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?
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.
Well, according to the addition problem, we have NR + RN = ABC. Since N and R are different digits, we can safely assume that N > R.
Now let's look at the units column where we add R + N. The greatest possible sum of two different digits is 9 + 8 = 17. In this case, we would carry the 1 to the tens column and have a sum of 7 in the units column.
Since the units column is BC, the greatest possible value of B is 7. And since we carried a 1, the greatest possible value of C is 8.
Therefore, the greatest possible value of B + C is 7 + 8 = 15.
So, the answer is 15. Bada bing, bada boom!
To find the greatest possible value of B + C, we can start by examining the digits N and R.
Since N and R are different digits, their sum (N + R) must be less than 10. This means that there will be no carrying from the ones digit to the tens digit in the final sum.
Now let's look at the hundreds digit. The sum of N + R is equal to C (since there is no carrying), so C is the greatest possible value for the hundreds digit.
Next, we need to consider the tens digit. Since N + R is less than 10, there will be no carrying from the tens digit to the hundreds digit. So the greatest possible value for the tens digit is 9 (as it is the largest single digit available).
Finally, let's consider the ones digit. The sum of N + R is equal to B (since there is no carrying), so B is the greatest possible value for the ones digit.
Putting it all together, the greatest possible value of B + C is 9 + C.
Therefore, the greatest possible value for B + C is 9 + C.
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