Question

if 31k5 is a multiple of 3 and k is a digit between 0 and 6, then k is :

Answer 1

As the divisibility test for 3 is that the sum of all the digits should be divisible by 3,
So here the sum of digits = 3+1+5+k = 9+k
for 9+k to be divisible by 3, k can take value of 3 or 6 as then the sum will become divisible by 3. The condition given is that k should be between 0 and 6 so we cant include 6 So the only value left is 3. Hence, the ans. is 3

Answer 2

here is your answer

we know that if the sum of all digit of the numbers are divisible by 3 iteans that the whole number is divisible by 3.
the digit is 31k5
s if digit = 3+1+5+k
9+k
the nearest number divisible by 3 is 12
9+k = 12
k=3

hope this helps you
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