1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 and 10?
Which is the smallest positive integer that yields zero remainder when divided by
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
It should be 0 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Log in to reply
Smallest positive integers doesn't include 0.
0 isn't positive
2520
Solution:
To find a positive integer take L.C.M of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.
Now, 6 is a multiple of 2 & 3. 8 is a multiple of 4. 10 is a multiple of 5.
So we can take LCM of 6 7 8 9 10.
Again 8 is a multiple of 2 and 9 is multiple of 3. So we can exclude 6 (3*2).
So now to take LCM of 7,8,9,10. Which is 2520.
it should be 1680!!!!
Log in to reply
why varsha its 2520 forgot LCM :p...... prove how???1680/9=18.66,you dont call it divisible!!!
1680 is not divisible by 9
it is not divisible by 9
0 it should be
Very Good!
forgot lowest common multiple how to perfrom already lol
what about 63 ? it gives zero remainde.
I've solved it using a simple c code. Here it is:
int main() { int i,cou; for(i=1;;) { if(i%1==0&&i%2==0&&i%3==0&&i%4==0&&i%5==0&&i%6==0&&i%7==0&&i%8==0&&i%9==0&&i%10==0) { cou=i; break; } else i++; } printf("%d",cou); }
take LCM of 1,2,3,4,5,6,7,8,9,10 which is 2520
LCM of numbers 1 2 3 4 5 6 7 8 9 10 which is equal to 2 2 2 3 3 5 7=2520
Problem Loading...
Note Loading...
Set Loading...
LCM(1,2,3,4,5,6,7,8,9,10)