Question 388635: hi,
when you look at positive three-digit whole numbers of which the middle digit is the average of the other two digits. one example is 741. how many of these numbers exist?
Found 2 solutions by richard1234, CharlesG2:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! It suffices to choose three (not necessarily distinct) digits between 0 and 9. I'll do it case by case:
Case 0: a, a, a
{0,0,0}, {1,1,1}, ..., {9,9,9} (9 cases)
Case 1: a, a+1, a+2
{0,1,2}, {1,2,3},...,{7,8,9} (8 cases)
Case 2: a, a+2, a+4
{0,2,4}, {1,3,5}, ..., {5,7,9} (6 cases)
Case 3: a, a+3, a+6
{0,3,6},...,{3,6,9} (4 cases)
Case 4: a, a+4, a+8
{0,4,8}, {1,5,9} (2 cases)
For cases 1 through 4, each set that does not have a zero can produce two different numbers (e.g. {4,5,6} produces 456, 654) and each set that has a zero can only produce one number (to account for leading zeroes). There are 16 sets with no zeroes and 4 sets with a zero, so the number of 3-digit numbers in cases 1-4 is 16*2 + 4 = 36.
For case 0, where all the digits are equal, there are nine set {1,1,1}...{9,9,9}. Each set implies one more number, so nine numbers. Therefore the total number of 3-digit numbers is 36+9 = 45.
Answer by CharlesG2(834)  (Show Source):
You can put this solution on YOUR website! hi,
when you look at positive three-digit whole numbers of which the middle digit is the average of the other two digits. one example is 741. how many of these numbers exist?
average = 1 --> sum 1st,3rd add to 2 --> 1 + 1, 2 + 0 --> 2 numbers
ave = 2 --> sum add to 4 --> 1 + 3, 2 + 2, 3 + 1, 4 + 0 --> 4 numbers
ave = 3 --> sum add to 6 --> 1 + 5, 2 + 4, 3 + 3, 4 + 2, 5 + 1, 6 + 0 --> 6
ave = 4 --> sum add to 8 --> 1 + 7, 2 + 6, 3 + 5, 4 + 4, 5 + 3, 6 + 2,
............................ 7 + 1, 8 + 0 --> 8 numbers
ave = 5 --> sum add to 10 --> 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5 + 5, 6 + 4,
............................ 7 + 3, 8 + 2, 9 + 1 --> 9
ave = 6 --> sum add to 12 --> 3 + 9, 4 + 8, 5 + 7, 6 + 6, 7 + 5, 8 + 4,
............................ 9 + 3 --> 7
ave = 7 --> sum add to 14 --> 5 + 9, 6 + 8, 7 + 7, 8 + 6, 9 + 5 --> 5
ave = 8 --> sum add to 16 --> 7 + 9, 8 + 8, 9 + 7 --> 3 numbers
ave = 9 --> sum add to 18 --> 9 + 9 --> 1 number
2 + 4 + 6 + 8 + 9 + 7 + 5 + 3 + 1
appears to be the sum of the digits from 1 to 9
sum digits = n(n+1)/2 = 9(10)/2 = 90/2 = 45 numbers
111, 210
123, 222, 321, 420
135, 234, 333, 432, 531, 630
147, 246, 345, 444, 543, 642, 741, 840
159, 258, 357, 456, 555, 654, 753, 852, 951
369, 468, 567, 666, 765, 864, 963
579, 678, 777, 876, 975
789, 888, 987
999
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