# DIVISIBILITY RULE FOR 5

Rule :

If the last digit is 0 or 5 then the number is divisible by 5.

Problem 1 :

Which of these numbers are divisible by 5?

a) 400    b) 628    c) 735    d) 21063    e) 384 005

Solution :

If a number ends with 0 or 5, then it is divisible by 5.

a) 400

The given number 400 ends with 0.

So, the given number 400 is divisible by 5.

b) 628

The given number 628 does not end with 0 or 5.

So, the given number 628 is not divisible by 5.

c) 735

The given number 735 ends with 5.

So, the given number 735 is divisible by 5.

d) 21063

The given number 21063 does not end with 0 or 5.

So, the given number 21063 is not divisible by 5.

e) 384 005

The given number 384005 ends with 5.

So, the given number 384005 is divisible by 5.

Problem 2 :

Which one of the following numbers is completely divisible by 45 ?

a)  181560    b) 331145     c)  202860      d) 2033550

Solution :

If the number is divisible by 45, it should be divisible by 5 and 9.

Divisibility rule for 5 :

The unit digit should be 0 or 5.

Divisibility rule for 9 :

The sum of the digits should be 9.

a) 181560 is divisible by 5

1 + 8 + 1 + 5 + 6 + 0 => 21 (not divisible by 9)

b) 331145 is divisible by 5

3 + 3 + 1 + 1 + 4 + 5 => 17 (not divisible by 9)

c)  202860 is divisible by 5

2 + 0 + 2+ 8 + 6 + 0 => 18 (divisible by 9)

Problem 3 :

The sum of all two digit numbers divisible by 5 ?

a) 1035   b) 1245   c) 1230    d) 945

Solution :

The first two digit number which is divisible by 5 is 10.

Two digit numbers are divisible by 5 are

10, 15, 20, 25, .................95

By observing the sequence, it is arithmetic progression.

Finding number of terms :

tn = a + (n - 1)d

tn = 95, a = 10 and d = 15 - 10 ==> 5

95 = 10 + (n - 1)5

95 - 10 = 5(n - 1)

85/5 = n - 1

n - 1 = 17

n = 17 + 1

n = 18

There are 18 two digit numbers are divisible by 5.

Sum of the numbers :

sn = (n/2) [a + l]

s18 = (18/2)[10 + 95]

s18 = 945

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