Seeing binary as ternary

In order to find whether a number is divisible by 3 or not, we need to check whether the sum of the digits of the number is divisible by 3 or not. If the sum of the digits is divisible by 3, then the number is divisible by 3. So, from the given values only $111,1011,1101,1110$ have the sum of their digits divisible by 3 and hence the numbers are divisible by 3. Thus, $\boxed{4}$ numbers are divisible by 3.

111,1011,1101,1110

multiple of 3 so the sum of the digits are divisible by 3. if like that so we just look for the number that has 3 digits one or multiple of 3. they are 111,1011,1101 and 1110. so there are 4 numbers of those lists that are a multiple of 3

If a number is multiple of 3, then the sum of the digits of that number will be divided by 3.

And, here just 1 and 0 are there. Then it is easily said that where there are just three 3's along with 0's or not with 0's, the number is a multiple of 3.

The answers are- 111 1011 1101 1110

Add up numbers and those whose sum will result is the multiple of 3 are the multiplies of 3 for example 1101 sum results in 1+1+0+1=3 as three is the multiple of 3 therefore 1101 will also be the multiple of 3

For a number to be a multiple of 3, all the digits must add up to be a multiple of 3. Here, 111,1011, 1101 and 1110 are hence multiples of 3.

simple divisibility rules must be impicated

simple follow the divisibility rule of 3

1. Find the sum of all digits of a number, if sum is divisible by 3 then the number is also divisible by 3.

here 111,1011,1101, 1110 are divisible by 3.

Thus answer is 4

First, we should know the divisibility test for 3. Simply add all the digits of a number. If the sum is divisible by 3, then the number itself is divisible by 3 (or you can add the digits of the solution and apply the test again - it works!) Then simply check each case by applying the rule. We see that there are four numbers which satisfy the test - 111 (since 1+1+1 = 3 which is divisible by 3), 1011 (since 1+0+1+1 = 3 is divisible by 3), 1101 (since 1+1+0+1 = 3 which is divisible by 3) and 1110 (which is again divisible by 3 since 1+1+1+0 = 3). :)

Add the digits of the numbers and if the sum is a multiple of 3, the numbers is a multiple of 3 e.g 110= 1+1+0=2 and 2 is not a multiple of 3 so nor is 110 however 111=1+1+1=3 and 3 is a multiple of 3 so 111 is a multiple of 3. Do this for all and you will find the answer 4

if the sum of total digits in the givin number is divisible by 3 then the number is divisible by three

A way of checking whether something is divisible by 3, is by adding the digits up. If the result adds up to a multiple of 3, then the original number is divisible by 3. As a result, 111, 1011, 1101 and 1110 are divisible by 3, giving the answer of4

Python: numbers = [1,10,11,100,101,110,111,1000,1001,101,1011,1100,1101,1110,1111] count = 0 for i in numbers: if i%3 == 0: count += 1

print count

This is just the application of divisiblity test of 3, which says that a number is divisible by 3 if the sum of digits of number is divisible by 3

In the list of 15 numbers, only 111, 1011, 1101, and 1110 are multiples of 3. The question is "How many numbers?" So, the answer is 4.

IF the digits in a number add up to a number divisible by 3 then it's a multiple of three: i.e. 100 is not a divisble of 3 as (1+0+0)/3 = 1/3. However 111 is a divisible of 3 as (1+1+1)/3 = 1

I solved it using programming in C++. You can get the code from here : http://pastebin.com/gtRuVY2k

the sum of the digits in a number must be divisible by 3 for the whole number to be divisible by 3

Ciri-ciri keterbagian 3 adalah jumlah digit-digitnya kelipatan 3, contoh : 81. 8+1=9 ( kelipatan 3 ) jadi dalam soal diatas kita cari bilangan yang mempunyai 3 buah bilangan 1. yaitu : 111,1011,1101, 1110. Jadi ada 4 buah bilangan yang merupakan kelipatan 3.

Only 4 numbers i.e. 111, 1011, 1101, 1110 are multiples of 3.

So, 4 .