Question
If the number 8357pt is divisible by 15, how many values
can 'p' take?Solution
ATQ,
Divisible by 15 ⇒ divisible by both 5 and 3
Divisible by 5 ⇒ t = 0 or 5
Divisible by 3 ⇒ sum divisible by 3
Case 1: t = 0
Sum = 8 + 3 + 5 + 7 + p + 0 = p + 23
Divisible by 3 ⇒ p = 1, 4, 7
Case 2: t = 5
Sum = 8 + 3 + 5 + 7 + p + 5 = p + 28
Divisible by 3 ⇒ p = 2, 5, 8
Total values of p: 3 + 3 = 6 values
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