Find the number of three digit numbers from 100 to 999 inclusive which have any one digit that is the average of the other two
Answers
121 three digit numbers from 100 to 999 inclusive which have any one digit that is the average of the other two
Step-by-step explanation:
Lets find triplets 1st in which 1 digit is average of other two
(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)
Triplet (a , a , a) will have only one number
Triplet ( a , b , c) will have 6 numbers
abc , acb , bac , bca , cab , cba
Total number = 9 * 1 + 6 * 16 = 105
Now numbers having 0
102 , 120 , 201 , 204 , 210 , 240 , 402 , 420 , 306 , 360 , 603 , 630 , 408 , 480 , 804 , 840 = 16 numbers
Total such numbers = 105 + 16 = 121
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