Not really rule of 5

For a number to be divisible by 5, the last digit must be 0 or 5. Thus we have two options.

For the digit before the last digit, we can choose from 0 to 9. As long as the last digit is 0 or 5. We have 10 options...

Therefore

10 $\times$ 2 = 20 possibilities

This is like an Arithmetic Progression.

The sets of numbers are:

00, 05, 10, 15 and so on till 95 because what we need are the las two digits.

Hence:

$a_{1}$ which is the first term of the AP = 0

$a_{n}$ which is the last term of the AP = 95 = $a_{1}$ + $(n-1) \times d$ where d is the common difference of 5.

Implies:

95 = 0 + (n-1)d = $(n-1) \times 5$

$\frac{95}{5}$ = n-1

19 = n-1

n =20 = No. of Possibilties

100/5 = 20

see we have to fill 2 places _ _
we can fill the first place means the tens place with 10 digits ( from 0 to 9) So, we have 10 possibilities to fill the 10's place.

And we can fill the ones place only with 0 or 5 as according to the divisibility rule of 5 ones place must be 0 or 5 . so we have only 2 possibilities for ones place.

so total possibilities are 10x2=20

00, 05,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95= 20 Possibilities

The last number of the multiple of 5 must be 0 or 5. we have 0 to 9 number. We can take it 10 times.. 10times2=20 possibilities

If the number is divisible by 5 the last digit should be 0 or 5. Anyway they ask for the last 2 digits.There are 10 one digit numbers (0,1,2,3,4,5,6,7,8,9).All these numbers can be the one before the last digit.All the numbers can go with 2 possible last digits namely 0 & 5.

       So,
           ✳ 10 ✖ 2 = 20

at second last digit 10 options at last only 2 as 0 and 5 so total 10*2=20

The divisibility test of 5 is whether it ends with 0 or 5.If it ends with 0 or 5, it has 5 as a factor. For the second last digit, it can be any digit (0,1,2,3,4,5,6,7,8,9). There are 2 cases for the last digit and 10 cases for the second last digit,so $2 \times 10$ =10= the answer.

QED

The number could end with 0 or 5. The second-to-last digit doesn't affect divisibility by 5, since 10 is a multiple of 5, so it can be any of the 10 digits. The answer is $2\cdot 10=\boxed{20}$ .

If the integer is divisible by 5, then the units digit must be 0 or 5. If we are looking for the last two digits of an integer that is divisible by 5, then the tens digit can any be number from 0 to 9. Therefore, the total number of possibilities is 10 x 2 which is 20.

All multiples of 5 must end in 0, or 5. The second to last digit can be anything. So we have 10*2=20

We know that a number is divisible by $5$ iff - the last digit is either $0$ or $5$ .

Therefore we have to choose for two spaces :

$\boxed{A} \boxed{B}$

Now for box A : We have $10$ possibilities : ${0,1,2,3,4,5,6,7,8,9}$

And for box B : We have $2$ possibilities : ${ 0 , 5 }$

Thus possibilities can be written as :

$\boxed{10} \boxed{2}$

Therefore total number of possibilities are : $10 \cdot 2 = \boxed{120}$