Questions on Properties of Numbers

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Question 1:
For an odd number n, find the highest number that always divides n × (n2 – 1)?

[1] 12

[2] 24

[3] 48

[4] 96

Answer & Solution
Option # 2

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.


Question 2:
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

Answer & Solution
Option # 2

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.


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
Option # 4

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.


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
Option # 2

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

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

From page number 100 to age 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 4th  such number?
Answer & Solution
Correct Answer: 288

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.


Question 6:
For how many integers n is n4 + 6n < 6n3 + n2?
Answer & Solution
Correct Answer: 4

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.


Question 7:
For which integer n is 28 + 211 + 2n is a perfect square?
Answer & Solution
Correct Answer: 12

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.


Question 8:
Find the smallest positive integer n for which (22 - 1)(32 - 1)(42 - 1)… (n2 - 1) is a perfect square.
Answer & Solution
Correct Answer: 8

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.


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 p2 + pq + q2 is

[1] 900

[2] 901

[3] 998

[4] 899

Answer & Solution
Option # 4

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


Question 10:
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

Answer & Solution
Option # 3

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.

à x2 + y2 = 512 + 502 = 5101.


Question 11:
For how many prime numbers p, is p2 + 3p - 1 a prime number?

[1] 0

[2] 1

[3] 2

[4] More than 2

Answer & Solution
Option # 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.


Question 12:
The number of positive integers n in the range 12 £ n £ 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
Option # 3

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
Correct Answer: 1

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


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
Correct Answer: 217

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
Correct Answer: 83

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 $x + y = 35 + 48 = 83$


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