CBSE Class 10 Real Numbers Sure Shot Questions Set C

Short Answer

Question. Prove that \( \sqrt{5} \) is irrational and hence show that \( 3 + \sqrt{5} \) is also irrational.
Answer: Let us assume, to the contrary, that \( \sqrt{5} \) is rational. So, we can find integers \( p \) and \( q \) (\( q \neq 0 \)), such that \( \sqrt{5} = \frac{p}{q} \), where \( p \) and \( q \) are coprime.
Squaring both sides, we get
\( 5 = \frac{p^2}{q^2} \Rightarrow 5q^2 = p^2 \) ...(i)
\( \Rightarrow 5 \) divides \( p^2 \therefore 5 \) divides \( p \)
So, let \( p = 5r \)
Putting the value of \( p \) in (i), we get
\( 5q^2 = (5r)^2 \Rightarrow 5q^2 = 25r^2 \)
\( \Rightarrow q^2 = 5r^2 \)
\( \Rightarrow 5 \) divides \( q^2 \therefore 5 \) divides \( q \)
So, \( p \) and \( q \) have atleast 5 as a common factor.
But this contradicts the fact that \( p \) and \( q \) have no common factor.
So, our assumption is wrong.
\( \therefore \sqrt{5} \) is irrational.
\( \sqrt{5} \) is irrational, 3 is a rational number.
So, we conclude that \( 3 + \sqrt{5} \) is irrational.

Question. Prove that \( 3 + 2\sqrt{3} \) is an irrational number.
Answer: Let us assume to the contrary, that \( 3 + 2\sqrt{3} \) is rational.
So that we can find integers \( a \) and \( b \) (\( b \neq 0 \)).
Such that \( 3 + 2\sqrt{3} = \frac{a}{b} \), where \( a \) and \( b \) are coprime.
Rearranging the equations, we get
\( 2\sqrt{3} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \)
\( \sqrt{3} = \frac{a - 3b}{2b} = \frac{a}{2b} - \frac{3b}{2b} \)
\( \sqrt{3} = \frac{a}{2b} - \frac{3}{2} \)
Since \( a \) and \( b \) are integers, we get \( \frac{a}{2b} - \frac{3}{2} \) is rational and so \( \sqrt{3} \) is rational.
But this contradicts the fact that \( \sqrt{3} \) is irrational.
So we conclude that \( 3 + 2\sqrt{3} \) is irrational.

Question. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together?
Answer: \( 9 = 3^2 \), \( 12 = 2^2 \times 3 \), \( 15 = 3 \times 5 \)
LCM = \( 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \) minutes or 3 hours.
\( \therefore \) They will next toll together after 3 hours.

Question. Two tankers contain 850 litres and 680 litres of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in exact number of times.
Answer: To find maximum capacity of a container which can measure the petrol of each tanker in exact number of times, we find the HCF of 850 and 680.
\( 850 = 2 \times 5^2 \times 17 \)
\( 680 = 2^3 \times 5 \times 17 \)
\( \therefore HCF = 2 \times 5 \times 17 = 170 \)
\( \therefore \) Maximum capacity of the container = 170 litres

Question. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly.
Answer: To find the length of the longest rod that can measure the dimensions of the room exactly, we have to find HCF.
L, Length = 8 m 50 cm = 850 cm = \( 2 \times 5^2 \times 17 \)
B, Breadth = 6 m 25 cm = 625 cm = \( 5^4 \)
H, Height = 4 m 75 cm = 475 cm = \( 5^2 \times 19 \)
\( \therefore HCF \) of L, B and H is \( 5^2 = 25 \) cm.
\( \therefore \) Length of the longest rod = 25 cm

Question. Three alarm clocks ring at intervals of 4, 12 and 20 minutes respectively. If they start ringing together, after how much time will they next ring together?
Answer: To find the time when the clocks will next ring together, we have to find LCM of 4, 12 and 20 minutes.
\( 4 = 2^2 \)
\( 12 = 2^2 \times 3 \)
\( 20 = 2^2 \times 5 \)
\( \therefore LCM \) of 4, 12 and 20 \( = 2^2 \times 3 \times 5 = 60 \) minutes.
So, the clocks will ring together again after 60 minutes or one hour.

Question. In a school there are two Sections A and B of class X. There are 48 students in Section A and 60 students in Section B. Determine the least number of books required for the library of the school so that the books can be distributed equally among all students of each Section.
Answer: Since the books are to be distributed equally among the students of Section A and Section B, therefore, the number of books must be a multiple of 48 as well as 60. Hence, required number of books is the LCM of 48 and 60.
\( 48 = 2^4 \times 3 \)
\( 60 = 2^2 \times 3 \times 5 \)
\( \therefore LCM = 2^4 \times 3 \times 5 = 16 \times 15 = 240 \)
Hence, required number of books is 240.

Question. By using Euclid’s algorithm, find the largest number which divides 650 and 1170.
Answer: Given numbers are 650 and 1170.
\( 1170 > 650 \)
\( 1170 = 650 \times 1 + 520 \)
\( 650 = 520 \times 1 + 130 \)
\( 520 = 130 \times 4 + 0 \)
\( \therefore HCF = 130 \)
\( \therefore \) The required largest number is 130.

Question. Find the HCF of 255 and 867 by Euclid’s division algorithm.
Answer: 867 is greater than 255. We apply the division lemma to 867 and 255, to get
\( 867 = 255 \times 3 + 102 \)
We continue the process till the remainder is zero
\( 255 = 102 \times 2 + 51 \)
\( 102 = 51 \times 2 + 0 \),
the remainder is zero.
\( \therefore HCF = 51 \)

Question. Using Euclid’s division algorithm, find whether the pair of numbers 847, 2160 are coprimes or not.
Answer: \( 2160 > 847 \)
\( 2160 = 847 \times 2 + 466 \)
\( 847 = 466 \times 1 + 381 \)
\( 466 = 381 \times 1 + 85 \)
\( 381 = 85 \times 4 + 41 \)
\( 85 = 41 \times 2 + 3 \)
\( 41 = 3 \times 13 + 2 \)
\( 3 = 2 \times 1 + 1 \)
\( 2 = 1 \times 2 + 0 \)
Since remainder = 0
\( \therefore HCF = 1 \)
Since, HCF is 1
\( \therefore \) 847 and 2160 are coprimes.

Long Answer

Question. Prove that \( 3 + 2\sqrt{5} \) is irrational.
Answer: Let us assume, to the contrary, that \( 3 + 2\sqrt{5} \) is rational
So that we can find integers \( a \) and \( b \) (\( b \neq 0 \)), such that \( 3 + 2\sqrt{5} = \frac{a}{b} \), where \( a \) and \( b \) are coprime.
Rearranging this equation, we get
\( \Rightarrow 2\sqrt{5} = \frac{a}{b} - 3 \Rightarrow \sqrt{5} = \frac{a - 3b}{2b} \)
\( \Rightarrow \sqrt{5} = \frac{a}{2b} - \frac{3b}{2b} \Rightarrow \sqrt{5} = \frac{a}{2b} - \frac{3}{2} \)
Since \( a \) and \( b \) are integers, we get that \( \frac{a}{2b} - \frac{3}{2} \) is rational and so \( \sqrt{5} \) is rational.
But this contradicts the fact that \( \sqrt{5} \) is irrational.
So we conclude that \( 3 + 2\sqrt{5} \) is irrational.

Question. There are 104 students in class X and 96 students in class IX in a school. In a house examination the students are to be evenly seated in parallel rows such that no two adjacent rows are of the same class.
(a) Find the maximum number of parallel rows of each class for the seating arrangement.
(b) Also find the number of students of class IX and also of class X in a row.
(c) What is the objective of the school administration behind such an arrangement?
Answer: \( 104 = 2^3 \times 13 \)
\( 96 = 2^5 \times 3 \)
\( \therefore HCF = 2^3 = 8 \)
(a) Number of rows of students of class X = \( \frac{104}{8} = 13 \)
Number maximum of rows of students of class IX = \( \frac{96}{8} = 12 \)
\( \therefore \text{Total number of rows} = 13 + 12 = 25 \)
(b) No. of students of class IX in a row = 8
No. of students of class X in a row = 8
(c) The objective of school administration behind such an arrangement is fair and clean examination, so that no student can take help from any other student of his/her class.

Question. Dudhnath has two vessels containing 720 ml and 405 ml of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled.
Answer: \( 1^{st} \text{ vessel} = 720 \text{ ml} \); \( 2^{nd} \text{ vessel} = 405 \text{ ml} \)
We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.
\( 405 = 3^4 \times 5 \)
\( 720 = 2^4 \times 3^2 \times 5 \)
\( HCF = 3^2 \times 5 = 45 \text{ ml} = \text{Capacity of glass} \)
No. of glasses filled from \( 1^{st} \text{ vessel} = \frac{720}{45} = 16 \)
No. of glasses filled from \( 2^{nd} \text{ vessel} = \frac{405}{45} = 9 \)
\( \therefore \text{Total number of glasses} = 25 \)

Question. Amita, Suneha and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?
Answer: To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.
\( 10 = 2 \times 5 \), \( 16 = 2^4 \)
\( 20 = 2^2 \times 5 \)
\( LCM = 2^4 \times 5 = 16 \times 5 = 80 \text{ minutes} \)
They will start preparing a new card together after 80 minutes.

Question. Find HCF of numbers 134791, 6341 and 6339 by Euclid’s division algorithm.
Answer: First we find HCF of 6339 and 6341 by Euclid’s division method.
\( 6341 > 6339 \)
\( 6341 = 6339 \times 1 + 2 \)
\( 6339 = 2 \times 3169 + 1 \)
\( 2 = 1 \times 2 + 0 \)
\( \therefore HCF \text{ of 6341 and 6339 is 1.} \)
Now, we find the HCF of 134791 and 1
\( 134791 = 1 \times 134791 + 0 \)
\( \therefore HCF \text{ of 134791 and 1 is 1.} \)
Hence, HCF of the given three numbers is 1.

Question. If two positive integers x and y are expressible in terms of primes as \( x = p^2q^3 \) and \( y = p^3q \), what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.
Answer: \( x = p^2q^3 \) and \( y = p^3q \)
\( LCM = p^3q^3 \); \( HCF = p^2q \) ...(i)
Now, \( LCM = p^3q^3 \Rightarrow LCM = pq^2 (p^2q) \)
\( \Rightarrow LCM = pq^2 (HCF) \)
Yes, LCM is a multiple of HCF.
Explanation: Let \( a = 12 = 2^2 \times 3 \)
\( b = 18 = 2 \times 3^2 \)
\( \therefore HCF = 2 \times 3 = 6 \) ...(ii)
\( LCM = 2^2 \times 3^2 = 36 \)
\( LCM = 6 \times 6 \)
\( LCM = 6 (HCF) \) ...[From (ii)]
Here LCM is 6 times HCF.

Question. Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer.
Answer: Let \( n, n + 1, n + 2 \) be three consecutive positive integers. We know that \( n \) is of the form \( 3q, 3q + 1, \) or \( 3q + 2 \).
Case I. When \( n = 3q \),
In this case, \( n \) is divisible by 3, but \( n + 1 \) and \( n + 2 \) are not divisible by 3.
Case II. When \( n = 3q + 1 \), in this case \( n + 2 \)
\( = (3q + 1) + 2 = 3q + 3 \)
\( = 3(q + 1) \), \( (n + 2) \) is divisible by 3, but \( n \) and \( n + 1 \) are not divisible by 3.
Case III. When \( n = 3q + 2 \), in this case, \( n + 1 \)
\( = (3q + 2) + 1 = 3q + 3 \)
\( = 3(q + 1) \), \( (n + 1) \) is divisible by 3, but \( n \) and \( n + 2 \) are not divisible by 3.
Hence, one and only one out of \( n, n + 1 \) and \( n + 2 \) is divisible by 3.

Question. Find the HCF and LCM of 306 and 657 and verify that \( LCM \times HCF = \text{Product of the two numbers.} \)
Answer: \( 306 = 2 \times 3^2 \times 17 \)
\( 657 = 3^2 \times 73 \)
\( HCF = 3^2 = 9 \)
\( LCM = 2 \times 3^2 \times 17 \times 73 = 22338 \)
L.H.S. \( = LCM \times HCF \)
\( = 22338 \times 9 = 201042 \)
R.H.S. \( = \text{Product of two numbers} \)
\( = 306 \times 657 = 201042 \)
\( \therefore \text{L.H.S. = R.H.S.} \)

Question. Show that any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \) where \( q \) is a positive integer.
Answer: Let \( a \) be a positive odd integer
By Euclid’s Division algorithm: \( a = 4q + r \)
[where \( q, r \) are positive integers and \( 0 \leq r < 4 \)]
\( \therefore a = 4q \)
or \( 4q + 1 \)
or \( 4q + 2 \)
or \( 4q + 3 \)
But \( 4q \) and \( 4q + 2 \) are both even
\( \therefore a \) is of the form \( 4q + 1 \) or \( 4q + 3 \).

MCQs

Question. For some integer m, every even integer is of the form
(a) m
(b) m + 1
(c) 2m
(d) 2m + 1
Answer: (c)
Answer: Explanation: We know that, an integer divisible by 2 is an even integer. Suppose m is any integer then, \( m = -3, -2, -1, 0, 1, 2, 3 ... \)
\( \Rightarrow 2m \text{ is } -6, -4, -2, 0, 2, 4, 6 ... \text{ is clearly divisible by 2.} \)
So, for some integer m, every even integer is of the form 2m.

Question. For some integer q, every odd integer is of the form
(a) q
(b) q + 1
(c) 2q
(d) 2q + 1
Answer: (d)
Answer: Explanation: Let q be an integer. Then, q is \( -3, -2, -1, 0, 1, 2, 3 ... \)
\( \Rightarrow 2q \text{ will be } -6, -4, -2, 0, 2, 4, 6 ... \)
Now, \( 2q + 1 \text{ will be } -5, -3, -1, 1, 3, 5, 7 ... \text{ which is clearly not divisible by 2.} \)
So, for some integer q, every odd integer is of the form \( 2q + 1 \).

Question. \( n^2 - 1 \) is divisible by 8, if n is
(a) an integer
(b) a natural number
(c) an odd integer
(d) an even integer
Answer: (c)
Answer: Explanation: In \( n^2 - 1 \), n can be either even or odd. Let \( x = n^2 - 1 \).
Case I: When n is even, i.e., \( n = 2k \). \( x = 4k^2 - 1 \). For \( k=1, x=3 \) (not div by 8).
Case II: When n is odd, i.e., \( n = 2k+1 \). \( x = (2k+1)^2 - 1 = 4k^2 + 4k = 4k(k+1) \).
At \( k=1, x=8 \) which is divisible by 8. Therefore, \( n^2 - 1 \) is divisible by 8 if n is an odd integer.

Question. If two positive integers a and b are written as \( a = x^3y^2 \) and \( b = xy^3 \), where x, y are prime numbers, then HCF (a, b) is
(a) xy
(b) \( xy^2 \)
(c) \( x^3y^3 \)
(d) \( x^2y^2 \)
Answer: (b)
Answer: Explanation: Here, \( a = x^3y^2 \) and \( b = xy^3 \). As HCF is the product of the smallest power of each common prime factor, \( HCF(a, b) = xy^2 \).

Question. If two positive integers p and q can be expressed as \( p = ab^2 \) and \( q = a^3b \); where a, b being prime numbers, then LCM (p, q) is
(a) ab
(b) \( a^2b^2 \)
(c) \( a^3b^2 \)
(d) \( a^3b^3 \)
Answer: (c)
Answer: Explanation: Given, \( p = ab^2 \) and \( q = a^3b \). LCM is the product of the greatest power of each prime factor involved. \( LCM(p, q) = a^3b^2 \).

Question. The decimal expansion of the rational number \( \frac{14587}{1250} \) will terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Answer: (d)
Answer: Explanation: \( \frac{14587}{1250} = \frac{14587}{2^1 \times 5^4} = \frac{14587 \times 2^3}{2 \times 5^4 \times 2^3} = \frac{116696}{10000} = 11.6696 \). Terminates after four decimal places.

Very Short Answer Type Questions

Question. Write whether every positive integer can be of the form \( 4q + 2 \), where q is an integer. Justify your answer.
Answer: No. Every positive integer cannot be of the form \( 4q + 2 \). By Euclid’s Lemma, \( b = aq + r, 0 \leq r < a \). For \( a=4 \), \( b = 4q + r \) for \( r = 0, 1, 2, 3 \). Thus it must be in the form \( 4q, 4q + 1, 4q + 2 \) or \( 4q + 3 \).

Question. Write whether the square of any positive integer can be of the form \( 3m + 2 \), where m is a natural number. Justify your answer.
Answer: No. Using Euclid’s lemma, any positive integer is of the form \( 3q, 3q + 1 \) or \( 3q + 2 \).
\( (3q)^2 = 9q^2 = 3m \)
\( (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2+2q)+1 = 3m+1 \)
\( (3q+2)^2 = 9q^2 + 12q + 4 = 3(3q^2+4q+1)+1 = 3m+1 \)
The square can be of the form \( 3m \) and \( 3m + 1 \) but not \( 3m + 2 \).

Question. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.
Answer: As the numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75 and the highest number among these is 75 so, the \( HCF(525, 3000) = 75 \).

Question. Without actually performing the long division find if \( 987/10500 \) will have terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.
Answer: On simplifying, \( \frac{987}{10500} = \frac{47}{500} \). Writing denominator in prime factors: \( 500 = 5^3 \cdot 2^2 \). As denominator is of type \( 2^m 5^n \), it is terminating. \( \frac{47 \times 2}{500 \times 2} = \frac{94}{1000} = 0.094 \).

Question. A rational number in its decimal expansion is 327.7081. What can you say about the prime factors of q, when this number is expressed in the form p/q? Give reasons.
Answer: As 327.7081 is a terminating decimal number hence, it can be written in the form \( p/q \) where \( q \) is of the form \( 2^m \times 5^n \). \( q = 10^4 = 2^4 \times 5^4 \). Prime factors are 2 and 5.

Short Answer Type Questions

Question. Show that cube or any positive integer is of the form \( 4m, 4m + 1 \) or \( 4m + 3 \), for some integer m.
Answer: Suppose \( a \) is any positive integer. \( a = 4k + r \), where \( 0 \leq r < 4 \).
\( a^3 = (4k+r)^3 = 64k^3 + 48k^2r + 12kr^2 + r^3 \)
When \( r=0, a^3 = 64k^3 = 4(16k^3) = 4m \).
When \( r=1, a^3 = 64k^3 + 48k^2 + 12k + 1 = 4(16k^3 + 12k^2 + 3k) + 1 = 4m + 1 \).
When \( r=3, a^3 = 64k^3 + 144k^2 + 108k + 27 = 4(16k^3 + 36k^2 + 27k + 6) + 3 = 4m + 3 \).

Question. Show that the square of any positive integer cannot be of the form \( 5q + 2 \) or \( 5q + 3 \) for any integer q.
Answer: \( a = 5k + r, 0 \leq r < 5 \).
\( a^2 = 5(5k^2 + 2kr) + r^2 \).
\( r=0, a^2 = 5q \).
\( r=1, a^2 = 5q + 1 \).
\( r=2, a^2 = 5q + 4 \).
\( r=3, a^2 = 5q + 4 \).
\( r=4, a^2 = 5q + 1 \).
Square cannot be in form \( 5q + 2 \) or \( 5q + 3 \).

Question. Show that the square of any odd integer is of the form \( 4m + 1 \), for some integer m.
Answer: Odd integers are of form \( 4q+1 \) and \( 4q+3 \).
\( (4q+1)^2 = 16q^2 + 8q + 1 = 4(4q^2+2q)+1 = 4m+1 \).
\( (4q+3)^2 = 16q^2 + 24q + 9 = 4(4q^2+6q+2)+1 = 4m+1 \).

Question. Prove that, if x and y are both odd positive integers, then \( x^2 + y^2 \) is even but not divisible by 4.
Answer: Let \( x = 2m+1, y = 2m+3 \).
\( x^2 + y^2 = (2m+1)^2 + (2m+3)^2 = 4m^2+4m+1 + 4m^2+12m+9 = 8m^2+16m+10 \).
\( = 2(4m^2+8m+5) \) which is even. But it is \( 4(2m^2+4m+2)+2 \), not divisible by 4.

Question. Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Answer: Subtract remainders: 1250, 9375, 15625.
HCF(15625, 9375): \( 15625 = 9375 \times 1 + 6250 \); \( 9375 = 6250 \times 1 + 3125 \); \( 6250 = 3125 \times 2 + 0 \).
HCF(3125, 1250): \( 3125 = 1250 \times 2 + 625 \); \( 1250 = 625 \times 2 + 0 \).
HCF is 625.

Question. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?
Answer: The minimum distance each should walk will be the LCM of 40 cm, 42 cm and 45 cm.
In terms of prime numbers,
\( 40 = 2 \times 2 \times 2 \times 5 \),
\( 42 = 2 \times 3 \times 7 \)
and \( 45 = 3 \times 3 \times 5 \)
Thus,
LCM (40, 42, 45) = \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520 \)
Minimum distance each should walk is 2520 cm.

Question. Write the denominator of rational number \( \frac{257}{5000} \) in the form \( 2^m \times 5^n \), where \( m = 3 \) and \( n = 4 \) are non-negative integers. Hence, write its decimal expansion, without actual division.
Answer: Denominator of the rational number \( \frac{257}{5000} \) is 5000.
In terms of prime factors,
\( 5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = (2)^3 \times (5)^4 \),
which is the same as \( 2^m \times 5^n \), with \( m = 3 \) and \( n = 4 \) and both are non-negative integers.
Now, Rational number, \( \frac{257}{5000} = \frac{257}{2^3 \times 5^4} \)
Multiplying the numerator and denominator by 2, we have
\( \frac{257}{5000} = \frac{257}{2^3 \times 5^4} \times \frac{2}{2} = \frac{514}{2^4 \times 5^4} = \frac{514}{(10)^4} \)
\( \Rightarrow \frac{257}{5000} = \frac{514}{10000} = 0.0514 \)
This is the required decimal expansion of the rational number \( \frac{257}{5000} \) and it is also a terminating decimal number.

Long Answer Type Questions

Question. Prove that one and only one out of \( n \), \( (n + 2) \) and \( (n + 4) \) is divisible by 3, where \( n \) is any positive integer.
Answer: Let \( q \) be the quotient and \( r \) be the remainder we get after dividing ‘\( n \)’ by 3
\( \Rightarrow n = 3q + r \), where, \( 0 \le r < 3 \)
or \( n = 3q + r \), where, \( r = 0, 1, 2 \)
So, \( n = 3q \) or \( n = 3q + 1 \) or \( n = 3q + 2 \)
When \( n = 3q \), then \( n \) is divisible by 3,
\( n + 2 = (3q + 2) \),
\( n + 4 = (3q + 4) \) are not divisible by 3.
When \( n = 3q + 1 \), then \( n \) is not divisible by 3,
\( n + 2 = (3q + 1 + 2) = 3q + 3 \), which is divisible by 3,
\( n + 4 = (3q + 1 + 4) = 3q + 5 \), which is not divisible by 3.
When \( n = 3q + 2 \), then \( n \) is not divisible by 3,
\( n + 2 = (3q + 2 + 2) = 3q + 4 \), which is not divisible by 3,
\( n + 4 = (3q + 2 + 4) = 3q + 6 \), which is divisible by 3.
Hence, one and only one out of \( n \), \( (n + 2) \) and \( (n + 4) \) is divisible by 3.

Question. For any positive integer \( n \), prove that \( n^3 - n \) is divisible by 6.
Answer: Let \( x = n^3 - n \)
\( \Rightarrow x = n(n^2 - 1) \)
\( \Rightarrow x = n (n - 1) \times (n + 1) \)
...[Using \( (a^2 - b^2) = (a - b)(a + b) \)]
\( x = (n - 1) \times n \times (n + 1) \) ...(i)
We know that, if a number is completely divisible by 2 and 3, then it is also divisible by 6.
Divisibility test for 3:
If the sum of digits of any number is divisible by 3, then it is divisible by 3.
Sum of the digits = \( (n - 1) + (n) + (n + 1) = 3n \)
\( \Rightarrow \) Number is divisible by 3.
Divisibility test for 2:
If \( n \) is odd then \( (n - 1) \) and \( (n + 1) \) will be even so, \( (n - 1) \times n \times (n + 1) \) will be divisible by 2.
If \( n \) is even then, \( (n - 1) \times n \times (n + 1) \) will be divisible by 2.
Therefore, for any positive integral value of \( n \), \( n^3 - n \) is divisible by 6.