Question
How many values can 'x' take if the number 7531xk must be
divisible by 15 for any valid digit 'k'?Solution
ATQ,
Divisible by 15 ⇒ divisible by both 5 and 3
Divisible by 5: last digit k must be 0 or 5
Divisible by 3: sum of digits must be divisible by 3
Case 1: k = 0
Sum = 7 + 5 + 3 + 1 + x + 0 = x + 16
Divisible by 3 ⇒ x = 2, 5, 8
Case 2: k = 5
Sum = 7 + 5 + 3 + 1 + x + 5 = x + 21
Divisible by 3 ⇒ x = 0, 3, 6, 9
Total values of x: 3 (case 1) + 4 (case 2) = 7 values
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