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>> $Fundamental Theorem of Arithmetic$

Fundamental Theorem of Arithmetic

Q: In a seminar, the number of participants in Hindi, English and Mathematics are \$60,\ 84\$ and \$108\$ respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.

The Number of room will be minimum if each room accomodates maximum number of participants. Since in each room the same number of participants are to be seated and all of them must be of the same subject. Therefore, the number of participants in each room must be the \$HCF\$ of \$60,84\$ and \$108\$. The prime factorisations of \$60,84\$ and \$108\$ are as under.\$60={2}^{2}\times 3\times 5,84={2}^{2}\times 3\times 7\$ and \$108={2}^{2}\times {3}^{3}\$\$\therefore\ HCF\$ of \$60,84\$ and \$108\$ is \${2}^{2}\times 3=12\$Therefore, in each room \$12\$ participants can be seated.\$\therefore\ Number\ of\ rooms\ required =\dfrac { Total\ number\ of\ participants }{ 12 }\$ \$=\dfrac { 60+84+108 }{ 12 } =\dfrac { 252 }{ 12 } =21\$

Q: Find the greatest possible length which can be used to measure exactly the lengths \$7\$ m\$, 3\$ m\$, 85\$ m\$, 12\$ m \$95\$ cm.

The three given lengths are \$7\$ m \$= 700\$ cm\$3\$ m \$85\$ cm \$= 385\$ cm and \$12\$ m \$95\$ m\$= 1295\$ cmnow the required length is HCF of \$700 , 385\$ and \$1295.\$\$ 700 = 2^2 \times 5^2 \times 7 \$\$ 385 = 5 \times 7 \times 11 \$\$1295 = 5 \times 7 \times 37 \$HCF \$(700 , 385\$ and \$1295 ) = 5 \times 7 = 35 \$The greatest possible length is 35 cm

Q: Using Prime factorization , find the HCF and LCM of \$8, 9, 25\$

prime factors of \$8, 9, 25\$\$ 8 = 2 \times 2 \times 2 = 2^3 \$\$ 9 = 3 \times 3=3^2 \$\$ 25 = 5 \times 5 =5^2 \$\$HCF(8,9,25) = 1\$\$LCM ( 8,9,25)=\$ \$ 2^3 \times 3^2 \times 5^2 = 1800 \$

Q: If a and b are two prime numbers, then find HCF (a,b) .

HCF of two primes is always \$1.\$HCF \$(a,b) = 1\$

Q: Using fundamental theorem of Arithmetic find L.C.M. and H.C.F of \$816\$ and \$170\$.

According to the fundamental theorem of arithmetic every composite number can be factorised as a product of primes and this factorization is unique apart from the order in which the prime factor occurs.Fundamental theorem of arithmetic is also called unique factorization theorem.Composite number = product of prime numbers.Any Integer greater than 1, either be a prime number or can be written as a product of prime factors.The prime factors of \$816=2\times2\times2\times2\times3\times17=2^4\times3\times17\$The prime factors of \$170=2\times5\times17\$LCM of \$816\$ and \$170=2^4\times3\times5\times17=4080\$HCF of \$816\$ and \$170=2\times17=34\$Option A.

Q: Express \$3825\$ as a product of prime factors

Factorise \$3825\$ as follows:\$3825=3×3×5×5×17=3^{ 2 }×5^{ 2 }×17\$Since, \$3,5\$ and \$17\$ are prime numbers as these are the numbers that are only divisible by themselves. Hence, \$3825=3^{ 2 }×5^{ 2 }×17\$

Q: Find the largest number which divides \$245\$ and \$1029\$ leaving remainder \$5\$ in each case.

It is given that the required number when divides \$245\$ and \$1029\$, the remainder is \$5\$ in each case, This means that \$245-5=240\$ and \$1029-5=1024\$ are completely divisible by the required number.It follows from this that the required number is a common factor of \$240\$ and \$1024\$. It is also given that the required number is the largest number satisfying the given property. Therefore, it is the \$HCF\$ of \$240\$ and \$1024\$.Let us now find the \$HCF\$ of \$240\$ and \$1024\$ by Prime factorization method(refer above image).Clearly, \$HCF\$ of \$240\$ and \$1024\$ is common divisor i.e., \$2\times2\times2\times2=16\$.Hence, required number \$=16\$.

Q: Find the HCF and LCM of the following integers by applying the prime factorization method\$40 , 36\$ and \$126\$

\$ 40 = 2 \times 2 \times 2 \times 5 = 2^{3} \times 5 \$\$ 36 = 2 \times 2 \times 3 \times 3 = 2^{2} \times 3^{2} \$and \$126 = 2 \times 3 \times 3 \times 7 = 2 \times 3^{2} \times 7 \$ For H.C.F taking minimum power of common factorH.C.F \$ = (2)^{1} = 2 \$For L.C.M, taking maximum power of prime factorsL.C.M \$ = 2^{3} \times 3^{2} \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520 \$ Hence, H.C.F \$ =2\$L.C.M \$ = 2520\$

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