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.

Solution:

Going upwards

165 × 2 = 330

330 × 2 = 660

Question:

Write the prime factorization of

(i) 72

(ii) 5005

Solution:

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.

Share

Leave a Reply

Your email address will not be published. Required fields are marked *

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>

*