What is the number of three digits number from 100 to 999 inclusive wh

Question

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

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unraveled · Answer

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.

GMATGuruNY · Answer

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

D

24Vikas · Answer

Hello
Could someone please explain the above question in a simpler way

sanya2711 · Answer

I did it by subtracting 25 from each answer option and then checking which option is divisible by 6 which is only option d. The number 25 is obtained by addind numbers from 111 to 999 which is 9 and 102,402,603,804 and the other possible combinations of these numbers which are 16. (16+9). the rest of the numbers will all have 6 possible combinations thus the number which remains after subtracting 25 should be divisible by 6.

Kinshook · Answer

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 ?

The 3-digits pairs from 100 to 999 inclusive which have any one digit that is the average of the other two = {(0,1,2),(0,2,4),(0,3,6),(0,4,8),,(1,1,1),(1,2,3),(1,3,5),(1,4,7),(1,5,9),(2,2,2),(2,3,4),(2,4,6),(2,5,8),(3,3,3),(3,4,5),(3,5,7),(3,6,9),(4,4,4),(4,5,6),(4,6,8),(5,5,5),(5,6,7),(5,7,9),(6,6,6),(6,7,8),(7,7,7),(7,8,9),(8,8,8),(9,9,9)}

The number of 3 digits numbers in each case respectively = {4,4,4,4,1,6,6,6,6,1,6,6,6,1,6,6,6,1,6,6,1,6,6,1,6,1,6,1,1}
Total number of three digits number from 100 to 999 inclusive which have any one digit that is the average of the other two = 4*4 + 1*9 + 6*16 = 121

IMO D

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What is the number of three digits number from 100 to 999 inclusive wh

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What is the number of three digits number from 100 to 999 inclusive wh

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