NCERT Class 10 Exercise 1.1 Solutions: HCF and LCM

Exer­cise 1.1 of the NCERT Class 10 text­book (Chap­ter 1, Real Num­bers) is where the Fun­da­men­tal The­o­rem of Arith­metic becomes a work­ing tool. You use it to write num­bers as prod­ucts of primes, to find the HCF and LCM of two or three num­bers, and to rea­son about what a num­ber can or can­not be divis­i­ble by. Every one of the seven ques­tions is solved below in the text­book's order, with each step shown and each answer checked.

Meth­ods you need

Prime fac­tori­sa­tion

The Fun­da­men­tal The­o­rem of Arith­metic says that every com­pos­ite num­ber can be writ­ten as a prod­uct of primes, and this fac­tori­sa­tion is unique apart from the order of the fac­tors. To find it:

1. Divide by the small­est prime that divides the num­ber exactly (try 2, then 3, 5, 7, 11, 13, and so on).
2. Repeat with the quo­tient.
3. Stop when the quo­tient is itself a prime. Col­lect repeated primes as pow­ers.

You only need to test primes up to the square root of the num­ber you are work­ing on: if none of them divides it, the num­ber is prime.

HCF and LCM from prime fac­tors

• HCF = prod­uct of the com­mon prime fac­tors, each taken with its small­est power.
• LCM = prod­uct of all prime fac­tors that appear, each taken with its great­est power.
• For any two pos­i­tive inte­gers $a$ and $b$: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$. This does not hold for three num­bers.

Prime fac­tori­sa­tion

Ques­tion 1 (i)

Express each num­ber as a prod­uct of its prime fac­tors: (i) 140.

140 is even, so divide by 2 twice, then by 5.

$$\begin{aligned} 140 &= 2 \times 70 \\ &= 2 \times 2 \times 35 \\ &= 2 \times 2 \times 5 \times 7 \end{aligned}$$

Check: $4 \times 5 \times 7 = 140$.

Answer: $140 = 2^2 \times 5 \times 7$

Ques­tion 1 (ii)

(ii) 156

$$\begin{aligned} 156 &= 2 \times 78 \\ &= 2 \times 2 \times 39 \\ &= 2 \times 2 \times 3 \times 13 \end{aligned}$$

Check: $4 \times 3 \times 13 = 4 \times 39 = 156$.

Answer: $156 = 2^2 \times 3 \times 13$

Ques­tion 1 (iii)

(iii) 3825

3825 is odd, so 2 does not divide it. Its digit sum is $3 + 8 + 2 + 5 = 18$, which is divis­i­ble by 3 (and by 9), so start with 3.

$$\begin{aligned} 3825 &= 3 \times 1275 \\ &= 3 \times 3 \times 425 \\ &= 3 \times 3 \times 5 \times 85 \\ &= 3 \times 3 \times 5 \times 5 \times 17 \end{aligned}$$

Check: $9 \times 25 \times 17 = 225 \times 17 = 3825$.

Answer: $3825 = 3^2 \times 5^2 \times 17$

Ques­tion 1 (iv)

(iv) 5005

5005 ends in 5, so start with 5. Then $1001 = 7 \times 143$ and $143 = 11 \times 13$.

$$5005 = 5 \times 1001 = 5 \times 7 \times 143 = 5 \times 7 \times 11 \times 13$$

Check: $5 \times 7 = 35$, $35 \times 11 = 385$, $385 \times 13 = 5005$.

Answer: $5005 = 5 \times 7 \times 11 \times 13$

Ques­tion 1 (v)

(v) 7429

7429 is not divis­i­ble by 2 (odd), 3 (digit sum 22) or 5 (last digit 9). Test­ing fur­ther primes, 7, 11 and 13 all leave remain­ders, but 17 divides exactly:

$$7429 = 17 \times 437 \qquad (17 \times 400 = 6800,\ 17 \times 37 = 629,\ 6800 + 629 = 7429)$$

Now $437 = 19 \times 23$ (since $19 \times 23 = 437$), and both 19 and 23 are prime.

Check, mul­ti­ply­ing in a dif­fer­ent order: $19 \times 23 = 437$ and $437 \times 17 = 4370 + 3059 = 7429$.

Answer: $7429 = 17 \times 19 \times 23$

HCF and LCM

Ques­tion 2 (i)

Find the LCM and HCF of the fol­low­ing pairs and ver­ify that LCM × HCF = prod­uct of the two num­bers: (i) 26 and 91.

$$26 = 2 \times 13, \qquad 91 = 7 \times 13$$

The only com­mon prime is 13, so $\text{HCF} = 13$. Tak­ing every prime at its high­est power, $\text{LCM} = 2 \times 7 \times 13 = 182$.

Ver­i­fi­ca­tion: $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$ and $26 \times 91 = 2340 + 26 = 2366$. The two are equal.

Answer: HCF = 13, LCM = 182, and $13 \times 182 = 26 \times 91 = 2366$.

Ques­tion 2 (ii)

(ii) 510 and 92

$$510 = 2 \times 3 \times 5 \times 17, \qquad 92 = 2^2 \times 23$$

The only com­mon prime is 2; its smaller power is $2^1$, so $\text{HCF} = 2$.

$$\text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 60 \times 17 \times 23 = 1020 \times 23 = 23460$$

Ver­i­fi­ca­tion: $2 \times 23460 = 46920$ and $510 \times 92 = 45900 + 1020 = 46920$. Equal.

Answer: HCF = 2, LCM = 23460, and $2 \times 23460 = 510 \times 92 = 46920$.

Ques­tion 2 (iii)

(iii) 336 and 54

$$336 = 2^4 \times 3 \times 7, \qquad 54 = 2 \times 3^3$$

Com­mon primes are 2 and 3. Smaller pow­ers: $2^1$ and $3^1$, so $\text{HCF} = 2 \times 3 = 6$.

Great­est pow­ers: $\text{LCM} = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 432 \times 7 = 3024$.

Ver­i­fi­ca­tion: $6 \times 3024 = 18144$ and $336 \times 54 = 16800 + 1344 = 18144$. Equal.

Answer: HCF = 6, LCM = 3024, and $6 \times 3024 = 336 \times 54 = 18144$.

Ques­tion 3 (i)

Find the LCM and HCF of the fol­low­ing inte­gers by apply­ing the prime fac­tori­sa­tion method: (i) 12, 15 and 21.

With three num­bers there is no prod­uct short­cut, so use the def­i­n­i­tions directly.

$$12 = 2^2 \times 3, \qquad 15 = 3 \times 5, \qquad 21 = 3 \times 7$$

The only prime com­mon to all three is 3 (power 1 each time), so $\text{HCF} = 3$. All primes at their great­est pow­ers: $\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420$.

Check: $420 \div 12 = 35$, $420 \div 15 = 28$, $420 \div 21 = 20$, all whole num­bers. Notice that $12 \times 15 \times 21 = 3780$, while $3 \times 420 = 1260$: the prod­uct rule really does fail for three num­bers.

Answer: HCF = 3, LCM = 420

Ques­tion 3 (ii)

(ii) 17, 23 and 29

All three are dif­fer­ent primes, so they have no com­mon prime fac­tor: $\text{HCF} = 1$. The LCM is their prod­uct.

$$17 \times 23 = 391, \qquad 391 \times 29 = 11730 - 391 = 11339$$

Check another way: $23 \times 29 = 667$ and $667 \times 17 = 6670 + 4669 = 11339$.

Answer: HCF = 1, LCM = 11339

Ques­tion 3 (iii)

(iii) 8, 9 and 25

$$8 = 2^3, \qquad 9 = 3^2, \qquad 25 = 5^2$$

No prime is shared, so $\text{HCF} = 1$ (the num­bers are pair­wise coprime even though none is prime). The LCM is $2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 72 \times 25 = 1800$.

Check: $9 \times 25 = 225$ and $225 \times 8 = 1800$.

Answer: HCF = 1, LCM = 1800

Ques­tion 4

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

For two num­bers, $\text{HCF} \times \text{LCM} =$ prod­uct of the num­bers, so

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

A quicker way to com­pute: $306 \div 9 = 34$, so the LCM is $34 \times 657 = 22338$. (Equiv­a­lently, $306 \times 657 = 201042$ and $201042 \div 9 = 22338$.)

Inde­pen­dent check by fac­tori­sa­tion: $306 = 2 \times 3^2 \times 17$ and $657 = 3^2 \times 73$. The com­mon part is $3^2 = 9$, which con­firms the given HCF, and $\text{LCM} = 2 \times 3^2 \times 17 \times 73 = 306 \times 73 = 22338$.

Answer: LCM(306, 657) = 22338

Rea­son­ing with prime fac­tors

Ques­tion 5

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

A num­ber ends in 0 exactly when it is divis­i­ble by 10. Since $10 = 2 \times 5$, such a num­ber must have 5 in its prime fac­tori­sa­tion.

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

The only primes here are 2 and 3. By the unique­ness part of the Fun­da­men­tal The­o­rem of Arith­metic, $6^n$ has no other prime fac­tori­sa­tion, so 5 can never be a fac­tor of $6^n$. Hence $6^n$ is never divis­i­ble by 10.

San­ity check: $6^1 = 6$, $6^2 = 36$, $6^3 = 216$, $6^4 = 1296$, all end­ing in 6.

Answer: No. $6^n$ can­not end with the digit 0 for any nat­ural num­ber $n$, because 5 is not a prime fac­tor of $6^n$.

Why this works: unique­ness of prime fac­tori­sa­tion means a prime that is miss­ing from one fac­tori­sa­tion is miss­ing for good.

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.

A com­pos­ite num­ber is one that has a fac­tor other than 1 and itself. The trick is to take out the fac­tor com­mon to both terms before mul­ti­ply­ing any­thing.

First num­ber.

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

So 1014 has the fac­tor 13, which is nei­ther 1 nor 1014. In fact $1014 = 2 \times 3 \times 13^2$. Check: $7 \times 11 \times 13 = 1001$ and $1001 + 13 = 1014$.

Sec­ond num­ber.

$$\begin{aligned} 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 &= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \\ &= 5 \times (1008 + 1) \\ &= 5 \times 1009 = 5045 \end{aligned}$$

So 5045 has the fac­tor 5, which is nei­ther 1 nor 5045. Check: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$ and $5040 + 5 = 5045$. (It does not mat­ter whether 1009 is prime; one fac­tor other than 1 and the num­ber is enough.)

Answer: $7 \times 11 \times 13 + 13 = 13 \times 78$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009$; each has a fac­tor other than 1 and itself, so both are com­pos­ite.

A word prob­lem

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?

Sonia is at the start­ing point after 18, 36, 54, … min­utes (mul­ti­ples of 18). Ravi is there after 12, 24, 36, … min­utes (mul­ti­ples of 12). They are there together at the first time com­mon to both lists, which is $\text{LCM}(18, 12)$.

$$18 = 2 \times 3^2, \qquad 12 = 2^2 \times 3$$

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

Check: in 36 min­utes Sonia com­pletes $36 \div 18 = 2$ rounds and Ravi com­pletes $36 \div 12 = 3$ rounds, both whole num­bers, and no smaller num­ber appears in both lists.

Answer: They meet again at the start­ing point after 36 min­utes.

Why this works: "first time both events hap­pen together" is always a least com­mon mul­ti­ple ques­tion; "largest equal group or piece" is an HCF ques­tion.

Key terms

Prime num­ber

A nat­ural num­ber greater than 1 whose only fac­tors are 1 and itself, such as 2, 3, 17 and 23.

Com­pos­ite num­ber

A nat­ural num­ber greater than 1 that has at least one fac­tor other than 1 and itself, such as 1014 or 5045.

Prime fac­tori­sa­tion

Writ­ing a num­ber as a prod­uct of primes, for exam­ple $3825 = 3^2 \times 5^2 \times 17$.

Fun­da­men­tal The­o­rem of Arith­metic

Every com­pos­ite num­ber can be expressed as a prod­uct of primes, and this fac­tori­sa­tion is unique apart from the order of the fac­tors.

HCF (high­est com­mon fac­tor)

The largest num­ber that divides each of the given num­bers; the prod­uct of com­mon primes at their small­est pow­ers.

LCM (low­est com­mon mul­ti­ple)

The small­est pos­i­tive num­ber that is a mul­ti­ple of each of the given num­bers; the prod­uct of all primes at their great­est pow­ers.

Coprime num­bers

Num­bers whose HCF is 1, such as 8, 9 and 25.

Com­mon ques­tions

Can I use HCF × LCM = prod­uct for three num­bers?

No. The iden­tity holds only for two num­bers. For 12, 15 and 21 the prod­uct is 3780, but HCF × LCM is only $3 \times 420 = 1260$.

How do I know when to stop test­ing primes?

Stop once the prime you are test­ing is larger than the square root of the num­ber left. For 437, $\sqrt{437}$ is a lit­tle under 21, so test­ing primes up to 19 is enough, and 19 does divide it.

Is 1 a prime num­ber?

No. A prime has exactly two dif­fer­ent fac­tors, and 1 has only one. That is also why 1 never appears in a prime fac­tori­sa­tion.

In Ques­tion 6, do I have to show that 1009 is prime?

No. To show a num­ber is com­pos­ite you only need one fac­tor other than 1 and itself. The fac­tor 5 already does that.

How can I tell whether a word prob­lem needs HCF or LCM?

If the answer must be a mul­ti­ple of the given num­bers (a later time, a longer length), it is LCM. If it must divide the given num­bers (the biggest equal share or the longest mea­sur­ing tape), it is HCF.

Ref­er­ences

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Bur­ton, D. M. Ele­men­tary Num­ber The­ory. McGraw-Hill Edu­ca­tion.