NCERT Class 10 Mathematics — Real Numbers, Exercise 1.1. All 7 questions solved.

This exercise is the Fundamental Theorem of Arithmetic put to work. The theorem says that every composite number can be written as a product of primes, and that the factorisation is unique apart from the order of the factors. Everything in the exercise runs on that and on two rules that follow from it:

• HCF = the product of the smallest power of each prime common to the numbers.
• LCM = the product of the greatest power of each prime that appears in any of them.

For two numbers only, these combine into

$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b.$$

Key insight. Once a number is written as a product of primes, every question here becomes a question about exponents. HCF takes the smaller exponent of each prime, LCM the larger; a number ends in $0$ only if both $2$ and $5$ appear among its primes; and a number is composite as soon as you can pull out a factor. The uniqueness of the factorisation is what makes “it has no factor of 5” a proof rather than a guess.

Question 1

Express each number as a product of its prime factors:

(i) $140$   (ii) $156$   (iii) $3825$   (iv) $5005$   (v) $7429$

Solution. Divide repeatedly by the smallest prime that goes in exactly, and stop when the quotient is itself prime. Using divisibility tests first saves trial and error.

(i) $140$ is even, and so is $70$:

$$140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$$

(ii) Halve twice, then $39 = 3 \times 13$:

$$156 = 2 \times 78 = 2 \times 2 \times 39 = 2^2 \times 3 \times 13$$

(iii) $3825$ ends in $25$, so it is divisible by $25$; and $153$ has digit sum $9$, so it is divisible by $9$:

$$3825 = 25 \times 153 = 5^2 \times 9 \times 17 = 3^2 \times 5^2 \times 17$$

(iv) It ends in $5$, and $1001 = 7 \times 143 = 7 \times 11 \times 13$:

$$5005 = 5 \times 1001 = 5 \times 7 \times 11 \times 13$$

(v) This is the one that needs patience. $7429$ is odd, its digit sum $22$ is not a multiple of $3$, and it does not end in $0$ or $5$. Trying $7$, $11$ and $13$ leaves remainders ($7 \times 1061 = 7427$, $13 \times 571 = 7423$, and the alternating digit sum $9 – 2 + 4 – 7 = 4$ rules out $11$). The next prime works:

$$7429 = 17 \times 437 = 17 \times 19 \times 23$$

Question 2

Find the LCM and HCF of the following pairs of integers and verify that LCM $\times$ HCF $=$ product of the two numbers.

(i) $26$ and $91$   (ii) $510$ and $92$   (iii) $336$ and $54$

Solution. Factorise both numbers, then read off the smallest power of each common prime for the HCF and the greatest power of every prime present for the LCM.

(i) $26 = 2 \times 13$ and $91 = 7 \times 13$. The only common prime is $13$:

$$\text{HCF} = 13, \qquad \text{LCM} = 2 \times 7 \times 13 = 182$$

Check: $182 \times 13 = 2366$ and $26 \times 91 = 2366$. ✓

(ii) $510 = 2 \times 3 \times 5 \times 17$ and $92 = 2^2 \times 23$. The common prime is $2$, and its smaller power is $2^1$:

$$\text{HCF} = 2, \qquad \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$$

Check: $23460 \times 2 = 46920$ and $510 \times 92 = 46920$. ✓

(iii) $336 = 2^4 \times 3 \times 7$ and $54 = 2 \times 3^3$. Both share $2$ and $3$; the smaller powers are $2^1$ and $3^1$, the greater are $2^4$ and $3^3$:

$$\text{HCF} = 2 \times 3 = 6, \qquad \text{LCM} = 2^4 \times 3^3 \times 7 = 3024$$

Check: $3024 \times 6 = 18144$ and $336 \times 54 = 18144$. ✓

The identity always holds for two numbers because, prime by prime, the HCF takes one exponent and the LCM takes the other, so between them they use every exponent exactly once.

Question 3

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) $12, 15$ and $21$   (ii) $17, 23$ and $29$   (iii) $8, 9$ and $25$

Solution. The same two rules, now across three numbers. A prime counts for the HCF only if it divides all three.

(i) $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$. Only $3$ is common to all three:

$$\text{HCF} = 3, \qquad \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420$$

(ii) All three numbers are prime, so they share no factor except $1$, and the LCM is simply their product:

$$\text{HCF} = 1, \qquad \text{LCM} = 17 \times 23 \times 29 = 11339$$

(iii) $8 = 2^3$, $9 = 3^2$, $25 = 5^2$. None of them is prime, but no prime appears in more than one of them, so again the HCF is $1$:

$$\text{HCF} = 1, \qquad \text{LCM} = 2^3 \times 3^2 \times 5^2 = 1800$$

Notice that the shortcut of question 2 does not carry over to three numbers: in (i), $\text{HCF} \times \text{LCM} = 1260$ but $12 \times 15 \times 21 = 3780$.

Question 4

Given that HCF $(306, 657) = 9$, find LCM $(306, 657)$.

Solution. There are exactly two numbers, so the identity from question 2 applies and there is no need to factorise:

$$\text{LCM}(306, 657) = \frac{306 \times 657}{\text{HCF}(306, 657)} = \frac{306 \times 657}{9} = 34 \times 657 = 22338$$

Dividing $306$ by $9$ first (it gives $34$) keeps the multiplication small. As a check, $306 = 2 \times 3^2 \times 17$ and $657 = 3^2 \times 73$, so the LCM is $2 \times 3^2 \times 17 \times 73 = 22338$.

Question 5

Check whether $6^n$ can end with the digit $0$ for any natural number $n$.

Solution. A number ends in $0$ exactly when it is divisible by $10 = 2 \times 5$, so its prime factorisation would have to contain the prime $5$.

But

$$6^n = (2 \times 3)^n = 2^n \times 3^n,$$

and by the uniqueness part of the Fundamental Theorem of Arithmetic, $2$ and $3$ are the only primes in the factorisation of $6^n$. There is no other way to factorise it that could smuggle in a $5$. So $5$ does not divide $6^n$, and $6^n$ cannot end in $0$.

(In fact every power of $6$ ends in $6$ — $6, 36, 216, 1296, \ldots$ — but checking a few cases is an observation, not a proof. The factorisation argument covers every $n$ at once.)

Question 6

Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers.

Solution. A composite number is one with a factor other than $1$ and itself. Both expressions have a number that appears in both terms, so it can be taken out as a common factor — there is no need to work out the value and hunt for divisors.

$$7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78 = 1014$$

So $1014$ has the factor $13$ (and $78$), besides $1$ and itself.

$$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times 1009 = 5045$$

So $5045$ has the factor $5$. It does not matter whether $1009$ is prime (it is): one factor other than $1$ and the number itself is enough.

Question 7

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution. Sonia is back at the start after $18, 36, 54, \ldots$ minutes — the multiples of $18$. Ravi is back after $12, 24, 36, \ldots$ minutes — the multiples of $12$. They are both at the start together at the first time that is a multiple of both, which is the least common multiple:

$$18 = 2 \times 3^2, \qquad 12 = 2^2 \times 3, \qquad \text{LCM}(18, 12) = 2^2 \times 3^2 = 36$$

By then Sonia has done $2$ rounds and Ravi $3$.

Common mistakes

• Question 1(v), declaring $7429$ prime. After $2, 3, 5, 7, 11, 13$ all fail it is tempting to stop, but you only need to go on until the prime squared exceeds the number — and $17$ divides it well before that.
• Question 2, swapping the rules. The HCF uses the smallest power of each common prime and the LCM the greatest. Using the greatest powers for the HCF of $336$ and $54$ gives $2^4 \times 3^3 = 432$ — a “common factor” larger than both numbers, which is impossible.
• Question 3, using HCF $\times$ LCM $=$ product for three numbers. It is true only for two. In 3(i) it would give an LCM of $3780 \div 3 = 1260$ instead of $420$.
• Question 3(ii) and (iii), writing the HCF as $0$. When no prime is shared, the highest common factor is $1$ — every integer is divisible by $1$, and nothing is divisible by $0$.
• Question 5, arguing from examples. Listing $6, 36, 216$ shows a pattern but does not prove it continues. The proof must say that $5$ is not among the primes of $2^n \times 3^n$, and that uniqueness rules out any other factorisation.
• Question 7, using the HCF. The HCF of $18$ and $12$ is $6$, but after $6$ minutes neither driver has completed a round. “When do both events happen together again” is always a common multiple.

Practise next

• Exercise 1.2 — proving that $\sqrt5$, $3 + 2\sqrt5$ and similar numbers are irrational, where the fact that a prime dividing $a^2$ must divide $a$ comes straight from the prime factorisation used here.