divisible problems - 1

If a number is divisible by $3$ then the sum of its digits must be divisible by $3$ .So let,s check all the options: $16708794\rightarrow1+6+7+0+8+7+9+4=7+7+15+13=14+28=42=3\times14$ So $(A)$ is divisible by $3$ .Let,s check $(B)$ : $14847486\rightarrow1+4+8+4+7+4+8+6=5+12+11+14=17+25=42=3\times14$ So $(B)$ is divisible by $3$ , Lets check $(C)$ : $18320282\rightarrow=1+8+3+2+0+2+8+2=9+5+2+10=14+12=26$ As $26$ is not divisible by $3$ so $(C)$ is not divisible by $3$ so only $\boxed{2}$ numbers are divisible by $3$

All numbers divisible by 3 must have their sums of the digits add up to a multiple of 3.

For example, the first one, the sum of the digits is 42. 42 is divisible by 3. Therefore, A is also divisible by 3.

You repeat this process for B and C, and you get that only A and B are divisible by 3.

The answer is 2

sum of digits should be divisible by 3 for a no to be divisible by 3. therefore a and b r divisible whereas c isn't.