Question
If 731x46 is a six-digit number that is divisible by 3, then find the maximum value of 7x.
Solution
We know that a number is divisible by 3 only when the sum of its digits is also divisible by 3. So,
7 + 3 + 1 + x + 4 + 6 = (21 + x) should be divisible by 3 So, possible values of x = 0, 3, 6, 9 Maximum possible value of x = 9 So, required value
= 7 × 9
= 63
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