Question
A six-digit number 4c68d2 is divisible by 12. Find the
sum of all possible values of 'c' for the largest possible value of 'd'.Solution
ATQ, A number is divisible by 12 when it is divisible by 3 and 4 both. A number is divisible by 4 when its last two digits are divisible by 4. Here 'd2' is divisible by 4 for 'd' = 0, 2, 4, 6, and 8. Largest possible value of 'd' = 8. A number is divisible by 3 when the sum of its digits is divisible by 3. Sum of digits of 4c6882 = 4 + c + 6 + 8 + 8 + 2 = 28 + c. So, possible values of 'c' = 2, 5, and 8. Therefore, the required sum = 2 + 5 + 8 = 15.
When the digits, which are odd in the number ‘84376529’ are decreased by 2 and the remaining digits are increased by 1, then what is the sum of 5
Statements: S ≤ T < I = M > R, A < Z ≥ O > I ≥ D
Conclusion:
I. S = D
II. R ≤ D
III. Z > T
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