**Question 1:**

For an odd number n, find the highest number that always divides n × (n

^{2}– 1)?

[1] 12

[2] 24

[3] 48

[4] 96

**Answer & Solution**

n × (n^{2} – 1) = (n – 1) × n × (n + 1), which is a product of three consecutive numbers.

Since n is odd, the numbers (n – 1) and (n + 1) are both even. As they are two consecutive even numbers one of these numbers will be a multiple of 2 and the other will be a multiple of 4.

Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three.

Hence, the product of three numbers will be a multiple of 8 × 3 = 24.

Therefore, the highest number that always divides n × (n^{2} – 1) is 24.

**Question 2:**

For every positive integer n, the highest number that n × (n

^{2}– 1) × (5n + 2) is always divisible by is

[1] 6

[2] 24

[3] 36

[4] 48

**Answer & Solution**

Case 1: If n is odd, n × (n^{2} – 1) is divisible by 24 as proved in the earlier question.

Case 2: If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive positive integers is always a multiple of 3 and n is even, the product n × (n^{2} – 1) is divisible by 6. Since n is even 5n is even.

If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n × (5n + 2) is a multiple of 8.

Hence, the product n × (n^{2} – 1) × (5n + 2) is a multiple of 24.

**Question 3:**

On writing first 252 positive integers in a straight line, how many times digit 4 appears?

[1] 50

[2] 52

[3] 54

[4] 55

**Answer & Solution**

In the 1^{st} 99 positive integers, digit 4 comes 20 times.

Similarly, from 100 to 199, digit 4 comes 20 times.

Now from 200 to 299, digit 4 comes again 20 times out of which we need to subtract 5 numbers (254, 264, 274, 284 and 294).

Therefore, total number of times that we write the digit 4 = 20 + 20 + 20 - 5 = 55.

**Question 4:**

If a book has 252 pages, how many digits have been used to number the pages?

[1] 650

[2] 648

[3] 660

[4] None of these

**Answer & Solution**

From page number 1 to page number 9, we will use 1 digit per page $\Rightarrow $ digits used = 9.

From page number 10 to page number 99, we will use 2 digits per page $\Rightarrow $ digits used = 2 × 90 = 180.

From page number 100 to page number 252, we will use 3 digits per page $\Rightarrow $ digits used = 3 × 153 = 459.

Therefore, total number of digits used = 9 + 180 + 459 = 648

**Question 5:**

1 and 8 are the first two positive integers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4

^{th}such number?

**Answer & Solution**

1 + 2 + 3 + … + n = $\frac{n(n+1)}{2}$= M^{2} (say) $\Rightarrow $ n(n + 1) = 2 M^{2}

Now n and n + 1 will have no factor in common. Since RHS is twice the square of a positive integer, one of n and n + 1 will be twice of a perfect square and the other will be a perfect square.

As twice of a perfect square will be even, the other square will be odd. We start investigating the odd squares and their neighbours. The fourth such numbers we get is 288.

**Question 6:**

For how many integers n is n

^{4}+ 6n < 6n

^{3}+ n

^{2}?

**Answer & Solution**

n^{4} + 6n - 6n^{3} - n^{2 }< 0 $\Rightarrow $ n (n^{2} - 1)(n - 6) < 0. n cannot be equal to 1 or 0 or 6 because LHS becomes 0. Now n^{2} - 1 will always be positive, therefore, n (n - 6) should be negative $\Rightarrow $ n = 2, 3, 4 and 5.

**Question 7:**

For which integer n is 2

^{8}+ 2

^{11}+ 2

^{n }is a perfect square?

**Answer & Solution**

In order to write the above expression in the form (a + b)^{2} = a^{2} + 2ab + b^{2}, we note that 2^{8} = (2^{4})^{2} and 2^{11} = 2 × 2^{4} × 2^{6}. Therefore, we need the square of 2^{6} $\Rightarrow $ 2^{n} = (2^{6})^{2} = 2^{12} $\Rightarrow $ n = 12.

**Question 8:**

Find the smallest positive integer n for which (2

^{2}- 1)(3

^{2}- 1)(4

^{2}- 1)… (n

^{2}- 1) is a perfect square.

**Answer & Solution**

Nth term = (n^{2} - 1) = (n + 1)(n - 1) $\Rightarrow $ series = 1 × 3 × 2 × 4 × 3 × 5 … × (n - 2) × (n) × (n - 1) × (n + 1) = 2 n (n + 1) × k^{2} because all the other terms are squared.

The first value of n for which 2 n (n + 1) is a perfect square is n = 8.

**Question 9:**

If p = $\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}-\sqrt{7}}$ and q = $\frac{\sqrt{8}-\sqrt{7}}{\sqrt{8}+\sqrt{7}}$, then the value of p

^{2}+ pq + q

^{2}is

[1] 900

[2] 901

[3] 998

[4] 899

**Answer & Solution**

p = $\frac{\sqrt{8}+\sqrt{7}}{\sqrt{8}-\sqrt{7}}$ = $\frac{{{(\sqrt{8}+\sqrt{7})}^{2}}}{(\sqrt{8}-\sqrt{7)}(\sqrt{8}+\sqrt{7})}$= ${{(\sqrt{8}+\sqrt{7})}^{2}}$ = $15+2\sqrt{56}$

Similarly, q = ${{(\sqrt{8}-\sqrt{7})}^{2}}$ = $15-2\sqrt{56}$

p^{2} + pq + q^{2} = ${{(15+2\sqrt{56})}^{2}}$+ ${{(15)}^{2}}-{{(2\sqrt{56})}^{2}}$+ ${{(15-2\sqrt{56})}^{2}}$ = 675 + 224 = 899

**Question 10:**

If x and y are positive integers and x

^{2}– y

^{2}= 101, find the value of x

^{2}+ y

^{2}.

[1] 5050

[2] 5150

[3] 5101

[4] None of these

**Answer & Solution**

x^{2} – y^{2} = (x + y)(x – y) = 101. But 101 is a prime number and cannot be written as product of two numbers unless one of the numbers is 1 and the other is 101 itself. Hence, x + y = 101 and x – y = 1. à x = 51, y = 50.

\(\Rightarrow\) x^{2} + y^{2} = 51^{2} + 50^{2} = 5101.

**Question 11:**

For how many prime numbers p, is p

^{2}+ 3p - 1 a prime number?

[1] 0

[2] 1

[3] 2

[4] More than 2

**Answer & Solution**

When p = 3, the expression gives a prime number (17).

When p is not equal to 3, p^{2} will be of the form 3k + 1 as every square number is of the form 3n or 3n + 1. Therefore, p^{2} + 3p - 1 = 3k + 1 + 3p - 1 $\Rightarrow $ a multiple of 3. Therefore, for only p = 3, do we get a prime number from the expression.

**Question 12:**

The number of positive integers n in the range \(12 \le n \le 40\) such that the product (n - 1) (n - 2) (n - 3)..3×2×1 is not divisible by n is

[1] 0

[2] 7

[3] 13

[4] None of these

**Answer & Solution**

The product (n - 1) (n - 2) (n - 3)..3×2×1 will not be divisible by n only when this product does not contain factors of n, i.e. n is a prime number. The prime numbers in the range given are 13, 17, 19, 23, 29, 31, and 37. 7 numbers in all.

**Question 13:**

The digits of the number (4)

^{24}are summed up continually till a single digit number is obtained. What is that number?

**Answer & Solution**

4^{3} = 64. Digital sum of 64 is = 1. 4^{24} = 4^{3} × 4^{3} × 4^{3} …× 4^{3} (8 times). Digital sums on both sides will be the same. $\Rightarrow $ digital sum of 4^{24} = digital sum of 1 × 1 × 1 × 1… (8 times) = 1

**Question 14:**

What is the value of $[ \sqrt { 1 } ] + [ \sqrt { 2 } ] + [ \sqrt { 3 } ] + \ldots + [ \sqrt { 49 } ] + [ \sqrt { 50 } ]$ where $[ \mathrm { x } ]$ denotes the greatest integer function?

**Answer & Solution**

It can be seen that

$[ \sqrt { 1 } ] = 1 , [ \sqrt { 2 } ] = [ 1.41 ] = 1 , [ \sqrt { 3 } ] = [ 1.73 ] = 1 , [ \sqrt { 4 } ] = 2$ and so on.

Therefore, from $[ \sqrt { 1 } ]$ to $[ \sqrt { 3 } ]$ , the value will be 1 , from $[ \sqrt { 4 } ]$ to $[ \sqrt { 8 } ]$ the value will be 2 , from $[ \sqrt { 9 } ]$ to $[ \sqrt { 15 } ]$ the value will be 3 and so on..

Therefore, the total value $= 3 \times 1 + 5 \times 2 + 7 \times 3 + \ldots + 13 \times 6 + 2 \times 7$

$= 217$ .

**Question 15:**

If $[ \sqrt { x } ] = 5 ]$ and $[ \sqrt { y } ] = 6 ,$ where $x$ and $y$ are natural numbers, what can be the greatest possible value of $x + y ?$

**Answer & Solution**

It is clear that $[ \sqrt { 25 } ] = 5 , [ \sqrt { 26 } ] = 5 , [ \sqrt { 27 } ] = 5$ and so on. The highest value of $x$ that we can take is 35 , since $[ \sqrt { 35 } ] = 5$ but $[ \sqrt { 36 } ] = 6$

Similarly, the highest value of y we can take is 48 , since

$[ \sqrt { 48 } ] = 6$ but $[ \sqrt { 49 } ] = 7$

Therefore, the greatest value of $x + y = 35 + 48 = 83$