HCF & LCM

Question: A factory manufactures products in batches of 12, 18, and 30 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

Solution:
To find the minimum number of units the factory needs to produce so that each batch size (12, 18, and 30) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is:
12 = 2 × 2 × 3 = 2² × 3¹

18 = 2 × 3 × 3 = 2¹ × 3²

30 = 2 × 3 × 5 = 2¹ × 3¹ × 5¹

Now,
The highest power of 2 is 2²
The highest power of 3 is 3²
The highest power of 5 is 5¹

So, the LCM of 12, 18, and 30 is:
= 2² × 3² × 5¹
= 4 × 9 × 5
= 180

∴ The minimum number of units the factory needs to produce is 180.

Question: The highest common factor (HCF) of x³ - 8, x⁴ - 16, and x³ - 2x² + 4x - 8 is:

Solution:
১ম রাশি = x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)

২য় রাশি = x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x - 2)(x + 2)(x² + 4)

৩য় রাশি = x³ - 2x² + 4x - 8
= (x³ - 2x²) + (4x - 8)
= x²(x - 2) + 4(x - 2)
= (x² + 4)(x - 2)
= (x - 2)(x² + 4)

∴ নির্ণেয় গ.সা.গু (HCF) = (x - 2)

Question: Three metal wires of lengths 90 cm, 126 cm, and 162 cm are provided. Calculate the maximum length of wire segments that can be cut, ensuring no waste.

Solution:
The maximum length of wire segments that can be cut (ensuring no waste) = (H.C.F. of 90, 126, 162) cm = 18 cm.

Elaborately,

The prime factorization of each number:
 90 = 2 × 3² × 5
126 = 2 × 3² × 7
162 = 2 × 3⁴

Common factors of all three numbers are: 2 (common), 3² = 9 (common)

∴ HCF = 2 × 3² = 18

Question: If the least common multiple of two numbers is twelve times their highest common factor, and their sum (HCF + LCM) equals 403, then what is the other number when one number is 93?

Solution:
Let HCF be h and LCM be l
Then l = 12h and
l + h = 403

∴12h + h = 403
⇒ h = 31

So, l = (403 - 31) = 372

Hence, the other number = (31 × 372)/93 = 124

Question: Find a number that is divisible by 7, but leaves remainder 5 when divided by 8, 12, and 20.

Solution:
Let the number be N.
It satisfies:
N ≡ 0 (mod 7)

If N leaves remainder 5 when divided by 8, 12, and 20, then N - 5 is divisible by all three numbers:
N - 5 ≡ 0 (mod 8, 12, 20)

Find LCM of 8, 12, 20:
8 = 2³
12 = 2² × 3
20 = 2² × 5
LCM = 2³ × 3 × 5 = 120

N must also be divisible by 7:
120k + 5 ≡ 0 (mod 7)  
⇒ 120k ≡ -5 ≡ 2 (mod 7)
(since -5 ≡ 2 mod 7)

Smallest positive k = 2 →
N = 120 × 2 + 5 = 240 + 5 = 245

Question: A bakery produces cakes in batches of 12, 18, and 30 units. What is the minimum number of units the bakery needs to produce so that each batch can be formed exactly?

Solution:
To find the minimum number of units the bakery needs to produce so that each batch size (12, 18, and 30) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is:
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²
30 = 2 × 3 × 5 = 2 × 3 × 5

Now,
The highest power of 2 is 2²
The highest power of 3 is 3²
The highest power of 5 is 5¹

So, the LCM of 12, 18, and 30 is:
= 2² × 3² × 5
= 4 × 9 × 5 = 180

∴ The minimum number of units the bakery needs to produce is 180.

Question: Find the H.C.F. of p(x) = x³ - 4x and q(x) = x² + 3x - 10

Solution:
Given that, p(x) = x³ - 4x and q(x) = x² + 3x - 10

Now,
The factors of p(x) = x³ - 4x
⇒ x(x² - 4)
⇒ x(x + 2)(x - 2)

And,
The factors of q(x) = x² + 3x - 10
⇒ x² + 5x - 2x - 10
⇒ x(x + 5) - 2(x + 5)
⇒ (x - 2)(x + 5)

∴ The required H.C.F. is (x - 2)

Question: What is the greatest number that will divide 96, 132, and 150 leaving remainders 6, 9, and 12 respectively? 

Solution:
We have to subtract the reminder first: 
96 - 6 = 90
132 - 9 = 123
150 - 12 = 138

Using prime factorization:
For 123:
123 = 3 × 41

For 90:
90 = 2 × 3² × 5

For 138:
138 = 2 × 3 × 23

Digit (123, 90, 138) have at least 3 in common. 

∴ The greatest number is 3

Question: Which one is the set of factors of 24?

Solution: 
1 × 24 
2 × 12
3 × 8
4 × 6

∴ The factors of 24 is 1, 2, 3, 4, 6, 8, 12, 24

So the set of factors of 24 is {1, 2, 3, 4, 6, 8, 12, 24}.

Question: What is the LCM of 8, 10, and 12?

Solution:
Prime factorization:
8 = 2×2×2 = 2³
10 = 2 × 5
12 = 2 × 2 × 3 = 2² × 3

LCM = take the highest power of each prime:
2³ × 3 × 5 = 8 × 3 × 5 = 120

Question: What is the H.C.F. of the following fractions? 6/7, 12/21, 18/35.

Solution:
আমরা জানি,
ভগ্নাংশের গসাগু = (লবের গসাগু)/(হরের লসাগু)

এখানে লবসমূহ = 6, 12 এবং 18
6 = 2 × 3
12 = 2² × 3
18 = 2 × 3²
∴ লবসমূহের গসাগু (H.C.F.) = 2 × 3 = 6

এখানে হরসমূহ = 7, 21 এবং 35
7 = 7 × 1
21 = 3 × 7
35 = 5 × 7
∴ হরসমূহের লসাগু (L.C.M.) = 3 × 5 × 7 = 105

ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 6/105
= 2/35 

Question: An officer was appointed on maximum daily wages on contract money of Tk. 6720. But on being absent for some days, he was paid Tk. 5600. For how many days was he absent?

Solution:
Maximum daily wages of the officers = H.C.F of Tk. 6720 and Tk. 5600

H.C.F of 6720 & 5600 = 1120

Maximum daily wage = Tk. 1120
Total days appointed = 6720 ÷ 1120 = 6 days

Days present = 5600 ÷ 1120 = 5 days
Absent days = 6 − 5 = 1

Question: What is the minimum number of chocolates that must be added to an existing stock of 966 chocolates, so the total stock can be equally distributed among 6, 7, 8 or 9 person?

Solution: 
LCM of 6, 7, 8 or 9 is,
6 = 2 × 3
7 = 7
8 = 2 × 2 × 2
9 = 3 × 3
∴ LCM = 2³ × 3² × 7 = 8 × 9 × 7 = 504

Double of the chocolates = 2 × 504 = 1008
The chocolate to be added = 1008 - 966 = 42

So the minimum number of chocolates that must be added to the stock of 966 is 42.

Question: What is the H.C.F. of the following fractions?
2/4, 4/6, 6/8.

Solution:
আমরা জানি, ভগ্নাংশের গসাগু = (লবের গসাগু)/(হরের লসাগু)

এখানে লব = 2, 4 এবং 6
2 = 2 × 1
4 = 2 × 2
6 = 2 × 3
∴ লবের গসাগু (H.C.F.) = 2

হর = 4, 6 এবং 8
4 = 2²
6 = 2 × 3
8 = 2³
∴ হরের লসাগু (L.C.M.) = 2³ × 3 = 8 × 3 = 24

ভগ্নাংশের গসাগু = লবের গসাগু/হরের লসাগু
= 2/24
= 1/12

Question: HCF and LCM of two numbers are 6 and 180 respectively. If the numbers are between 30 and 80, what is the sum of the numbers? 

Solution:
Given that,
HCF and LCM of two numbers are 6 and 180.

Let the numbers be 6x and 6y (where x and y are co-prime)
∴ LCM = 6xy
∴ 6xy = 180
⇒ xy = 180/6 = 30

Possible co-prime factor pairs of 30: (1, 30), (2, 15), (3, 10), (5, 6)

The numbers must lie between 30 and 80:
6 × 5 = 30 and 6 × 6 = 36, both between 30 and 80

Other pairs give at least one number ≤ 18 or ≥ 90, not valid
∴ Required numbers are 30 and 36
∴ Sum of the numbers = 30 + 36 = 66

So, the sum of the numbers is 66.

Question: What is the H.C.F. of 6/15, 12/20, and 18/25?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M. of denominators)

H.C.F. of numerators:
H.C.F.(6, 12, 18) = 6

L.C.M. of denominators:
15 = 3 × 5
20 = 2² × 5
25 = 5²

∴ L.C.M. = 2² × 3 × 5² = 4 × 3 × 25 = 300

∴ Required H.C.F. = 6/300 = 1/50

প্রশ্ন: দুইটি সংখ্যার অনুপাত ৫ : ৮ এবং লসাগু ২০০। সংখ্যা দুইটি কত?

সমাধান:
ধরি,
সংখ্যা দুইটি যথাক্রমে = ৫ক এবং ৮ক
∴ ৫x এবং ৮x এর লসাগু = ৪০ক

প্রশ্নানুসারে,
৪০ক = ২০০
⇒ ক = ২০০/৪০
∴ ক = ৫

∴ সংখ্যা দুইটি  হলো ৫ক = ৫ × ৫ = ২৫
এবং ৮ক = ৮ × ৫ = ৪০

Question: The H.C.F. of two number is 11 and their L.C.M. is 693. If one of the numbers is 77, find the other.

Solution:
Given that,
H.C.F. of two numbers = 11
L.C.M. of two numbers = 693
One number = 77
Let the other number = x


We know the relation between H.C.F., L.C.M., and two numbers.
⇒ Product of the two numbers = H.C.F. × L.C.M.
⇒ 77 × x = 11 × 693 
⇒ x = (11 × 693)/77
⇒ x = 693/7
∴ x = 99

So the other number is 99.

Question: What is the largest four-digit number that is exactly divisible by 12, 18, 24, and 36?

Solution:
Greatest number of 4-digits is 9999.

L.C.M. of 12, 18, 24 and 36 is 72.

On dividing 9999 by 72, the remainder is 63.

Required number = (9999 - 63) = 9936.

Question: Four bells ring at intervals of 6, 8, 12 and 18 minutes respectively. If they all ring together at 2 : 00 noon, when will they ring together again?

Solution:
চারটি ঘণ্টা দুপুর 2 টায় একত্রে বাজলে 6, 8, 12, 18 এর ল.সা.গুর সমান সময়ের পর ঘণ্টাগুলো পুনরায় একত্রে বাজবে। 

সংখ্যা গুলোর মৌলিক উৎপাদক
6 = 2 × 3
8 = 2 × 2 × 2
12 = 2 × 2× 3
18 = 2 × 3 × 3

6, 8, 12, 18 এর ল.সা.গু. = 2 × 2 × 2 × 3 × 3 = 72
অর্থাৎ 72 মিনিট

সুতরাং, ঘণ্টা গুলো একবার দুপুর 2 টায় বাজার পর পুনরায় বাজবে = 2 টা + 72 মিনিটে
= 2 টা + (60 + 12) মিনিটে
= 2 টা + 1 ঘণ্টা 12 মিনিটে
= 3 টা 12 মিনিটে

Question: The greatest number that divides 86, 182, and 366 leaving the same remainder of 6 in each case is:

Solution:
Given that,
Numbers are 86, 182, 366
And remainder is 6

Now, numbers leaving remainder 6:
86 - 6 = 80
182 - 6 = 176
366 - 6 = 360

HCF of 80, 176, and 360:
Prime factorization:
80 = 2⁴ × 5
176 = 2⁴ × 11
360 = 2³ × 3² × 5
∴ Common factor = 2³ = 8

So the greatest number is 8.

Question: What is the H.C.F. of 8/15, 12/25 and 16/30?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M of denominators)

H.C.F of numerators:
H.C.F. of 8, 12 and 16 = 4

& L.C.M of denominators:
L.C.M. of 15, 25 and 30 = 150

∴ Required H.C.F. = 4/150
= 2/75

Question: Find the H.C.F. of p(x) = x³ - 9x and q(x) = x² + 2x - 15.

Solution:
Given that, p(x) = x³ - 9x and q(x) = x² + 2x - 15

Now, the factors of p(x) = x³ - 9x
⇒ x(x² - 9)
⇒ x(x + 3)(x - 3)

And, the factors of q(x) = x² + 2x - 15
⇒ x² + 5x - 3x - 15
⇒ x(x + 5) - 3(x + 5)
⇒ (x - 3)(x + 5)

∴ The required H.C.F. is (x - 3)

Question: What is the least number which when doubled is exactly divisible by 8, 12, 16, and 20?

Solution:
Let the number be x.
Doubled the number is 2x.

8 = 2 × 2 × 2
12 = 2 × 2 × 3
16 = 2 × 2 × 2 × 2
20 = 2 × 2 × 5

∴ LCM = 2 × 2 × 2 × 2 × 3 × 5 = 240
∴ x = 240/2 = 120

Question: The least perfect square number divisible by 3, 4, 5, 6 and 8 is:

Solution:
LCM of 3, 4, 5, 6, 8 is 120
Now, 120 = 2 × 2 × 2 × 3 × 5
To make it a perfect square, it must be multiplied by 2 × 3 × 5
So, required number
= 2² × 2² × 3² × 5²
= 4 × 4 × 9 × 25
= 3600

Question: Find the HCF of 4/9, 8/15, 12/25 is-

Solution:
We know,
HCF = (HCF of all numerators)/(LCM of all denominators)

Now, numerators: 4, 8, 12
4 = 2 × 2
8 = 2 × 2 × 2
12 = 2 × 2 × 3
∴ HCF = 2 × 2 = 4

And, denominators: 9, 15, 25
9 = 3 × 3
15 = 3 × 5
25 = 5 × 5
∴ LCM = 3 × 3 × 5 × 5 = 9 × 25 = 225

∴ HCF = (HCF of all numerators)/(LCM of all denominators) 
= 4/225

Question: HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is ?

Solution:
Given that, 
HCF and LCM of two numbers are 7 and 140

Let,
The numbers are 7x and 7y
∴ LCM = 7xy

∴ 7xy = 140
⇒ xy = 20
⇒ Possible co-prime factors of xy = (1, 20), (4, 5)
⇒ Numbers are between 20 and 45
∴ Required number are = 4 × 7 = 28 and 5 × 7 = 35
∴ Sum of numbers are = 28 + 35 = 63

So the sum of the numbers is 63.

Question: What is the largest number that divides 43, 91, and 183, leaving the same remainder in all three cases?

Solution:
এখানে বৃহত্তম সংখ্যা হচ্ছে সংখ্যাগুলোর পার্থক্যগুলোর H.C.F. (গ.সা.গু)।

প্রথম দুটি সংখ্যার পার্থক্য: 91 - 43 = 48
পরের দুটি সংখ্যার পার্থক্য: 183 - 91 = 92

48 এবং 92 এর মৌলিক উৎপাদক:
48 = 2 × 2 × 2 × 2 × 3 = 2⁴ × 3
92 = 2 × 2 × 23 = 2² × 23

H.C.F. = 2² = 4

∴ নির্ণীত বৃহত্তম সংখ্যা = 4

Question: Identify the smallest integer that, when doubled, can be evenly divided by 18, 24, 28, and 36.

Solution:
LCM of 18, 24, 28, and 36 is = 504
So, the number will be half of 504 = 504/2 = 252

Question: Find the smallest number divisible by 7, 8, and 9.

Solution:
The smallest number divisible by several numbers is their LCM.
Factorize the numbers:
7 = 7
8 = 2³
9 = 3²
LCM = product of highest powers of all prime factors:
LCM = 2³ × 3² × 7
= 8 × 9 × 7
= 504

Question: How many positive integers less than 100 are multiples of both 2 and 3?

Solution:
A number that is a multiple of both 2 and 3 is a multiple of their LCM:
LCM(2, 3) = 6

Find how many multiples of 6 are less than 100: 
Multiples of 6: 6, 12, 18, …

Largest multiple less than 100: 96
Number of terms in this sequence (arithmetic progression with first term a₁=6, common difference d = 6, last term a_n = 96):
n = (a_n - a₁)/d + 1
= (96 - 6)/6+1
= 90/6 + 1
= 15 + 1
= 16

Question: The ratio of two numbers is 2 : 5 and their H.C.F is 6. Their L.C.M is -

Solution:
ধরি, সংখ্যা দুটি হলো 2x এবং 5x
∴ গসাগু (H.C.F) = x = 6

∴ সংখ্যা দুটি হলো: 2 × 6 = 12 এবং 5 × 6 = 30

∴ সংখ্যাদ্বয়ের গুণফল = 12 × 30 = 360
এবং H.C.F = 6

আমরা জানি,
L.C.M = (Product of two numbers)/H.C.F
= 360/6
= 60

∴ সংখ্যা দুটির লসাগু (L.C.M) = 60

Question: What is the least number that, when tripled, is exactly divisible by 9, 12, 15, and 18?

Solution:
Let the number be x.
Tripled the number is 3x.

9 = 3 × 3
12 = 2 × 2 × 3
15 = 3 × 5
18 = 2 × 3 × 3

LCM = 2 × 2 × 3 × 3 × 5 = 180

∴ 3x = 180
∴ x = 180/3 = 60

Question: If the ratio of two numbers is 3 : 4, and their Least Common Multiple (LCM) is 180, what are the two numbers?

Solution:
ধরি,
সংখ্যা দুটি = 3x এবং 4x
∴ সংখ্যা দুটির লসাগু (LCM) = 12x

প্রশ্নমতে,
12x = 180
⇒ x = 180/12
∴ x = 15

∴ সংখ্যা দুটি = 3 × 15 = 45 এবং 4 × 15 = 60

Question: What is the H.C.F. of 20/48, 28/72, and 36/90?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators)/(L.C.M of denominators)

H.C.F of numerators: H.C.F. of 20, 28 and 36 = 4
& L.C.M of denominators: L.C.M. of 48, 72 and 90 = 720

∴ Required H.C.F. = 4/720 = 1/180

Question: Three lights blink at intervals of 18 sec, 24 sec, and 32 sec. If they blink together now, how many times will they blink together in 6 hours?

Solution:
The lights will blink together at LCM of their intervals.
Intervals: 18, 24, 32 seconds

Factorize:
18 = 2 × 3²
24 = 2³ × 3
32 = 2⁵

LCM = product of highest powers of all primes:
LCM = 2⁵ × 3² =3² × 9 = 288 seconds
Convert 6 hours to seconds:
6 hours = 6 × 60 × 60 = 21600 seconds

Number of times they blink together:
Times = 21600/288 + 1 (+1 because they blink together now, at time 0)
= 75 + 1
= 76

Note: If you only count after the first blink at time 0, it is 75 times; including the initial moment, it is 76 times.

Question: Find the smallest number that is a multiple of 11 and leaves a remainder of 5 when divided by 8, 12, 16, and 24.

Solution:
L.C.M. of 8, 12, 16 and 24 is 48.
Let required number be 48k + 5, which is multiple of 11.

Least value of k for which (48k + 5) is divisible by 11 is k = 7.

Required number = (48 × 7) + 5 = 336 + 5 = 341.

∴ নির্ণেয় ক্ষুদ্রতম সংখ্যাটি হলো 341

Question: A merchant has three different types of oil: 240 litres, 300 litres, and 420 litres. Find the minimum number of containers of equal size that can store the oil without mixing.

Solution:
The size of each container should be the greatest possible that divides all three quantities, i.e., H.C.F of 240, 300, and 420.

H.C.F(240, 300, 420) = 60 litres

Total oil = 240 + 300 + 420 = 960 litres

Number of containers required = 960/60 = 16

Question: The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is- 

Solution:
If each student gets the same number of pens and same number of pencils, the maximum number of students will be the HCF of the total numbers.

∴ Maximum number of students = HCF(1001, 910)

Now, 
Factorize the numbers,
1001 = 7 × 11 × 13 and 910 = 2 × 5 × 7 × 13
Common factors are 7 and 13

∴ Multiply them = 7 × 13 = 91

So the maximum number of students = 91

Question: Find the greatest number which on dividing 1567 and 1979, leaves remainders 3 and 7 respectively.

Solution:
1567 - 3 = 1564
1979 - 7 = 1972

বৃহত্তম সংখ্যাটি হবে 1564 এবং 1972 এর গসাগু।

ইউক্লিডীয় পদ্ধতিতে গসাগু বের করি,

∴ গসাগু = 68

∴ নির্ণেয় বৃহত্তম সংখ্যা = 68

Question: What is the H.C.F. of 9/20, 15/28, and 21/35?

Solution:
We know, H.C.F. of fractions = (H.C.F. of numerators) / (L.C.M. of denominators)
H.C.F of numerators: H.C.F of 9, 15 and 21
9 = 3 × 3 
15 = 3 × 5
21 = 3 × 7
H.C.F = 3

L.C.M of denominators: L.C.M of 20, 28 and 35
20 = 2² × 5
28 = 2² × 7
35 = 5 × 7
L.C.M = 2² × 5 × 7 = 140

∴ Required H.C.F. = 3/140

Question: Two numbers are in the ratio 4 : 5 and their greatest common divisor (GCD) is 14. Find their least common multiple (LCM).

Solution:
Let the two numbers are,
First number = 4k
Second number = 5k
Since the numbers are in the ratio 4 : 5, and 4 and 5 are co-prime gcd(4,5) = 1,
the HCF of the two numbers is exactly k.

Given that, HCF = 14
Therefore, k = 14

So the actual numbers are:
First number = 4 × 14 = 56
Second number = 5 × 14 = 70

We know,
HCF × LCM = Product of the two numbers
⇒ 14 × LCM = 56 × 70
⇒ LCM = (56 × 70)/14
∴ LCM = 4 × 70 = 280

So the least common multiple (LCM) of the two numbers is 280.