# Divisibility Rule for 3

Rule :

In the given number, if the sum of the digits is divisible by 3, then the original number is also divisible by 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

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.

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

tn = 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.

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