When the sum of the numbers from 9 to 33 is divided by 9 the remainder is


TCS SkillRack on 16-Mar-2016 11:35

When the sum of the numbers from 9 to 33 is divided by 9 the remainder is
on 16-Mar-2016 19:32

7
on 17-Mar-2016 17:12

3
on 18-Mar-2016 17:23

3
on 20-Mar-2016 13:13

Sn=1/2*n(2a+(n-1)d) 472/9=3
on 21-Mar-2016 11:53

6
on 21-Mar-2016 14:01

Sm=n/2(2a+(n-1)d) where n=24, a=9,d=1 thn sm=492 492/9=6remndr
on 28-Mar-2016 15:25

Sm=(33*34)/2-(8*9)/2=525 525/9=58+(3/9) where 3 is the remainder
on 29-Mar-2016 17:42

number of terms=25(i.e 33-9=24 as both tems are involved total terms=24+1=25) sum=25/2[2*9+24*1]=525 remainder=525/9=3
on 25-Apr-2016 12:22

7
Solution:


Sm=(33*34)/2-(8*9)/2=525 525/9=58+(3/9) where 3 is the remainder