if 31p5 is a multiple of 3,where p is a digit,what might be the value of p?
Answers
Answer:
Explanation:
The divisibility test for a number to be divisible by 3 is the sum of digits constituting number Sould be a multiple of 3
3z5 is divisible by 3
Hence ( 3 + 5 + z ) i.e. ( z + 8 ) is a multiple of 3 .
For z = 1 , sum is 9 .
Similarly for z = 4 , z = 7 the sum will be divisible by 3 .
Question :- If 31p5 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,
→ 31p5 ÷ 3 = Remainder 0
or,
→ 3 + 1 + p + 5 ÷ 3 = Remainder 0
→ 9 + p ÷ 3 = Remainder 0
then, Possible values of p are :-
- if p = 0 => 9 + 0 = 9 ÷ 3 = Remainder 0 .
- if p = 1 => 9 + 1 = 10 ÷ 3 ≠ Remainder 0 .
- if p = 2 => 9 + 2 = 11 ÷ 3 ≠ Remainder 0 .
- if p = 3 => 9 + 3 = 12 ÷ 3 = Remainder 0 .
- if p = 4 => 9 + 4 = 13 ÷ 3 ≠ Remainder 0 .
- if p = 5 => 9 + 5 = 14 ÷ 3 ≠ Remainder 0 .
- if p = 6 => 9 + 6 = 15 ÷ 3 = Remainder 0 .
- if p = 7 => 9 + 7 = 16 ÷ 3 ≠ Remainder 0 .
- if p = 8 => 9 + 8 = 17 ÷ 3 ≠ Remainder 0 .
- if p = 9 => 9 + 9 = 18 ÷ 3 = Remainder 0 .
hence,
- Possible value p are = 0, 3, 6 and 9. (Ans.)
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