NCERT Solutions Class 10 Maths Exercise 1.1

Solution

We need to express each number as a product of prime numbers only. We will divide the given number, step by step, by the smallest prime factor each time and continue until all the remaining numbers are prime.

(i) 140

First, look at 140. It is even, so it is divisible by \(2\), which is a prime number.

Divide 140 by \(2\): \(140 \div 2 = 70\).

Now we have 70. This is also even, so again it is divisible by \(2\).

Divide 70 by \(2\): \(70 \div 2 = 35\).

Now we have 35. It is not even, so it is not divisible by \(2\). Next check if it is divisible by the prime \(3\). A number is divisible by 3 if the sum of its digits is divisble by 3. The sum of digits of 35 is \(3 + 5 = 8\), which is not divisible by \(3\), so 35 is not divisible by \(3\).

Try the next prime, \(5\). The last digit of 35 is 5, so it is divisible by \(5\).

Divide 35 by \(5\): \(35 \div 5 = 7\).

Now we get 7. The number 7 is a prime number.

So the prime factors of 140 are \(2, 2, 5, 7\). Writing them as a product, we get \(2 \times 2 \times 5 \times 7\).

In index form, \(2 \times 2 = 2^2\), so we write: \(140 = 2^2 \times 5 \times 7\).

(ii) 156

Look at 156. It is even, so it is divisible by \(2\).

Divide 156 by \(2\): \(156 \div 2 = 78\).

Now we have 78. This is also even, so again it is divisible by \(2\).

Divide 78 by \(2\): \(78 \div 2 = 39\).

Now we have 39. It is not even, so not divisible by \(2\). Check divisibility by \(3\). The sum of digits of 39 is \(3 + 9 = 12\), and 12 is divisible by \(3\), so 39 is divisible by \(3\).

Divide 39 by \(3\): \(39 \div 3 = 13\).

Now we get 13. The number 13 is a prime number.

So the prime factors of 156 are \(2, 2, 3, 13\). As a product, \(156\) is \(2 \times 2 \times 3 \times 13\).

We can write \(2 \times 2 = 2^2\), so we get \(2^2 \times 3 \times 13\).

(iii) 3825

Start with the smallest prime, \(2\). Since 3825 is odd, it is not divisible by \(2\). So we will not check \(2\) again for any of the later numbers.

Next prime is \(3\). The sum of digits is \(3 + 8 + 2 + 5 = 18\), and 18 is divisible by \(3\). So 3825 is divisible by \(3\).

Divide 3825 by \(3\): \(3825 \div 3 = 1275\).

Now we continue checking with the same prime, \(3\). The sum of digits of 1275 is \(1 + 2 + 7 + 5 = 15\), which is divisible by \(3\). So 1275 is divisible by \(3\).

Divide 1275 by \(3\): \(1275 \div 3 = 425\).

Now check 425 for \(3\). The sum of digits is \(4 + 2 + 5 = 11\), which is not divisible by \(3\). So we will not check \(3\) again for any later numbers.

The next prime is \(5\). The last digit of 425 is 5, so 425 is divisible by \(5\).

Divide 425 by \(5\): \(425 \div 5 = 85\).

Continue checking with \(5\). The last digit of 85 is 5, so 85 is divisible by \(5\).

Divide 85 by \(5\): \(85 \div 5 = 17\).

Now check 17 for \(5\). The last digit is 7, so it is not divisible by \(5\). So we will not check \(5\) again for further numbers.

The next prime is \(7\). 17 is not divisible by \(7\), and since \(7 > \sqrt{17}\), 17 must be prime.

Thus, the prime factors of 3825 are \(3, 3, 5, 5, 17\).

As a product, we write this as \(3 \times 3 \times 5 \times 5 \times 17\).

Grouping equal primes: \(3^2 \times 5^2 \times 17\).

(iv) 5005

Take 5005. The last digit is 5, so it is divisible by \(5\).

Divide 5005 by \(5\): \(5005 \div 5 = 1001\).

Now consider 1001. It is not even, and the sum of its digits is \(1 + 0 + 0 + 1 = 2\), so it is not divisible by \(2\) or \(3\). Check for \(5\) next: it does not end in 0 or 5, so it is not divisible by \(5\).

Try the next prime, \(7\). We test and find that \(1001 \div 7 = 143\), so 1001 is divisible by 7.

Now we have 143. It is not even, and the sum of digits is \(1 + 4 + 3 = 8\), so it is not divisible by \(2\) or \(3\). It does not end in 0 or 5, so not divisible by \(5\).

Try the next prime, \(7\), but \(143 \div 7\) is not a whole number.

Try \(11\). We find that \(143 \div 11 = 13\).

Now we get 13, which is prime.

So the prime factors of 5005 are \(5, 7, 11, 13\).

We write this as a product: \(5 \times 7 \times 11 \times 13\).

(v) 7429

Now take 7429. It is not even, so it is not divisible by \(2\). Check the sum of digits: \(7 + 4 + 2 + 9 = 22\). Since 22 is not divisible by \(3\), 7429 is not divisible by \(3\).

The last digit is 9, so it is not divisible by \(5\).

Try dividing by the prime \(7\). The value \(7429 \div 7\) is not a whole number, so 7 is not a factor.

Try the next prime, \(11\). The division \(7429 \div 11\) is also not a whole number.

Try \(13\). Again, \(7429 \div 13\) does not give a whole number.

Next, try the prime \(17\). We get \(7429 \div 17 = 437\), which is a whole number, so 17 is a factor.

Now consider 437. It is not even, and the sum of its digits is \(4 + 3 + 7 = 14\), so it is not divisible by \(3\). It does not end in 0 or 5, so not divisible by \(5\).

Try dividing 437 by \(17\); it does not give a whole number.

Try the next prime, \(19\). We get \(437 \div 19 = 23\), which is a whole number.

Now we have 23, which is a prime number.

So the prime factors of 7429 are \(17, 19, 23\).

Writing them as a product, we get \(17 \times 19 \times 23\).

Solution

To find the HCF and LCM using prime factorisation, we first write each number as a product of prime numbers. We test divisibility by primes in the correct order (2, 3, 5, 7, …), and once a number is not divisible by a smaller prime, we do not check that prime again for the same number.

HCF (Highest Common Factor) is the product of all common prime factors with the smallest power.

LCM (Least Common Multiple) is the product of all prime factors that appear in any of the numbers, using the highest power.

After that, we verify:

\( \text{LCM} \times \text{HCF} = \text{product of the two numbers} \)

(i) 26 and 91

Prime factorisation of 26:

26 is even, so it is divisible by 2.

\( 26 = 2 \times 13 \)

13 is prime.

Prime factorisation of 91:

91 is not divisible by 2.

Sum of digits = 10, not divisible by 3, so 91 is not divisible by 3.

Last digit is not 0 or 5, so not divisible by 5.

Check next prime 7: \(91 \div 7 = 13\).

So, \( 91 = 7 \times 13 \).

So we have:

\(26 = 2 \times 13\)

\(91 = 7 \times 13\)

HCF

Common prime = 13.

So, \( \text{HCF} = 13 \).

LCM

All primes used: 2, 7, 13

\( \text{LCM} = 2 \times 7 \times 13 = 182 \)

Verification

\( \text{LCM} \times \text{HCF} = \text{product of the two numbers} \)

\(182 \times 13 = 2366\)

\(26 \times 91 = 2366\)

Hence verified.

(ii) 510 and 92

Prime factorisation of 510:

510 is even ⇒ divide by 2.

\(510 = 2 \times 255\)

255 ends in 5 ⇒ divisible by 5.

\(255 = 5 \times 51\)

51 digit sum = 6 ⇒ divisible by 3.

\(51 = 3 \times 17\)

So, \(510 = 2 \times 5 \times 3 \times 17\)

Prime factorisation of 92:

92 is even ⇒ divisible by 2.

\(92 = 2 \times 46\)

46 is even again.

\(46 = 2 \times 23\)

23 is prime.

So, \(92 = 2^2 \times 23\)

So we have:

\(510 = 2^1 \times 3^1 \times 5^1 \times 17\)

\(92 = 2^2 \times 23\)

HCF

Common prime = 2.

Lowest power = \(2^1\).

So, HCF = 2.

LCM

All primes: 2, 3, 5, 17, 23

\( \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \)

Verification

\( \text{LCM} \times \text{HCF} = \text{product of the two numbers} \)

\(23460 \times 2 = 46920\)

\(510 \times 92 = 46920\)

Hence verified.

(iii) 336 and 54

Prime factorisation of 336:

336 is even:

\(336 = 2 \times 168\)

\(168 = 2 \times 84\)

\(84 = 2 \times 42\)

\(42 = 2 \times 21\)

21 is not even ⇒ check 3.

Digit sum = 3 ⇒ divisible by 3.

\(21 = 3 \times 7\)

So, \(336 = 2^4 \times 3 \times 7\)

Prime factorisation of 54:

54 is even:

\(54 = 2 \times 27\)

27 is not divisible by 2.

Digit sum = 9 ⇒ divisible by 3.

\(27 = 3^3\)

So, \(54 = 2 \times 3^3\)

So we have:

\(336 = 2^4 \times 3^1 \times 7\)

\(54 = 2^1 \times 3^3\)

HCF

Common primes = 2, 3

For 2: smallest power = 1

For 3: smallest power = 1

\( \text{HCF} = 2 \times 3 = 6 \)

LCM

Highest powers:

2 → \(2^4\)

3 → \(3^3\)

7 → \(7^1\)

\( \text{LCM} = 2^4 \times 3^3 \times 7 = 3024 \)

Verification

\( \text{LCM} \times \text{HCF} = \text{product of the two numbers} \)

\(3024 \times 6 = 18144\)

\(336 \times 54 = 18144\)

Hence verified.

Solution

We will use the prime factorisation method. First, we write each number as a product of prime numbers.

HCF (Highest Common Factor) is the product of all prime factors that are common to all the given numbers, taken with their smallest powers.

LCM (Least Common Multiple) is the product of all prime factors that appear in any of the numbers, taken with their highest powers.

We now apply this to each group of numbers.

(i) 12, 15 and 21

First, write prime factorisation of each number.

For 12:

\(12 = 2^2 \times 3\)

For 15:

\(15 = 3 \times 5\)

For 21:

\(21 = 3 \times 7\)

Finding HCF

Look for prime factors that are common to all three numbers.

The prime factors are:

12 has \(2, 3\)

15 has \(3, 5\)

21 has \(3, 7\)

The only prime factor common to all three is \(3\).

So,

\(\text{HCF} = 3\)

Finding LCM

Now take all distinct prime factors appearing in any of the numbers.

These are \(2, 3, 5, 7\).

The highest powers among the three numbers are:

\(2^2\) from 12

\(3^1\) from each (all have at most one 3)

\(5^1\) from 15

\(7^1\) from 21

So,

\(\text{LCM} = 2^2 \times 3 \times 5 \times 7\)

Now multiply step by step.

\(2^2 = 4\)

\(4 \times 3 = 12\)

\(12 \times 5 = 60\)

\(60 \times 7 = 420\)

So,

\(\text{LCM} = 420\)

Final answers for (i)

\(\text{HCF} = 3\)

\(\text{LCM} = 420\)

(ii) 17, 23 and 29

All three numbers 17, 23 and 29 are prime numbers.

So their prime factorisations are:

\(17 = 17\)

\(23 = 23\)

\(29 = 29\)

Finding HCF

For the HCF, we look for prime factors that are common to all three numbers.

Since each number is a different prime, they have no common prime factor greater than 1.

So,

\(\text{HCF} = 1\)

Finding LCM

For the LCM of distinct primes, we simply multiply them together.

So,

\(\text{LCM} = 17 \times 23 \times 29\)

First multiply 17 and 23.

\(17 \times 23 = 391\)

Now multiply this result by 29.

\(391 \times 29 = 11339\)

So,

\(\text{LCM} = 11339\)

Final answers for (ii)

\(\text{HCF} = 1\)

\(\text{LCM} = 11339\)

(iii) 8, 9 and 25

Now factor each number into primes.

For 8:

Divide by 2 again and again.

\(8 = 2^3\)

For 9:

\(9 = 3^2\)

For 25:

\(25 = 5^2\)

Finding HCF

We look for prime factors that are common to all three numbers.

8 has only prime factor \(2\).

9 has only prime factor \(3\).

25 has only prime factor \(5\).

There is no prime factor that is common to all three.

So,

\(\text{HCF} = 1\)

Finding LCM

Now take all prime factors that appear in any of the numbers.

These are \(2, 3, 5\).

The highest powers are:

\(2^3\) from 8

\(3^2\) from 9

\(5^2\) from 25

So,

\(\text{LCM} = 2^3 \times 3^2 \times 5^2\)

Calculate step by step.

\(2^3 = 8\)

\(3^2 = 9\)

\(5^2 = 25\)

Now multiply these results.

\(8 \times 9 = 72\)

\(72 \times 25 = 1800\)

So,

\(\text{LCM} = 1800\)

Final answers for (iii)

\(\text{HCF} = 1\)

\(\text{LCM} = 1800\)

Solution

We are given two numbers, 306 and 657, and their HCF (Highest Common Factor) is 9.

There is a standard relation between LCM and HCF of two numbers:

\( \text{LCM} \times \text{HCF} = \text{product of the two numbers} \)

We can use this relation to find the LCM when the HCF and the two numbers are known.

Step 1: Write the relation for our numbers

Here, the numbers are 306 and 657, and

\( \text{HCF} = 9 \)

So,

\( \text{LCM}(306, 657) \times 9 = 306 \times 657 \)

We want to find \( \text{LCM}(306, 657) \).

Step 2: Rearrange the formula to make LCM the subject

From

\( \text{LCM} \times 9 = 306 \times 657 \)

we get

\( \text{LCM}(306, 657) = \dfrac{306 \times 657}{9} \)

Step 3: Simplify the fraction

It is easier to first divide 306 by 9, instead of multiplying first and then dividing.

Compute:

\( 306 \div 9 = 34 \)

So the expression becomes:

\( \text{LCM}(306, 657) = 34 \times 657 \)

Step 4: Multiply 34 and 657

We multiply step by step using simple break-up.

Write 34 as \( 30 + 4 \).

First, calculate \( 657 \times 30 \).

\( 657 \times 30 = 657 \times 3 \times 10 \)

\( 657 \times 3 = 1971 \)

So,

\( 657 \times 30 = 1971 \times 10 = 19710 \)

Next, calculate \( 657 \times 4 \).

\( 657 \times 4 = 2628 \)

Now add these two results:

\( 19710 + 2628 = 22338 \)

So,

\( 34 \times 657 = 22338 \)

Step 5: Conclude the value of LCM

We have found:

\( \text{LCM}(306, 657) = 22338 \)

Therefore, the LCM of 306 and 657 is 22338.

Solution

We are asked to check whether the number \(6^n\) can ever end with the digit 0, where \(n\) is any natural number.

Step 1: Understand what it means for a number to end with 0

If a number ends with 0, it must be divisible by 10.

Being divisible by 10 means the number must contain both prime factors:

\(2\) and \(5\)

This is because:

\(10 = 2 \times 5\)

So, for \(6^n\) to end in 0, the number must include the factor 5.

Step 2: Look at the structure of \(6^n\)

We know:

\(6 = 2 \times 3\)

So,

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

Using exponents, this becomes:

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

Step 3: Check whether any power of 6 contains the factor 5

The prime factors in \(6^n\) are only \(2\) and \(3\).

There is no factor of \(5\) in \(6^n\) for any natural number \(n\).

Since a factor 5 never appears, \(6^n\) can never be divisible by 10.

Step 4: Observe the last digit pattern of powers of 6

Let us check the first few powers:

\(6^1 = 6\)

\(6^2 = 36\)

\(6^3 = 216\)

\(6^4 = 1296\)

In each case, the last digit is 6.

This pattern continues for every higher power because multiplying a number ending in 6 by 6 again always gives a number ending in 6.

Step 5: Conclude the result

Since no power of 6 contains the factor 5, \(6^n\) can never be divisible by 10 and therefore cannot end with 0.

Thus:

\(6^n\) never ends with 0 for any natural number \(n\).

Solution

We have to show that each given number is composite. A composite number is a natural number greater than 1 which has at least one factor other than 1 and itself. So our aim is to factorise each expression.

1) Number: \(7 \times 11 \times 13 + 13\)

Start with the expression:

\(7 \times 11 \times 13 + 13\)

We can see that both terms contain the number \(13\). So, we factor \(13\) out as a common factor.

First term: \(7 \times 11 \times 13\)

Second term: \(13\)

Take \(13\) common:

\(7 \times 11 \times 13 + 13 = 13 \big(7 \times 11 + 1\big)\)

Now simplify inside the bracket:

\(7 \times 11 = 77\)

So we get:

\(13 \big(7 \times 11 + 1\big) = 13 \big(77 + 1\big)\)

\(13 \big(77 + 1\big) = 13 \times 78\)

Here, both \(13\) and \(78\) are greater than 1.

So the number \(7 \times 11 \times 13 + 13\) has factors \(13\) and \(78\), apart from 1 and itself.

Therefore, \(7 \times 11 \times 13 + 13\) is a composite number.

2) Number: \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)

First observe the product part:

\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)

This is the same as \(7!\) (factorial 7), but we will work it out step by step.

Multiply step by step:

\(7 \times 6 = 42\)

\(42 \times 5 = 210\)

\(210 \times 4 = 840\)

\(840 \times 3 = 2520\)

\(2520 \times 2 = 5040\)

\(5040 \times 1 = 5040\)

So the product is:

\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)

Now add 5:

\(5040 + 5 = 5045\)

So the number is \(5045\).

Now show that 5045 is composite.

Notice that 5045 ends in the digit 5. Any number ending in 5 is divisible by \(5\).

So 5 is a factor of 5045.

Let us divide 5045 by 5:

\(5045 \div 5 = 1009\)

Therefore, we can write:

\(5045 = 5 \times 1009\)

Here, both \(5\) and \(1009\) are greater than 1.

So, 5045 has at least two factors other than 1 and itself, namely \(5\) and \(1009\).

Hence, \(5045\) is a composite number.

Conclusion

We have factorised each expression as a product of two numbers greater than 1:

\(7 \times 11 \times 13 + 13 = 13 \times 78\)

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

Therefore, both given numbers are composite.

Solution

We are told that Sonia and Ravi both start from the same point on a circular path at the same time. Sonia takes \(18\) minutes for one round, and Ravi takes \(12\) minutes for one round.

We want to find after how many minutes they will both be together again at the starting point.

This type of question is solved using the LCM (Least Common Multiple) of their times.

Idea: Each time Sonia completes a round, a multiple of \(18\) minutes has passed. Each time Ravi completes a round, a multiple of \(12\) minutes has passed. They will both be at the starting point together when the time passed is a common multiple of \(18\) and \(12\).

The earliest such time is the LCM of \(18\) and \(12\).

Step 1: Write prime factorisation

Factorise \(18\) into primes.

\(18 = 2 \times 9\)

\(9 = 3 \times 3\)

So,

\(18 = 2 \times 3^2\)

Now factorise \(12\) into primes.

\(12 = 2 \times 6\)

\(6 = 2 \times 3\)

So,

\(12 = 2^2 \times 3\)

Step 2: Find the LCM using highest powers

List all distinct prime factors: \(2\) and \(3\).

For \(2\):

Highest power is \(2^2\) (from 12).

For \(3\):

Highest power is \(3^2\) (from 18).

So,

\(\text{LCM} = 2^2 \times 3^2\)

Now calculate step by step.

\(2^2 = 4\)

\(3^2 = 9\)

Multiply these:

\(4 \times 9 = 36\)

So,

\(\text{LCM}(18, 12) = 36\)

Step 3: Interpret the result

The LCM \(36\) tells us that after \(36\) minutes:

– Sonia will have completed a whole number of rounds.

– Ravi will have completed a whole number of rounds.

So both will be back together at the starting point.

Final answer: They will meet again at the starting point after \(36\) minutes.