Question

from integers 1 to 30,how many ways are there to add 3 integers that is divisible by 3?
help!!

Answer 1

are we allowed to add one integer twice

Answer 2

yup.

Answer 3

1. add to 3 one 3 = 6 the first nr again 3 = 9 - the second and again 3 = 12 the 3th number
2. multiply 3 with 3 = 9 this 9 with 2 = 18 the second and to this assum 3=21 the 3th

Answer 4

but you can us the exponential possibillity

3*1 =3
3*2 =9
3*3 =27

Answer 5

or you can us the numbers multiply of 9 like 9 and 18 and 27

Answer 6

or check thos numbers like 12 --- 1+2=3 --- divisidily with 3
15 --- 1+5=6 --- divisibily with 3
18 --- 1+8=9 --- div with 3
21
24
27

Answer 7

Every natural number can be represented as
3n
3n+1 or
3n+2
here n is any number(1,2,3......)
in our question we need n from 0 to 10 (to get 1to 30)
for any number to be divisible by 3
it has to be of the form
3n+3n or 3n+1 + 3n+2
Extending this concept to three numbers
it has to be of the form
3n+3n+3n or 3n+ 3n+1 +3n +2
taking the first case
there are 10 numbers of the form 3n (3,6,9......30)
as repeatition is allowed it can be got by
10 10 10
C + C + C =10 +10 + 10 =30
1 1 1
Second case
3n +(3n+1) +(3n+2)
3n---10 numbers of this type (3,6,9...30)
3n+1-----9 numbers of this type (1,4,7....28)
3n+2 ------9 numbers of this type (2,5,8,....29)

this can be got by
10 9 9
C + C + C = 10 + 9 + 9 =28
1 1 1

therefore total ways possible =30+28=58
jst check wheather it is correct coz i could have made some mistakes.
sorry couldn't explain it better on comp.

Answer 8

thanks SRIRAM!!

Answer 9

welcome