Divisibility Rule for 3

Problem 1 :

Which of these numbers are divisible by 3?

a) 84   b) 123  c) 437  d) 111114  e) 707052

Solution :

In a number, if the sum of all the digits is divisible by 3 or a multiple of 3, then the number is divisible by 3.

a) 84.

Add all the digits in the number 84.

8 + 4 = 12

Since 12 is divisible by 3, the given number 84 is also divisible by 3.

b) 123

Adding all the digits in the number 123.

1 + 2 + 3 = 6

The sum of the digits in the given number 123 is 6, which is a multiple of 3.

So, 123 is divisible by 3.

c) 437

Adding all the digits in the number 437.

4 + 3 + 7 = 14

The sum of the digits in the given number 437 is 14 which is not a multiple of 3.

So, 437 is not divisible by 3.

d)  111114

Adding all the digits in the number 111114.

1 + 1 + 1 + 1 + 1 + 4 = 9

The sum of the digits in the given number 111114 is 9 which is a multiple of 3.

So, 111114 is divisible by 3.

e) 707052

Add all the digits in the number 707052.

7 + 0 + 7 + 0 + 5 + 2 = 21

The sum of the digits in the given number 707052 is 21 which is a multiple of 3.

So, 707052 is divisible by 3.

Problem 2 :

If the number 517__324 is completely divisibly by 3, then the smallest whole number is the place of ___ will be 

a) 0   b)   1     c)  2      d) none

Solution :

Let x be the unknown. Using the divisibility rule for 3.

= 5 + 1 + 7 + x + 3 + 2 + 4

= 22 + x

If x = 0, we get 22, which is not divisible by 3

If x = 1, we get 23, which is not divisible by 3.

If x = 2, we get 24, which is divisible by 3

So, the answer is 2.

Problem 3 :

How many 3 digit numbers are divisible by 3.

Solution :

3 digit numbers starts with 100.

The first 3 digit number which is divisible by 3 is 102.

keep adding 3, we get

102, 105, 108, .............999

By observing the above sequence, it is arithmetic progression.

To find number of terms of the arithmetic sequence, we use the formula

t_n = a + (n - 1)d

tn = 999, a = 102, d = 3

999 = 102 + (n - 1)3

999 - 102 = 3(n - 1)

897 = 3(n - 1)

n - 1 = 897/3

n - 1 = 299

n = 299 + 1

n = 300

So, 300 three digit numbers are divisible by 3.

Problem 4 :

What digits could replace ___so that these numbers are divisible by 3?

a) 3_8      b)  8__5     c)  3__14     d) __229

Solution :

In a number, if the sum of all the digits is divisible by 3 or a multiple of 3, then the number is divisible by 3.

a) 3___8 

Let x be the unknown.

3 x 8

Sum of the digits = 3 + x + 8

= 11 + x

If x = 1, then 3 + 8 + 1 = 12

It is divisible by 3. So, the required number is 1.

b) 8___5 

Let x be the unknown.

8x5

Add all the digits in the number 8x5.

Sum of the digits = 8 + x + 5

= 13 + x

If x = 2, then 13 + 2 ==> 15 (divisible by 3)

So, the required number is 2.

c) 3__14

Let x be the unknown.

3x14

Add all the digits in the number 3x14.

Sum of the digits = 3 + x + 1 + 4

= 8 + x

= 8 + 1

= 9

So, the required number is 1.

d) ___229

Let x be the unknown.

x229

Add all the digits in the number x229.

Sum of the digits = x + 2 + 2 + 9

= 13 + x

If x = 2, then 2 + 2 + 2 + 9 = 15 (divisible by 3)

So, the required number is 2.