Question

There are 28 students in section A,36 students in section B and 32 students in section C of class 6th in a school . What is the minimum number of books required for their class library, so that can be distributed equally among students and of the three sections equally .

Find the greatest number of 4-digits exactly divisible by 12,16,24,28and 36

For each of the following pairs of numbers, show that the product of their HCF and LCM equals their product:-
14,21. 27,90


by using the HCF, find the PCM of:-
299,391. 1333,1767

Answer 1

i.Book required is equals to LCM of 28,32 and 36

28=2×2×7
32=2×2×2×2×2
36=2×2×3×3

LCM(28,32,26)=2×2×2×2×2×3×3×7
=2,016

Hence, total numbers of book required is 2,016.

12=2×2×3 = 2^3×3
16=2×2×2×2 = 2^4
24=2×2×2×3 = 2^3×3
36=2×2×3×3 =2^2×3^2

LCM=(12,16,24,36)=product of greatest power of each factor
=>2^4×3^2
=>16×9
=> 144

Greatest 4-digit number is 9,999

If any number is divisible by 12,16,24 and 36 it is also divisible by 144.

On dividing 9,999 by 144 we get 63 as remainder.

Hence, the required number is (9,999-63)=9,936 Ans.

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iii. For each of the following pairs of numbers,show that product of ther LCM and HCF is equals to their products.

A.14 and 21

14=2×7
21=3×7

HCF(14,21)=poduct of smallest power of each common factor
=>7
Hence ,HCF=7

LCM(21,14)=Product of greatest power of each common factor.
=>2×3×7
=>42

Hence, LCM=42

Now,
LCM×HCF=42×7=>294

Products of the numbers=14×21=>294

Thus,proved that LCM×HCF=product of the numbers

B.27 and 90

27=3×3×3 =3^3
90=2×5×3×3 =2×5×3^2

HCF(27,90)=poduct of smallest power of each common factor
=>3^2
=>9
Hence ,HCF=9

LCM(27,90)=Product of greatest power of each common factor.
=>2×5×3^3
=>270

Hence, LCM=270

Now,
LCM×HCF=9×270=>2,430

Products of the numbers=90×27=>2,430

Thus,proved that LCM×HCF=product of the numbers.

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3.By using HCF find LCM
i.299 and 391

299=23×13
391=23×17

HCF(299,391)=Product of smallest power of each factor
=>23

Hence, HCF of 299 and 391 is 23

Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=299×391
HCF×LCM=1,16,909
23×LCM=1,16,909
LCM=1,16,909 /23
LCM=5,083 Ans.

ii.1333 and 1767

1,333=31×43
1,767=31×57

Clearly,31 is the HCF

Now, we know that
HCF×LCM=product of the numbers
HCF×LCM=1,333×1,767
HCF×LCM= 23,55,411
31×LCM= 23,55,411
LCM= 23,55,411/31
LCM= 75,981 Ans.