Question
Determine the sum of all three-digit numbers which are
divisible by 8 and the sum of their digits is divisible by '3'.Solution
ATQ, Since, the sum of digits of the number is divisible by '3', the number must be divisible by '3' too. LCM (8 and 3) = 24 So, we need to find the sum of all three-digit numbers which are divisible by '24'. So, first three digit multiple of 24 = 120 And, last three-digit number that is divisible by 24 = 984 General term of an arithmetic progression = a + (n - 1) × d {Where 'a' is the first term, 'n' is the number of terms and 'd' is the common difference}. 984 = 120 + (n - 1) × 24 Or, 864 = 24n - 24 Or, 888 = 24n So, n = 37 So, required sum = (37/2) × (120 + 984) = 20,424
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