show that, if the 3 digit number in the form of 906 (i.e. digit at tens place is 0) is subtracted from the sum of its digits, we always get a number divisible by 11
the number is 906.
so sum of its digit =9+6=15
906-15=891
now if we divide 891 by 11, we get 81 as the answer.
hope it helps you out!!!
3
javalam619... answered this
The three digit no. is 906.
And the sum is 9+0+6=15
We have to show that if a three digit no. is subtracted from the sum of the digits then we always get a number divisible by 11.
After subtracting it ,we get
906-15=891
891/11=81
I hope this will help you
7
Kumkum answered this
Show that if three digit number is in the form of 906 is subtracted from the sum of its digit we always get a number divisible by 11
-2
Rudransh Tiwari answered this
Let us take 906 ... So, sum of its digit = 9+0+6 = 15 ,
So, if we subtract it by 906 we get, 906-15 = 891,
And, 891/11 = 81
So, proved that when a three digit number having 0 at tens place is divided by 11 always...
1
Daniel Isaac Joy . answered this
Here, the 3 digit number will be in the form (100x + y).
so ,if we subtract the sum of digits from the number , it would be (100x + y) - ( x + y) .
(100x + y) - ( x + y) = 99x.
99x is divisible by 11. (11 x 9x = 99x).
hence, it is proved that when the sum of the digits is subtracted from the 3 digit number having 0 at tens place, the number got is divsible by 9.
