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12 Feb 2021, 07:00
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:48) correct
67%
(03:05)
wrong
based on 90
sessions
History
Date
Time
Result
Not Attempted Yet
What is the number of three digits number from 100 to 999 inclusive which have any one digit that is the average of the other two ?
A. 50
B. 60
C. 120
D. 121
E. 242
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 50
B. 60
C. 120
D. 121
E. 242
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Bunuel
Since there are three digits and that the average of any two should be the third digit or one of the digits, there are 9 possible averages i.e. digits(1,2,3 ...9). Lets see how many numbers each of these averages form to get us total required numbers.
1. 1 : 1+1 has 1 three digit number
2. 2 : 1 + 3 | 2 + 2 has (3! + 1) three digit number
3. 3 : 1 + 5 | 2 + 4 | 3 + 3 has (2*3! + 1) three digit number
4. 4 : 1 + 7 | 2 + 6 | 3 + 5 | 4 + 4 has (3*3! + 1) three digit number
5. 5 : 1 + 9 | 2 + 8 | 3 + 7 | 4 + 6 | 5 + 5 has (4*3! + 1) three digit number
6. 6 : 3 + 9 | 4 + 8 | 5 + 7 | 6 + 6 has (3*3! + 1) three digit number
7. 7 : 5 + 9 | 6 + 8 | 7 + 7 has (2*3! + 1) three digit number
8. 8 : 7 + 9 | 8 + 8 has (3! + 1) three digit number
9. 9 : 9 + 9 has 1 three digit number
Edit:
Additionally, 1, 2, 3 and 4 can be averages with other two being digits from 0 to 8.
1. 1 : 0 + 2 has 4 three digit number
2. 2 : 0 + 4 has 4 three digit number
3. 3 : 0 + 6 has 4 three digit number
4. 4 : 0 + 8 has 4 three digit number
Total = 4 * 4 = 16
G. Total = 1 + 3! + 1 + 2*3! + 1 + 3*3! + 1 + 4*3! + 1 + 3*3! + 1 + 2*3! + 1 + 3! + 1 + 1 + 16 = 16*6 + 9*1 + 16 = 121.
Answer D.
12 Feb 2021, 13:41
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Bunuel
Consider integer \(ABC\).
If A is equal to the average of B and C, we get:
\(A = \frac{B+C}{2}\)
\(2A = B+C\)
Implication:
Digits B and C must sum to twice the value of A.
A --> options for B and C
1 --> 0,2...............................................1,1
2 --> 0,4..........1,3................................2,2
3 --> 0,6..........1,5...2,4........................3,3
4 --> 0,8..........1,7...2,6...3,5................4,4
5 --> ...............1,9...2,8...3,7...4,6........5,5
6 --> ...............3,9...4,8...5,7................6,6
7 --> ...............5,9...6,8........................7,7
8 --> ...............7,9................................8,8
9 --> ....................................................9.9
Case 1: Digit combinations implied just to the right of the arrows
Number of cases in the column just to the right of the arrows = 4
Number of options for the hundreds digit = 2 (Cannot be 0)
Number of options for the tens digit = 2 (Either of the two remaining digits)
Number of options for the units digit = 1 (Only 1 digit left)
To combine these options, we multiply:
4*2*2*1 = 16
Case 2: Digit combinations implied in the middle of the chart
Number of cases in the middle of the chart = 16
Number of ways to arrange the 3 digits in each case = 3! = 6
To combine these options, we multiply:
16*6 = 96
Case 3: Digit combinations implied all the way to the right
Number of cases in the rightmost column = 9
Since the three digits in each case are identical, each case yields only one possible arrangement, for a total of 9 options.
Number of viable integers = Case 1 + Case 2 + Case 3 = 16 + 96 + 9 = 121
