If n is a positive integer, prove that n + n + 1 = 4 n + 2 , where · denoted greatest integer function.

Since, n + n + 1 2 = 2 n + 1 + 2 n n + 1 We know that, if n is a positive integer, then n n + 1 > n 2 ⇒ n + n + 1 2 = 2 n + 1 + 2 n n + 1 > 2 n + 1 + 2 n 2 ⇒ n + n + 1 2 > 4 n + 1 … i Also, we know that n n + 1 < n + 1 2 ⇒ n + n + 1 2 < 2 n + 1 + 2 n + 1 2 ⇒ n + n + 1 2 < 4 n + 3 … ii Using, i and ii , we get 4 n + 1 < n + n + 1 2 < 4 n + 3 … iii Let k = 4 n + 2 , then by using the property of G . I . F . we know that x ≤ x < x + 1 , ⇒ k ≤ 4 n + 2 < k + 1 , ⇒ k 2 ≤ 4 n + 2 < k + 1 2 . Since 4 n + 2 and 4 n + 3 both are not perfect square numbers ( a perfect square number cannot have a remainder 2 or 3 when when divided by 4 ) , thus between two consecutive perfect squares we can have two numbers of the form 4 n + 2 and 4 n + 3 . Therefore, k 2 < 4 n + 2 < 4 n + 3 < k + 1 2 Also, a perfect square can be of the form 4 n or 4 n + 1 . ⇒ k 2 ≤ 4 n + 1 < 4 n + 2 < 4 n + 3 < k + 1 2 ⇒ k 2 ≤ 4 n + 1 < 4 n + 3 < k + 1 2 . . . iv Using, the equations iii and iv , we get ⇒ k 2 < n + n + 1 2 < k + 1 2 ⇒ k < n + n + 1 < k + 1 Thus, n + n + 1 = k = 4 n + 2 .

For every positive integer n , show that n + n + 1 = 4 n + 1 = 4 n + 2 = 4 n + 3 , where · denotes the greatest integer function.

Since, n + n + 1 2 = 2 n + 1 + 2 n n + 1 We know that, if n is a positive integer, then n n + 1 > n 2 ⇒ n + n + 1 2 = 2 n + 1 + 2 n n + 1 > 2 n + 1 + 2 n 2 ⇒ n + n + 1 2 > 4 n + 1 … i Also, we know that n n + 1 < n + 1 2 ⇒ n + n + 1 2 < 2 n + 1 + 2 n + 1 2 ⇒ n + n + 1 2 < 4 n + 3 … ii Using, i and ii , we get 4 n + 1 < n + n + 1 2 < 4 n + 3 … iii Let k = 4 n + 2 , then by using the property of G . I . F . , we know that x ≤ x < x + 1 , ⇒ k ≤ 4 n + 2 < k + 1 , ⇒ k 2 ≤ 4 n + 2 < k + 1 2 . Since, 4 n + 2 and 4 n + 3 both are not perfect square numbers ( a perfect square number cannot have a remainder 2 or 3 when when divided by 4 ) , thus between two consecutive perfect squares we can have two numbers of the form 4 n + 2 and 4 n + 3 . Therefore, k 2 < 4 n + 2 < 4 n + 3 < k + 1 2 Also, a perfect square can be of the form 4 n or 4 n + 1 . ⇒ k 2 ≤ 4 n + 1 < 4 n + 2 < 4 n + 3 < k + 1 2 ⇒ k 2 ≤ 4 n + 1 < 4 n + 3 < k + 1 2 . . . iv Using, the equations iii and iv , we get ⇒ k 2 < n + n + 1 2 < k + 1 2 ⇒ k < n + n + 1 < k + 1 Thus, n + n + 1 = k = 4 n + 2 . Also, from the equation iv , we can observe that k 2 ≤ 4 n + 1 < k + 1 2 ⇒ 4 n + 1 = k Thus, we get k = 4 n + 1 = 4 n + 2 . Again, from the equation iv , we can observe that k 2 < 4 n + 1 < k + 1 2 ⇒ k = 4 n + 3 Hence, finally we get n + n + 1 = 4 n + 1 = 4 n + 2 = 4 n + 3 .

Solve the equation x x + x = 2 x + 10 , where · denotes the greatest integer function and · denotes the fractional part function.

By using x = x + x , the given equation x x + x = 2 x + 10 becomes x x + x + x = 2 x + 10 ⇒ x x + x - x = 10 , ⇒ x x - x + x - 1 = 10 - 1 , ⇒ x - 1 x + x - 1 = 9 , ⇒ x - 1 x + 1 = 9 . Since, x - 1 is an integer, so x + 1 is a rational number. Let x = m n , where m and n are integers and we know that 0 ≤ x < 1 ⇒ 0 ≤ m n < 1 , ⇒ 0 ≤ m < n , Then, we get x - 1 m n + 1 = 9 ⇒ x - 1 m + n = 9 n . . . i When, m = 0 , then x = 0 , so x = 10 . When, m > 0 , let H . C . F . m , n = 1 , then H . C . F . m + n , n = 1 , So, from equation i m + n be a divisor of 9 i.e., m + n can take values 1 , 3 or 9 . But m + n = 1 is not possible because of m > 0 and m < n , then the possible values for m = 0 , n = 1 and x = 0 , which we had already taken. i When, m + n = 3 , then m = 1 , n = 2 , So x = m n = 1 2 , Put this value in the equation i , to get x - 1 3 = 9 × 2 ⇒ x - 1 = 6 ⇒ x = 7 , ⇒ x = x + x = 7 + 1 2 = 7 1 2 . i When m + n = 9 , then there are three possible cases m = 1 , n = 8 ⇒ x = 1 8 , Put this value in the equation i , to get x - 1 9 = 9 × 8 ⇒ x - 1 = 8 ⇒ x = 9 , ⇒ x = 9 + 1 8 = 9 1 8 . And, m = 2 , n = 7 ⇒ x = 2 7 , ⇒ x - 1 9 = 9 × 7 ⇒ x - 1 = 7 ⇒ x = 8 , ⇒ x = 8 + 2 7 = 8 2 7 And, m = 4 , n = 5 ⇒ x = 4 5 , ⇒ x - 1 9 = 9 × 5 ⇒ x - 1 = 5 ⇒ x = 6 ⇒ x = 6 + 4 5 = 6 4 5 . Thus, there are five solutions for x : 10 , 7 1 2 , 9 1 8 , 8 2 7 , 6 4 5 .

E M B I B E

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.6>Q 13

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Find the sum of all the digit of largest integer $N$ such that both $N + 496$ and $N + 224$ are perfect squares.

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Important Questions on Number System

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Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.7>Q 1

If

n

is a positive integer, prove that

[\sqrt{n} + \sqrt{n + 1}] = [\sqrt{4 n + 2}],

where

[\cdot]

denoted greatest integer function.

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IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.7>Q 2

The value of

x

satisfying the equation

[x^{3}] + [x^{2}] + [x] = \{x\} - 1,

where

[\cdot]

denotes the greatest integer function and

\{\cdot\}

denotes the fractional part function, is

m

. Then

|m|

is

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IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.7>Q 3

Solve equation

[3 x + 1] = 2 x - \frac{1}{2},

and find the absolute value of the sum of all roots, where

[\cdot]

denotes the greatest integer function.

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Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.7>Q 4

Find the minimum natural number

n,

such that the equation

[\frac{10^{n}}{x}] = 1989

has integer solution

x,

where

[\cdot]

denoted greatest integer function

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IMPORTANT

Non-Routine Mathematics Resource Book-1 for PRMO>Chapter 2 - Number System>EXERCISE - 2.7>Q 6

The number of different non-negative integers that are there in the sequence $[\frac{1^{2}}{1980}], [\frac{2^{2}}{1980}], [\frac{3^{2}}{1980}], \dots, [\frac{1980^{2}}{1980}]$ be $m$ , then what is the sum of digits of $m$ ?