when two three digit numbers 6r3 and 2s5 are added together, the answer is a number divisible by 9.the largest

Question

possible value of r+s is

Answer 1

The largest number that r+s could be is 11, and here's why:
A number is divisible by 9 if the sum of its digits is divisible by 9. So we add up the two three digit numbers and get 8 (r+s) 8 for its three digits. Adding thise digits together gives us r+s+16 or r+s+7, adding the 1 and 6. So r+s must equal 2, 11, 20, 29, and so on for the number to be divisible by 9.
Now, since we know that r and s are both digits between 0 and 9, we can eliminate anything higher than 18 as the sum or r and s. Hence, the largest possible value of r+s is 11.

Answer 2

The sum of 6r3 and 2s5 is a multiple of 9. It is possible only if by adding all digits of 6r3 and 2s5 the result is a multiple of 9. That means (6+r+3+2+s+5)= a multiple of 9.
That is 16+r+s = a multiple of 9

After 16 the multiples of 9 are 18, 27, 36 etc
To get 18, r+s should be (18-16)=2
To get 27, r+s should be (27-16)=11
To get 36, r+s should be (36-16)=20

But 20 is not a possible value of (r+s) as sum of two digits cannot be more than 18.
Therefore the largest possible value of (r+s) is 11.

Answer 3

11. The sum of the digits of a number divisible by 9 is a multiple of 9. 6+3+2+5=16, so r+s can be 2 to make the digits sum to 18, or 11 to make them sum to 27.

Answer 4

11

Answer 5

11

Answer 6

11

Answer 7

no div by 9, => sum f all digits is div by 9
=> 8+r+s+8 div by 9,
possible sums:
18 => (r+s)=2
27 => (r+s)=11
36 => (r+s)=20 not posible as max value of (r+s) = 18, wen r=s=9.
Hence, largest value f (r+s) = 11.

Answer 8

6r3=18r,2s5=10s
18r+10s+2+3= divisible by 9
6+18r+10s=2(3+9r+5s)
2(3+9+5)=34if s+r=2
34+2=36 that divided by 9

s+r=2,in largest 47

Answer 9

The possible value of r+s are 2 and 11

Let me explain why?

(r,s) or (s,r) are

0,2) or (2,0) r+s=2

(1,1) r+s=2

(2,9) or (9,2) r+s=11

(3,8) or (8,3) r+s=11

(4,7) or (7,4) r+s=11

(5,6) or (6,5) r+s=11