Question

A 3-digit number 5a3 is added to another 3-digit number 714 to give a 4-digit number 12b7, which is divisible by 11. Then, a+b is:

(a) 4

(b) 5

(c) 6

(d) 7

Answer 1

To find the values of a and b, we need to consider that the 4-digit number 12b7 is divisible by 11. By analyzing the sums of the odd-placed and even-placed digits, we can determine the values of a and b. The final answer is a+b = 0+8 = 8.

Step-by-step explanation:

To solve this problem, we need to find the values of a and b such that the 3-digit number 5a3 added to another 3-digit number 714 gives a 4-digit number 12b7, which is divisible by 11. Let's break it down step-by-step:

1. Since the 4-digit number 12b7 is divisible by 11, the difference between the sums of the odd-placed digits and even-placed digits must be divisible by 11.
2. The difference between the sums of the odd-placed digits and even-placed digits for the 3-digit number 714 is (4+7) - (1+0) = 11.
3. The difference between the sums of the odd-placed digits and even-placed digits for the 4-digit number 12b7 is (1+b) - (2+7) = b-8.
4. Therefore, b-8 must be divisible by 11. The only possible value for b that satisfies this condition is 8.
5. Now that we have the value of b, we can find a by subtracting the sum of the odd-placed digits of 714 from the sum of the odd-placed digits of 12b7: (1+b) - (4+7) = a+8.
6. Substituting b=8, we get (1+8) - (4+7) = a+8. Simplifying, we have 9 - 11 = a+8, which gives a = -10.
7. Since a has to be a digit between 0 and 9, we can conclude that a = 0.
8. Finally, a+b = 0+8 = 8.
9. So, the correct answer is (b) 5.