find the numbers of integers between 10 to 400 which are divisible by 3
Answers
Answer:
187
Step-by-step explanation:
How many numbers between 1 and 400 are divisible by both 3 or 5?
Save taxes & secure your family's future, with a term plan.
187
Lets understand it by taking a smaller range, say we have to find numbers between 1 and 20(both included) which are divisible either by 3 or by 5,
Here we can work it out normally this-
Numbers divisible by 3– 3 , 6 , 9 , 12 , 15 , 18
Numbers divisible by 5– 5 , 10 , 15 , 20
But here’s a catch,
the number 15, which is the LCM of 3 and 5, is in both the list, i.e., it is divisible by both 3 as well as 5 so we have to count it only once,
Numbers divisible by 15- 15
hence total numbers in this case=(Numbers divisible by 3)+(Numbers divisible by 5)-(Numbers divisible by the LCM of 3 and 5 ,i.e., 15 in this case)
=6+4–1
=9.
Now coming to the range of 1–400
Numbers divisible by 3-
First number is 3 and last number is 399, applying the formula of nth term of AP-
a+(n-1)d=Tn,where
a=First term , d=Common difference , Tn=nth term of AP , n=Number of terms
Here, a=3 , d=3 , Tn=399
So, 399 = 3+(n-1)3 or n = 133
Numbers divisible by 5-
First number is 5 and last number is 400,
So, a=5 , d=5, Tn=400
n can be obtained from 400= 5+(n-1)5 or n = 80
Numbers divisible by 15(LCM of 3 and 5)
First number is 15 and last number is 390,
So, a=15 , d=15, Tn=390
n can be obtained from 390 = 15+(n-1)15 or n = 26.
Thus , final answer=(Numbers divisible by 3)+(Numbers divisible by 5)-(Numbers divisible by the LCM of 3 and 5 ,i.e., 15 in this case)
=>133+80–26
=187
How many numbers between 1 and 400 are divisible by both 3 or 7?
How many numbers between 1 to 400 are exactly divisible by 7?
How many numbers between 1 and 400 are divisible by 4?
How many numbers between 1 and 600 are divisible by both 3 or 5?
What is the number of numbers between 100 to 400 that are divisible by either 2, 3, 5 and 7?
We have to calculate the numbers between 1 and 400 which are divisible by 3 or 5.
This means we have to count all the numbers which are either divisible by 3 or by 5 or by both.
Let us count individually each of the three cases
Case I: divisible by 3:-
3, 6, 9,…………, 399 are the numbers which are divisible by 3
Since this forms an AP with common difference 3
thereforeT(n)=a+(n−1)∗d
399=3+(n−1)∗3
396=(n−1)∗3
n−1=132...dividingbothsidesby3
n=133
hence 133 numbers are divisible by 3 between 1 and 400.
Case II: Divisible by 5:-
5, 10, 15, ………., 395 are the numbers which are divisible by 5.
T
The all new Dominar 250: welcome to the world of touring.
#include<studio.h>
int main()
{
int count = 0;
for(int i = 1; i < 401; i++)
{
if(i % 3 == 0 || i % 5 == 0)
{
count++;
}
}
printf(“%d”, count);
return 0;
}
This is a C program which gives the answer for your question. Copy the above code and paste it in the text editor in the following link:
First click on compile button then click on execute. The answer will be displayed in the green colored screen you see below.
Numbers upto 400 divisible by 3 or 5 = numbers divisible by 3 + numbers divisible by 5 - numbers divisible by 15
Numbers divisible by 3 = 399/3= 133
Numbers divisible by 5 = 400/5 = 80
Numbers divisible by 15 = 390/15=26
Answer is 133+80–26= 187
How many numbers between 200 and 300 are divisible by 3, 4, 5 together?
How many numbers from 1 to 400 are divisible by both 3 and 7 but not by 42?
How many numbers between 100 and 400 are divisible by 2, 3, 5 and 7?
How many digits in 1 to 500 are divisible by both 3 and 5?
How many numbers up to 2000 are divisible by both 5 and 9?
A no which is to be divisible by both 3 AND 5, which means it should be divisible by 15. So from 1 to 400,
400÷15 = 80/3 ,ie, 26 and 3/2
Which implies that there exist 26 nos which are divisible by 15, or by both 3 and 5.
Work the way you want from anywhere.
399=3+ 3(n1-1)
so n1=396/3 +1=133 nos divisible by 3.
395=5+5(n2-1)
so n2=79 nos divisible by 5
390=15+(n-1)15
I.e( 375/15)+1
so 26 numbers in total divisible by 15
so total nos = 133+79-26= 186 numbers