Question

If 265p32 is a multiple of 3, where p is a digit, what might be the values of p?

Answer 1

If 265p32 is a multiple of 3, where p is a digit, we have to find value of p.

According to Divisibility rule of 3 :-
➡If sum of all digits of a number is divisible by 3 then the whole number is divisible by 3.

To check that whether 265p32 we have add all its digits ;
➡2 + 6 + 5 + p + 3 + 2 = multiple of 3.
➡18 + p = multiple of 3.

➡Here “18” is multiple of 3.
So, value of p should be multiple of 3.

➡Values of p might be = 0, 3, 6 or 9.

Answer 2

Question :- If 265p32 is a multiple of 3, where p is a digit, what might be the values of p ?

Solution :-

we know that,

• if a number is a multiple of 3 , then sum of its digits is also a multiple of 3.

so,

→ 265p32 ÷ 3 = Remainder 0

or,

→ (2 + 6 + 5 + p + 3 + 2) ÷ 3 = Remainder 0

→ 18 + p ÷ 3 = Remainder 0

then, Possible values of p are :-

• if p = 0 => 18 + 0 = 18 ÷ 3 = Remainder 0 .
• if p = 1 => 18 + 1 = 19 ÷ 3 ≠ Remainder 0 .
• if p = 2 => 18 + 2 = 20 ÷ 3 ≠ Remainder 0 .
• if p = 3 => 18 + 3 = 21 ÷ 3 = Remainder 0 .
• if p = 4 => 18 + 4 = 22 ÷ 3 ≠ Remainder 0 .
• if p = 5 => 18 + 5 = 23 ÷ 3 ≠ Remainder 0 .
• if p = 6 => 18 + 6 = 24 ÷ 3 = Remainder 0 .
• if p = 7 => 18 + 7 = 25 ÷ 3 ≠ Remainder 0 .
• if p = 8 => 18 + 8 = 26 ÷ 3 ≠ Remainder 0 .
• if p = 9 => 18 + 9 = 27 ÷ 3 = Remainder 0 .

hence,

• Possible value p are = 0, 3, 6 and 9. (Ans.)

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