NCERT Exemplar Solutions Class 10 Maths Real Numbers — Exercises 1.3 and 1.4

NCERT Exemplar Problems, Class 10 Mathematics — Real Numbers, Exercises 1.3 and 1.4. All 19 questions solved.

These are the two written-proof exercises of the unit, and the exemplar prints no answer for fifteen of the nineteen — they are “show that” and “prove that” questions, so the working is the answer. Exercise 1.4’s five questions are collected here with Exercise 1.3’s fourteen because five is too few for a page of its own, and because the two exercises run on the same single idea.

That idea is the remainder-class argument. To prove something about every positive integer, write the integer as $bq + r$ for a well-chosen $b$, list the finitely many possible values of $r$, and check each. Choosing $b$ is the only creative step, and the question usually hands it to you: a claim about the form $4q+1$ wants $b = 2$ or $b = 4$; a claim about $5q+2$ wants $b = 5$.

The rest of the exercise is HCF and LCM by Euclid’s algorithm (Questions 8, 9 and 12), the denominator test (Question 13), and two irrationality proofs by contradiction (Questions 10 and 14).

A note on the current syllabus. Questions 8 and 9 name Euclid’s division algorithm explicitly, and Question 13 is the terminating-decimal test — both dropped from the rationalised textbook, which now runs Introduction → The Fundamental Theorem of Arithmetic → Revisiting Irrational Numbers and sets only Exercises 1.1 and 1.2. Question 8’s HCF can equally be found by prime factorisation, which is on the syllabus, and the working below gives both. The division-form arguments in every other question here are unaffected: they are elementary, they are not named results, and they are exactly what a board proof question still expects.

Key insight. A statement about all positive integers is proved by splitting them into $b$ remainder classes and checking $b$ cases. The whole skill is picking $b$ — and the modulus that makes a proof short is almost always the number appearing in the claim.

Exercise 1.3

Exercise 1.3, Question 1

Show that the square of any positive integer is either of the form $4q$ or $4q + 1$ for some integer $q$.

Solution. Take $b = 2$: every positive integer is even or odd.

If $n = 2m$, then $n^2 = 4m^2 = 4q$ with $q = m^2$.

If $n = 2m + 1$, then

$$n^2 = 4m^2 + 4m + 1 = 4\left(m^2 + m\right) + 1 = 4q + 1 \quad \text{with } q = m^2 + m$$

Those two cases are exhaustive, so every square is of one of the two forms.

Exercise 1.3, Question 2

Show that the cube of any positive integer is of the form $4m$, $4m + 1$ or $4m + 3$, for some integer $m$.

Solution. Here the claim is about division by $4$, so take $b = 4$ and check all four classes.

• $n = 4k$: $\;n^3 = 64k^3 = 4\left(16k^3\right) = 4m$.
• $n = 4k+1$: $\;n^3 = 64k^3 + 48k^2 + 12k + 1 = 4\left(16k^3 + 12k^2 + 3k\right) + 1 = 4m + 1$.
• $n = 4k+2$: $\;n^3 = 8(2k+1)^3 = 4\left[2(2k+1)^3\right] = 4m$.
• $n = 4k+3$: $\;n^3 = 64k^3 + 144k^2 + 108k + 27 = 4\left(16k^3 + 36k^2 + 27k + 6\right) + 3 = 4m + 3$.

The last line uses $27 = 24 + 3$. So the only remainders a cube can leave on division by $4$ are $0$, $1$ and $3$ — never $2$.

Exercise 1.3, Question 3

Show that the square of any positive integer cannot be of the form $5q + 2$ or $5q + 3$ for any integer $q$.

Solution. The claim concerns division by $5$, so write $n = 5k + r$ with $r = 0, 1, 2, 3, 4$. Squaring,

$$n^2 = 25k^2 + 10kr + r^2 = 5\left(5k^2 + 2kr\right) + r^2$$

so the remainder of $n^2$ on division by $5$ is the remainder of $r^2$. Running through the five cases:

$$r^2 = 0,\; 1,\; 4,\; 9,\; 16 \quad\longrightarrow\quad \text{remainders } 0,\; 1,\; 4,\; 4,\; 1$$

The set of possible remainders is $\{0, 1, 4\}$. Neither $2$ nor $3$ appears, so no square is of the form $5q+2$ or $5q+3$.

Exercise 1.3, Question 4

Show that the square of any positive integer cannot be of the form $6m + 2$ or $6m + 5$ for any integer $m$.

Solution. Same method with $b = 6$. Writing $n = 6k + r$, $r = 0, \dots, 5$,

$$n^2 = 6\left(6k^2 + 2kr\right) + r^2$$

so the remainder is that of $r^2$:

$$r^2 = 0,\; 1,\; 4,\; 9,\; 16,\; 25 \quad\longrightarrow\quad \text{remainders } 0,\; 1,\; 4,\; 3,\; 4,\; 1$$

The possible remainders are $\{0, 1, 3, 4\}$. Both $2$ and $5$ are missing, which is what was to be shown.

Exercise 1.3, Question 5

Show that the square of any odd integer is of the form $4q + 1$, for some integer $q$.

Solution. An odd integer is $n = 2m + 1$ for some integer $m$, so

$$n^2 = 4m^2 + 4m + 1 = 4\left(m^2 + m\right) + 1 = 4q + 1, \qquad q = m^2 + m$$

This is the odd half of Question 1, and it says something a little stronger than it appears: since $m^2 + m = m(m+1)$ is a product of consecutive integers and therefore even, $q$ is even, and the square of an odd integer is in fact of the form $8k + 1$. Question 6 is that observation.

Exercise 1.3, Question 6

If $n$ is an odd integer, then show that $n^2 – 1$ is divisible by $8$.

Solution. Write $n = 2m+1$. Then

$$n^2 – 1 = (2m+1)^2 – 1 = 4m^2 + 4m = 4m(m+1)$$

Now $m(m+1)$ is a product of two consecutive integers, so it is even: $m(m+1) = 2k$. Therefore

$$n^2 – 1 = 4 \times 2k = 8k$$

which is divisible by $8$.

The same thing seen through the factorisation $n^2 – 1 = (n-1)(n+1)$: for odd $n$ these are consecutive even numbers, so one of them is a multiple of $4$ and the other of $2$.

Exercise 1.3, Question 7

Prove that if $x$ and $y$ are both odd positive integers, then $x^2 + y^2$ is even but not divisible by $4$.

Solution. Let $x = 2m+1$ and $y = 2n+1$. Then

$$x^2 + y^2 = \left(4m^2 + 4m + 1\right) + \left(4n^2 + 4n + 1\right) = 4\left(m^2 + m + n^2 + n\right) + 2$$

Write $t = m^2 + m + n^2 + n$, so $x^2 + y^2 = 4t + 2 = 2(2t + 1)$.

• It is even, because of the factor $2$.
• It is not divisible by $4$, because $2t+1$ is odd: the expression carries exactly one factor of $2$, so dividing by $4$ leaves remainder $2$.

The second half is the part that gets dropped. “Even” only needs one factor of $2$; the question asks you to show there is no more than one.

Exercise 1.3, Question 8

Use Euclid’s division algorithm to find the HCF of $441$, $567$, $693$.

Solution. Take the numbers two at a time. Start with $567$ and $693$:

$$693 = 567 \times 1 + 126$$ $$567 = 126 \times 4 + 63$$ $$126 = 63 \times 2 + 0$$

The last non-zero remainder is $63$, so $\operatorname{HCF}(567, 693) = 63$. Now bring in $441$:

$$441 = 63 \times 7 + 0$$

so $\operatorname{HCF}(63, 441) = 63$, and hence $\operatorname{HCF}(441, 567, 693) = 63$.

By prime factorisation — the route the current textbook takes — the same answer, taking the smallest power of each shared prime:

$$441 = 3^2 \times 7^2, \qquad 567 = 3^4 \times 7, \qquad 693 = 3^2 \times 7 \times 11$$

$$\operatorname{HCF} = 3^2 \times 7 = 63$$

Exercise 1.3, Question 9

Using Euclid’s division algorithm, find the largest number that divides $1251$, $9377$ and $15628$ leaving remainders $1$, $2$ and $3$, respectively.

Solution. Strip the remainders first, so the divisions become exact:

$$1251 – 1 = 1250, \qquad 9377 – 2 = 9375, \qquad 15628 – 3 = 15625$$

The number wanted is the HCF of these three. Applying the algorithm to the first two:

$$9375 = 1250 \times 7 + 625$$ $$1250 = 625 \times 2 + 0$$

so $\operatorname{HCF}(1250, 9375) = 625$. Then

$$15625 = 625 \times 25 + 0$$

so the HCF of all three is $625$.

Check the remainders are admissible: $625$ is comfortably larger than $1$, $2$ and $3$, so it genuinely leaves those remainders rather than dividing further.

Exercise 1.3, Question 10

Prove that $\sqrt{3} + \sqrt{5}$ is irrational.

Solution. By contradiction. Suppose $\sqrt{3} + \sqrt{5}$ is rational, say

$$\sqrt{3} + \sqrt{5} = a, \qquad a \in \mathbb{Q}$$

Note $a > 0$, so in particular $a \ne 0$. Isolate one surd and square, which is the move that makes the other one disappear:

$$\sqrt{5} = a – \sqrt{3} \implies 5 = a^2 – 2a\sqrt{3} + 3$$

Rearranging,

$$2a\sqrt{3} = a^2 – 2 \implies \sqrt{3} = \frac{a^2 – 2}{2a}$$

The right-hand side is built from rationals by arithmetic, so it is rational; but $\sqrt{3}$ is irrational. This contradiction shows the original assumption is false.

Exercise 1.3, Question 11

Show that $12^n$ cannot end with the digit $0$ or $5$ for any natural number $n$.

Solution. A number ends in $0$ or $5$ exactly when it is divisible by $5$. So the question is whether $5$ can ever be a factor of $12^n$.

$$12 = 2^2 \times 3 \implies 12^n = 2^{2n} \times 3^n$$

The only primes in that factorisation are $2$ and $3$. By the Fundamental Theorem of Arithmetic a positive integer has exactly one prime factorisation, so $5$ is not a factor of $12^n$ for any $n$ — there is nowhere for it to have come from.

Since $12^n$ is not divisible by $5$, it cannot end in $0$ or $5$.

Exercise 1.3, Question 12

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?

Solution. “Complete steps” means the distance must be a whole number of steps for each walker — that is, a common multiple of $40$, $42$ and $45$. “Minimum” makes it the least common multiple.

$$40 = 2^3 \times 5, \qquad 42 = 2 \times 3 \times 7, \qquad 45 = 3^2 \times 5$$

Take the highest power of each prime that appears anywhere:

$$\operatorname{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520 \text{ cm}$$

So each should walk $2520$ cm, which is $25.2$ m — that is $63$ steps for the first walker, $60$ for the second and $56$ for the third.

Exercise 1.3, Question 13

Write the denominator of the rational number $\dfrac{257}{5000}$ in the form $2^m \times 5^n$, where $m$, $n$ are non-negative integers. Hence write its decimal expansion, without actual division.

Solution. Factorise the denominator:

$$5000 = 5 \times 1000 = 5 \times 10^3 = 5 \times 2^3 \times 5^3 = 2^3 \times 5^4$$

so $m = 3$ and $n = 4$.

Since the denominator is of the form $2^m5^n$, the expansion terminates, after $\max(3, 4) = 4$ places. To write it without dividing, make the denominator a power of ten by supplying the missing factor — here one more $2$:

$$\frac{257}{5000} = \frac{257 \times 2}{2^4 \times 5^4} = \frac{514}{10^4} = \frac{514}{10000} = 0.0514$$

Exercise 1.3, Question 14

Prove that $\sqrt{p} + \sqrt{q}$ is irrational, where $p$, $q$ are primes.

Solution. This is Question 10 with the particular primes replaced by general ones, and the proof is unchanged. Suppose, for contradiction, that

$$\sqrt{p} + \sqrt{q} = a, \qquad a \in \mathbb{Q}$$

Since $p, q \ge 2$, the left-hand side is positive, so $a \ne 0$. Isolate and square:

$$\sqrt{q} = a – \sqrt{p} \implies q = a^2 – 2a\sqrt{p} + p$$

$$\implies \sqrt{p} = \frac{a^2 + p – q}{2a}$$

The right-hand side is rational, being built from the rationals $a$, $p$, $q$ by arithmetic. But $\sqrt{p}$ is irrational for every prime $p$ — the standard argument: if $\sqrt p = \frac{c}{d}$ in lowest terms then $c^2 = p\,d^2$, so $p \mid c^2$, so $p \mid c$ (because $p$ is prime), so $p^2 \mid c^2 = p d^2$, so $p \mid d^2$ and $p \mid d$, contradicting $\gcd(c,d) = 1$.

The contradiction proves the claim. Note that the argument does not require $p \ne q$: if $p = q$ the last display reads $\sqrt p = \frac{a}{2}$, which is rational, and is just as impossible.

Exercise 1.4

Exercise 1.4, Question 1

Show that the cube of a positive integer of the form $6q + r$, where $q$ is an integer and $r = 0, 1, 2, 3, 4, 5$, is also of the form $6m + r$.

Solution. Expand the cube and collect everything that is visibly a multiple of $6$:

$$(6q+r)^3 = 216q^3 + 108q^2r + 18qr^2 + r^3 = 6\left(36q^3 + 18q^2 r + 3qr^2\right) + r^3$$

So the cube is a multiple of $6$ plus $r^3$, and the claim reduces to showing that $r^3$ itself is of the form $6k + r$ for each of the six values of $r$:

$r$	$0$	$1$	$2$	$3$	$4$	$5$
$r^3$	$0$	$1$	$8$	$27$	$64$	$125$
$r^3$ as $6k+r$	$6(0)+0$	$6(0)+1$	$6(1)+2$	$6(4)+3$	$6(10)+4$	$6(20)+5$

In every case $r^3 = 6k + r$. Substituting back,

$$(6q+r)^3 = 6\left(36q^3 + 18q^2r + 3qr^2 + k\right) + r = 6m + r$$

Exercise 1.4, Question 2

Prove that one and only one out of $n$, $n + 2$ and $n + 4$ is divisible by $3$, where $n$ is any positive integer.

Solution. Write $n = 3q + r$ with $r = 0$, $1$ or $2$, and check each case. The phrase “one and only one” means both halves have to be shown: that one of them is divisible, and that the other two are not.

• $n = 3q$. Then $n$ is divisible by $3$. And $n+2 = 3q+2$ leaves remainder $2$; $n + 4 = 3(q+1) + 1$ leaves remainder $1$. Neither is divisible.
• $n = 3q+1$. Then $n$ leaves remainder $1$. But $n + 2 = 3q + 3 = 3(q+1)$ is divisible. And $n+4 = 3(q+1)+2$ leaves remainder $2$.
• $n = 3q+2$. Then $n$ leaves remainder $2$ and $n+2 = 3(q+1)+1$ leaves remainder $1$, but $n + 4 = 3q + 6 = 3(q+2)$ is divisible.

In each of the three cases exactly one of the three numbers is a multiple of $3$, which is the claim.

The reason it works: $n$, $n+2$, $n+4$ leave remainders $r$, $r+2$, $r+4$, and modulo $3$ those are $r$, $r+2$, $r+1$ — the three different remainder classes in some order. Exactly one class is $0$.

Exercise 1.4, Question 3

Prove that one of any three consecutive positive integers must be divisible by $3$.

Solution. Let the three be $n$, $n+1$, $n+2$, and write $n = 3q + r$ with $r = 0, 1, 2$.

• $r = 0$: $n = 3q$ is divisible by $3$.
• $r = 1$: $n + 2 = 3q + 3 = 3(q+1)$ is divisible by $3$.
• $r = 2$: $n + 1 = 3q + 3 = 3(q+1)$ is divisible by $3$.

The cases are exhaustive, so in every case at least one of the three is a multiple of $3$.

More directly: three consecutive integers leave the remainders $r$, $r+1$, $r+2$, which modulo $3$ are the three distinct classes $0$, $1$, $2$ in some order — so one of them is exactly $0$, and only one.

Exercise 1.4, Question 4

For any positive integer $n$, prove that $n^3 – n$ is divisible by $6$.

Solution. Factorise, and the problem turns into the previous two questions:

$$n^3 – n = n\left(n^2 – 1\right) = (n-1)\,n\,(n+1)$$

which is the product of three consecutive integers.

• Divisible by $2$: among any two consecutive integers one is even, so certainly among three.
• Divisible by $3$: by Question 3 above, one of any three consecutive integers is a multiple of $3$.

Since $2$ and $3$ are coprime, a number divisible by both is divisible by their product $6$. Hence $6 \mid n^3 – n$.

The coprimality matters. Divisibility by $2$ and by $4$ would not give divisibility by $8$; it is only because $\gcd(2,3) = 1$ that the two factors can be multiplied together.

Exercise 1.4, Question 5

Show that one and only one out of $n$, $n + 4$, $n + 8$, $n + 12$ and $n + 16$ is divisible by $5$, where $n$ is any positive integer.

[Hint: Any positive integer can be written in the form $5q$, $5q+1$, $5q+2$, $5q+3$, $5q+4$.]

Solution. Write $n = 5q + r$ with $r = 0, 1, 2, 3, 4$, and in each case find which of the five numbers is a multiple of $5$.

• $n = 5q$: $n$ itself.
• $n = 5q+1$: $n + 4 = 5q + 5 = 5(q+1)$.
• $n = 5q+2$: $n + 8 = 5q + 10 = 5(q+2)$.
• $n = 5q+3$: $n + 12 = 5q + 15 = 5(q+3)$.
• $n = 5q+4$: $n + 16 = 5q + 20 = 5(q+4)$.

That is one from each case, and the cases are exhaustive, so at least one of the five is always divisible by $5$.

For “only one”, look at the remainders. The five numbers are $n + 4k$ for $k = 0,1,2,3,4$, and

$$4k \equiv 0,\; 4,\; 3,\; 2,\; 1 \pmod 5$$

which are the five different remainder classes. Adding $r$ to five distinct classes gives five distinct classes again, so exactly one of them is $0$ — no two of the five numbers can be divisible by $5$ at once.

Common mistakes

• Choosing the wrong modulus. Question 3 is about the form $5q+2$, so split the integers by $5$; Question 4 is about $6m+2$, so split by $6$. Students who reach for $b = 2$ out of habit get an argument that never closes. The number in the claim tells you the modulus.
• Proving “at least one” when the question says “one and only one”. Questions 2 and 5 of Exercise 1.4 both ask for uniqueness as well as existence. Showing that some member is divisible is half the answer; you must also show the other members are not.
• Stopping at “even” in Exercise 1.3 Question 7. $x^2 + y^2 = 2(2t+1)$ — the second half of the question is that the remaining factor is odd, which is exactly why $4$ does not divide it.
• Forgetting to strip the remainders in Question 9. The HCF wanted is that of $1250$, $9375$ and $15625$, not of $1251$, $9377$ and $15628$ — whose HCF is $1$.
• Squaring without isolating in Questions 10 and 14. Squaring $\sqrt3+\sqrt5 = a$ directly leaves $2\sqrt{15}$ behind and gets you nowhere useful. Move one surd across first, so that squaring eliminates it.
• Assuming $p \ne q$ in Question 14. The question says $p$, $q$ are primes, not distinct primes. The proof given covers $p = q$ as well, and a proof that quietly assumes distinctness is incomplete.
• Adding the exponents in Question 13. $5000 = 2^3 \times 5^4$ terminates after $\max(3,4) = 4$ places, not $3 + 4 = 7$. The number of places is the larger exponent, because that is the power of ten you have to build.
• Treating divisibility by $2$ and $3$ as automatically giving $6$ in Exercise 1.4 Question 4. It works here only because $2$ and $3$ are coprime. The same reasoning applied to $2$ and $4$ would be wrong.

Practise next

• Exemplar Exercise 1.2 — the same remainder-class arguments in short-answer form, where the marks go to the justification rather than the algebra.
• Exemplar Exercise 1.1 for the multiple-choice versions of Questions 6, 8 and 13 above.
• NCERT textbook Exercise 1.2 — the irrationality proofs of $\sqrt5$, $3+2\sqrt5$ and $6+\sqrt2$, which are Questions 10 and 14 above stripped to a single surd.
• NCERT textbook Exercise 1.1 for the LCM word problem of the same shape as Question 12 — Sonia and Ravi on the circular path.