## TCS Sample Divisibility Problems

Dear Reader, Below are three problems based on divisibility of numbers.

**Question 1**

Mylapore Times a local free newspaper conducted a test for students studying in corporation schools falling within the jurisdiction of Chennai. The test was aimed at ascertaining the level of mathematics knowledge among the school students. The following question was given :

“How many numbers are there between 133 to 294 both included, which are divisible by 7?”

a) 23 b) 24 c) 26 d) none of these.

**Answer : ** b) 24.

Solution :

In this problem both the boundary numbers 133 and 294 are divisible by 7. Also the boundary numbers are inclusive for calculation. In such scenarios solving is very simple.

**Quotient 1 (when upper boundary number is divided by the divisor, in our case 7)** : When 294 is divided by 7 we get 42. (1)

**Quotient 2 (when lower boundary number is divided by the divisor, in our case 7)** : When 133 is divided by 7 we get 19.(2)

Answer can be obtained using the formula, Quotient 1 - Quotient 2 + 1 = 42 - 19 + 1 = 24

42 - 19 = 23 + 1 = 24

**Question 2**

Porur Times is a newspaper which is distributed free to residents in and near Porur. It carries advertisement on different aspects such as rental, real estate, tuition, business deals etc. Porur Times gave a puzzle and wanted answers to be emailed to them within a day. The question read as under:

“How many three digit numbers can be formed using the digits 1,2,3,4,5 (but with repetition) that are divisible by 4?”

a) 12 b) 20 c) 60 d) 10

**Answer : ** b) 20

Solution :

To solve this problem, we are going to utilize the simple rule that for a number to be divisible by 4, its last two digits must be divisible by 4.

Using digits present in 1,2,3,4,5 the two digit combination's that are divisible by 4 include

12, 24, 32, 44

Now placing any of the digits from 1,2,3,4,5 before the 2 digit numbers that we arrived in previous step, we can actually find the total number of 3 digit numbers formed from 1,2,3,4,5 that are divisible by 4.

They are,

112, 124, 132, 144 212, 224, 232, 244 312, 324, 332, 344 412, 424, 432, 444 512, 524, 532, 544

Therefore there are 20 numbers which is our answer.

**Question 3**

Kumaresh attended a placement examination online and the following question was posed to him:

“ The difference between the squares of two consecutive odd integers is always divisible by" which of the following:

a) 6 b) 8 c) 7 d) 3

**Answer : ** b) 8

Let the two consecutive odd integers be (2x + 1) and (2x + 3).

Then, (2x + 3)^{2} - (2x +1)^{2}

4x^{2} + 12x + 9 - 4x^{2} - 4x -1 = 8x + 8 = 8(x +1).

Now for any value of x, 8(x +1). is divisible by 8. Therefore, answer is 8.

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