For an odd number n, find the highest number that always divides n × (n2 – 1)?
[1] 12
[2] 24
[3] 48
[4] 96
n × (n2 – 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 × (n2 – 1) is 24.
For every positive integer n, the highest number that n × (n2 – 1) × (5n + 2) is always divisible by is
[1] 6
[2] 24
[3] 36
[4] 48
Case 1: If n is odd, n × (n2 – 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 × (n2 – 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 × (n2 – 1) × (5n + 2) is a multiple of 24.
On writing first 252 positive integers in a straight line, how many times digit 4 appears?
[1] 50
[2] 52
[3] 54
[4] 55
In the 1st 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.
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
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
1 and 8 are the first two positive integers for which 1 + 2 + 3 + ... + n is a perfect square. Which number is the 4th such number?
1 + 2 + 3 + … + n = $\frac{n(n+1)}{2}$= M2 (say) $\Rightarrow $ n(n + 1) = 2 M2
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.
For how many integers n is n4 + 6n < 6n3 + n2?
n4 + 6n - 6n3 - n2 < 0 $\Rightarrow $ n (n2 - 1)(n - 6) < 0. n cannot be equal to 1 or 0 or 6 because LHS becomes 0. Now n2 - 1 will always be positive, therefore, n (n - 6) should be negative $\Rightarrow $ n = 2, 3, 4 and 5.
For which integer n is 28 + 211 + 2n is a perfect square?
In order to write the above expression in the form (a + b)2 = a2 + 2ab + b2, we note that 28 = (24)2 and 211 = 2 × 24 × 26. Therefore, we need the square of 26 $\Rightarrow $ 2n = (26)2 = 212 $\Rightarrow $ n = 12.
Find the smallest positive integer n for which (22 - 1)(32 - 1)(42 - 1)… (n2 - 1) is a perfect square.
Nth term = (n2 - 1) = (n + 1)(n - 1) $\Rightarrow $ series = 1 × 3 × 2 × 4 × 3 × 5 … × (n - 2) × (n) × (n - 1) × (n + 1) = 2 n (n + 1) × k2 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.
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 p2 + pq + q2 is
[1] 900
[2] 901
[3] 998
[4] 899
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 = ${{(\sqrt{8}-\sqrt{7})}^{2}}$ = $15-2\sqrt{56}$
p2 + pq + q2 = ${{(15+2\sqrt{56})}^{2}}$+ ${{(15)}^{2}}-{{(2\sqrt{56})}^{2}}$+ ${{(15-2\sqrt{56})}^{2}}$ = 675 + 224 = 899
If x and y are positive integers and x2 – y2 = 101, find the value of x2 + y2.
[1] 5050
[2] 5150
[3] 5101
[4] None of these
x2 – y2 = (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.
\(\Rightarrow\) x2 + y2 = 512 + 502 = 5101.
For how many prime numbers p, is p2 + 3p - 1 a prime number?
[1] 0
[2] 1
[3] 2
[4] More than 2
When p = 3, the expression gives a prime number (17).
When p is not equal to 3, p2 will be of the form 3k + 1 as every square number is of the form 3n or 3n + 1. Therefore, p2 + 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.
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
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.
The digits of the number (4)24 are summed up continually till a single digit number is obtained. What is that number?
43 = 64. Digital sum of 64 is = 1. 424 = 43 × 43 × 43 …× 43 (8 times). Digital sums on both sides will be the same. $\Rightarrow $ digital sum of 424 = digital sum of 1 × 1 × 1 × 1… (8 times) = 1
What is the value of $[ \sqrt { 1 } ] + [ \sqrt { 2 } ] + [ \sqrt { 3 } ] + \ldots + [ \sqrt { 49 } ] + [ \sqrt { 50 } ]$ where $[ \mathrm { x } ]$ denotes the greatest integer function?
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 $= 3 \times 1 + 5 \times 2 + 7 \times 3 + \ldots + 13 \times 6 + 2 \times 7$
$= 217$ .
If $[ \sqrt { x } ] = 5 ]$ and $[ \sqrt { y } ] = 6 ,$ where $x$ and $y$ are natural numbers, what can be the greatest possible value of $x + y ?$
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 $x + y = 35 + 48 = 83$