 and are divisible by 3,5 and 7.
If you want the individual number of terms separately for each, the here is the answer as follows:
Number of terms between 1 - 100 divisible by 3:
=> a = 3, d = 3, n = ?, = 99
=> 99 = 3 + ( n - 1 ) 3
=> 99 - 3 = ( n - 1 )3
=> 96 = 3 ( n - 1 )
=> 96 / 3 = ( n - 1 )
=> 32 = ( n - 1 )
=> n = 33 terms
Hence the number of terms divisible by 3 is 33 terms
Number of terms between 1 to 100 divisible by 5:
=> a = 5, d = 5, n = ?, = 100
=> 100 = 5 + ( n - 1 )5
=> 100 - 5 = ( n - 1 ) 5
=> 95 = ( n - 1 ) 5
=> 95 / 5 = ( n - 1 )
=> 19 = ( n - 1 )
=> 19 + 1 = n => n = 20 terms
Hence there are 20 terms between 1 - 100 which are divisible by 5.
Number of terms between 1 to 100 divisible by 7:
a = 7, d = 7, n = ?, = 98
=> 98 = 7 + ( n - 1 ) 7
=> 98 - 7 = ( n - 1 ) 7
=> 91 = ( n - 1 ) 7
=> 91 / 7 = ( n - 1 )
=> 13 = ( n - 1 )
=> n = 13 + 1 = 14
Hence there are 14 terms between 1 to 100 which are divisible by 7.
Hope my answer helped !
Answer:
20 positive integers are divisble by 5 and 2 ( including 0 )
Step-by-step explanation:
There are 2 numbers less than 100 divisble by 5 and 2
The numbers are :
0 , 5 , 10 , 15 , 20 , 25 , 30 , 35 , 40 , 45 ,50 , 55 , 60 , 65 , 70 , 75 , 80 , 85 , 90 , 95 .
Thank you...