A four digit number abcd where a is greater than b and none of the dig...
Since A+B=17 and A is greater than B, so A=9 and B=8. Here the maximum possible multiple of 9 can be 9 itself. According to the question none of the digits repeat, so C =7,2,5,4,3&6 and D =2,7,4,5,6&3 So the numbers can be formed are•9872•9827•9854•9845•9836•9863Already mentioned that ABCD is divisible by 11So 9845 is completely divisible by 11.And hence the digit at tens place ie C = 4
A four digit number abcd where a is greater than b and none of the dig...
Given Conditions
- We have a four-digit number represented as abcd.
- The digits are a, b, c, d, where a > b and all digits are unique.
- The sum of the digits in the thousands (a) and hundreds (b) places equals 17.
- The number is divisible by 11.
- The sum of the tens (c) and units (d) places is a multiple of 9.
Step 1: Find Possible Values for a and b
- Since a + b = 17 and a > b, the possible pairs (a, b) are:
- (9, 8)
- (8, 7)
- (7, 6)
- (6, 5)
Step 2: Check Divisibility by 11
- For a number to be divisible by 11, the difference between the sum of the digits in odd positions and even positions must be a multiple of 11.
- The odd positions are a and c, while the even positions are b and d:
- |(a + c) - (b + d)| must be a multiple of 11.
Step 3: Explore Values for c and d
- Since d + c must be a multiple of 9, we can explore combinations based on the previous pairs of (a, b).
Step 4: Analyze Possible Combinations
- Let's take (9, 8):
- Then, c + d must be a multiple of 9.
- Possible pairs: (0, 9), (1, 8), (2, 7), (3, 6), (4, 5), (5, 4), (6, 3), (7, 2), (8, 1), (9, 0).
- However, since we already have used 8 and 9, c and d must be among (0, 1, 2, 3, 4, 5, 6, 7).
Conclusion
- After checking for valid combinations, the only possible tens place digit c that satisfies these conditions while ensuring no digit repetition is 0.
- Thus, the possible tens place digit is 0.