RN cannot be 99, since R≠N
If RN=98 then
98+89 = 187
So, B+C=15
That's my final answer.
If B+C=16, then since B≠C, the digits must be 79. 97+79=176, B+C=13
it's also easy to show that B+C≠17 or 18
NR
+ RN
--------
ABC
The addition problem above is correct. If N, R, A, B, and C are different digits, what is the greatest possible value of B+C?
4 answers
It is 11
There is only possibility of six addition without repeating digits i.e. 48+84=132 57+75=132 58+85=143
67+76=143 68+86=154 and 78+87=165
Some of them have same answer.In 1st and 2nd case the answer is 5.In 3rd and fourth case answer is 7 then fifth and sixth case have 9 and 11 answer respectively.
So the greatest value of B+C is 11.
67+76=143 68+86=154 and 78+87=165
Some of them have same answer.In 1st and 2nd case the answer is 5.In 3rd and fourth case answer is 7 then fifth and sixth case have 9 and 11 answer respectively.
So the greatest value of B+C is 11.
98+89=187 AND ANS IS 15