# CBSE 10th Mathematics | Fundamental Theorem of Arithmetic and Solved Examples

## Fundamental Theorem of Arithmetic

Every composite number can be expressed as a product of primes and this expression is unique, except from the order in which the prime factors occur.

By taking the example of prime factorization of 140 in different orders

## Solved Examples Based On Fundamental Theorem of Arithmetic

### Question:

Is 7 × 11 × 13 + 11 a composite number?

### Solution:

$\displaystyle 11\times (7\times 13+1)=11\times (91+1)$ = 11 × 92 = 1012

It is a composite number which can be factorized into primes.

### Question:

Find the missing numbers.

Going upwards

165 × 2 = 330

330 × 2 = 660

### Question:

Write the prime factorization of

(i) 72

(ii) 5005

### Question:

Find HCF and LCM of 45, 75 and 125.

(Note: HCF is the lowest power of common prime and LCM is the highest power of primes.)

### Solution:

45 = 3 × 3 × 5 = 32 × 5

75 = 3 × 5 × 5 = 3 × 52

125 = 5 × 5 × 5 = 53

HCF = 51 = 5

LCM = 32 × 53 = 9 × 125 = 1125

### Question:

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

(Hint: HCF (a, b) × LCM (a, b) = a × b.)

### Solution:

HCF (306, 657) × LCM (306, 657) = 306 × 657

Þ    9 × LCM (306, 657) = 306 × 657

Þ    LCM (306, 657) = $\displaystyle \frac{306\times 657}{9}$ = $\displaystyle 34\times 657$

Þ    LCM (306, 657) = 22338

### Question:

Check whether 6n can end with the digit 0, where n is a natural number.

Hint: Any number ending with zero must have a factor of 2 and 5.

### Solution:

Let the number 6n ends with the digit 0, where n is a natural number.

Þ    6n is divisible by 5.

Þ    5 is a prime factor of 6n.

Þ    But 2 and 3 are the only prime factors of 6n.

Þ    Hence, the number 6n is divisible by 5 is not possible. As the prime factorization of a number is always unique. Our supposition is wrong.

Hence, 6n does not end with the digit 0, where n is a natural number.

### Question:

Find the largest positive integer that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively.

### Solution:

Given condition is that on dividing 398 by the required number, there is a remainder of 7 so that 398 – 7 = 391 is exactly divisible by the required number or the required number is a factor of 391.

In the same way, required positive integer is a factor of 436 – 11 = 425 and 542 – 15 = 527 also

Clearly, required number is the HCF of 391, 425 and 527.

Using the factor tree for the prime factorisation of 391, 425 and 527 are as under:

391 = 17 ´ 23, 425 = $\displaystyle {{5}^{2}}$ ´ 17 and 527 = 17 ´ 31

\    HCF of 391, 425 and 527 is 17

Hence, required number = 17

### Question:

Find the HCF and LCM of 26 and 91 and verify that L.C.M. × H.C.F. = product of two numbers.

### Solution:

By prime factorization

26 = 2 × 13

91 = 7 × 13

HCF (26, 91) = 13

LCM (26, 91) = 13 × 2 × 7 = 26 × 7 = 182

LCM × HCF = 13 × 182 = 2366

Product of two numbers = 26 × 91 = 2366.

Hence, L.C.M. × H.C.F. = Product of two numbers.

### Question:

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

### Solution:

In order to find the maximum number of column in which army contingent can march. We have to find the largest number that divides 616 and 32.

Clearly, such a number is the HCF.

616 = 2 × 2 × 2 × 7 × 11 = 23 × 7 × 11

32 = 2 × 2 × 2 × 2 × 2 = 25

H.C.F. = 23 = 8

Hence, 8 is the maximum number of columns in which they can march.

### Question:

Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.

### Solution:

Greatest number of 6 digits is 999999. In order to find the greatest 6 digit number divisible by 24, 15, 36, we find their LCM.

L.C.M. of 24, 15 and 36 = 360

Subtracting remainder i.e., 279 from 999999, we get

999999 – 279 = 999720

Hence, 999720 is the largest 6 digit number divisible by 24, 15 and 36.

### Question:

If d is the HCF of 56 and 72, find x, y satisfying d = 56x + 72y. Also, show that x and y are not unique.

### Solution:

Applying Euclid’s division lemma to 56 and 72, we get

72 = 56 ´ 1+16        …(i)

56 = 16 ´ 3 + 8        …(ii)

16 = 8 ´ 2 + 0            …(iii)

\    Last divisor 8 is the HCF of 56 and 72.

From (ii), we get

8 = 56 – 16 ´ 3

Þ         8 = 56 – (72 – 56 ´ 1) ´ 3        [From equation (i)]

Þ         8 = 56 – 3 ´ 72 + 56 ´ 3

Þ         8 = 56 ´ 4 + (–3) ´ 72

Comparing it with d = 56x + 72y, we get x = 4 and y = –3.

Now,          8 = 56 ´ 4 + (–3) ´ 72

8 = 56 ´ 4 + (–3) ´ 72 – 56 ´ 72 + 56 ´ 72

Þ         8 = 56 ´ 4 – 56 ´ 72 + (–3) ´ 72 + 56 ´ 72

Þ         8 = 56 ´ (4 – 72) {(–3) + 56} ´ 72

Þ         8 = 56 ´ (–68) + (53) ´ 72

Þ         x = –68 and y = 53.

Hence, x and y are not unique.