# Problems on H.C.F and L.C.M

244 Problems on H.C.F and L.C.M জানি এই টপিকটি মোটামুটি সোজা, তথাপিও আমরা আসার উপযোগী কিছু ম্যাথ দিয়ে দিলাম। বিভিন্ন ওয়েবসাইট থেকে গুরুত্বপূর্ণ ম্যাথ গুলোই তুলে দেয়া হল। প্র্যাকটিস করতে থাকুন আশা করি কমন পাবেন। আমাদের দেয়া পোষ্ট গুলোতে যদি ভুল থাকে কমেন্টের মাধ্যমে জানাবান, আর প্রচুর শেয়ার চাই। আপনারা উপকৃত হলেই আমরা সার্থক ।

ল.সা.গু ও গ. সা. গু এর নিয়মঃ
১. ভগ্নাংশের লসাগু = লবগুলোর লসাগু / হরগুলো গসাগু
২. ভগ্নাংশের গসাগু = লবগুলোর গসাগু/হরগুলোর লসাগু
৩. দুটি সংখ্যার গুনফল = দুটি সংখার লসাগু x গসাগু
৪. লসাগু = সংখ্যাদুটির গুনফল / গসাগু
৫. গসাগু = সংখ্যাদুটির গুনফল / লসাগু
৬. একটি সংখ্যা = (লসাগু x গসাগু) / প্রদত্ত সংখ্যা

More Important Formulas:

1. Factors and Multiples: If a number ‘a’ divided another number b exactly, we say that a is a factor of b. , In this case, b is called a multiple of a.

2. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.): The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly. There are two methods of finding the H.C.F. of a given set of numbers:

1. Factorization Method: Express each one of the given numbers as the product of prime factors. The product of the least powers of common prime factors gives H.C.F.

2. Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F.

Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.

Similarly, the H.C.F. of more than three numbers may be obtained.

3. Least Common Multiple (L.C.M.): The least number which is exactly divisible by each one of the given numbers is called their L.C.M. There are two methods of finding the L.C.M. of a given set of numbers:

1. Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of the highest powers of all the factors.

2. Division Method (short-cut): Arrange the given numbers in a row in any order. Divide by a number which divided exactly at least two of the given numbers and carries forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

4. Product of two numbers = Product of their H.C.F. and L.C.M.

5. Co-primes: Two numbers are said to be co-prime if their H.C.F. is 1.

6. H.C.F. and L.C.M. of Fractions:

1. H.C.F. =

2. L.C.M. =

7. H.C.F. and L.C.M. of Decimal Fractions: In a given number, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without a decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

এখন H.C.F. and L.C.M. নিয়ে গুরুত্বপূর্ণ কিছু ম্যাথ দেখে নেয়া যাকঃ

1. প্রশ্নঃ Two numbers are in the ratio 2 : 3. If their L.C.M. is 48. what is sum of the numbers?

A. 64
B. 42
C. 28
D. 40

Let the numbers be 2x and 3x
LCM of 2x and 3x =6x (∵ LCM of 2 and 3 is 6. Hence LCM of 2x and 3x is 6x)
Given that LCM of 2x and 3x is 48
⇒ 6x=48
⇒ x=48/6=8

∴ Sum of the numbers = 2x + 3x = 5x
= 5 × 8 = 40

2. প্রশ্নঃ What is the greatest number of four digits which is divisible by 15, 25, 40 and 75?

A. 9200
B. 9600
C. 9800
D. 9400

The greatest number of four digits = 9999
LCM of 15, 25, 40 and 75 = 600
9999 ÷ 600 = 16, remainder = 399
Hence, greatest number of four digits which is divisible by 15, 25, 40 and 75 = 9999 – 399 = 9600

3. প্রশ্নঃ Three numbers are in the ratio of 2 : 3: 4 and their L.C.M. is 240. Their H.C.F. is:

A. 40
B. 20
C. 10
D. 30

Let the numbers be 2x, 3x and 4x
LCM of 2x, 3x and 4x = 12x
⇒ 12x = 240
⇒ x = 240/12
⇒ x = 20

∴ H.C.F of 2x, 3x and 4x = x =20

4. প্রশ্নঃ What is the lowest common multiple of 12, 36, and 20?

A. 120
B. 160
C. 220
D. 180 LCM = 2 × 2 × 3 × 1 × 3 × 5 = 180

5. প্রশ্নঃ What is the least number which when divided by 5, 6, 7 and 8 leave a remainder 3, but when divided by 9 leaves no remainder?

A. 1683
B. 1108
C. 2007
D. 3363

LCM of 5, 6, 7 and 8 = 840
Hence the number can be written in the form (840k + 3) which is divisible by 9.
If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9.
If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9.
Hence 1683 is the least number which when divided by 5, 6, 7, and 8 leave a remainder 3, but when divided by 9 leaves no remainder.

Solution-2:

Just see which of the given choices satisfy the given conditions.

Take 3363. This is not even divisible by 9. Hence this is not the answer.
Take 1108. This is not even divisible by 9. Hence this is not the answer.
Take 2007. This is divisible by 9.
2007 ÷ 5 = 401, remainder = 2 . Hence this is not the answer

Take 1683. This is divisible by 9.
1683 ÷ 5 = 336, remainder = 3
1683 ÷ 6 = 280, remainder = 3
1683 ÷ 7 = 240, remainder = 3
1683 ÷ 8 = 210, remainder = 3
Hence 1683 is the answer

6. প্রশ্নঃ Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

A. 4
B. 7
C. 9
D. 13

Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)
= H.C.F. of 48, 92 and 140 = 4.

7. প্রশ্নঃ The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

A. 276
B. 299
C. 322
D. 345

Clearly, the numbers are (23 x 13) and (23 x 14).
∴ Larger number = (23 x 14) = 322.

8. প্রশ্নঃ Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10, and 12 seconds respectively. In 30 minutes, how many times do they toll together?

A. 4
B. 10
C. 15
D. 16

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
∴ In 30 minutes, they will toll together 30/2 + 1 = 16 times.

9. প্রশ্নঃ The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

A. 101
B. 107
C. 111
D. 185

Let the numbers be 37x and 37y.
Then, 37x x 37y = 4107
xy = 3.
Now, co-primes with product 3 are (1, 3).
So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111)
∴ Greater number = 111.

10. প্রশ্নঃ Three number are in the ratio of 3: 4: 5 and their L.C.M. is 2400. Their H.C.F. is:

A. 40
B. 80
C. 120
D. 200

Let the numbers be 3x, 4x, and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
∴ The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, the required H.C.F. = 40.

11. প্রশ্নঃ The G.C.D. of 1.08, 0.36, and 0.9 is

A. 0.03
B. 0.9
C. 0.18
D. 0.108

The given numbers are 1.08, 0.36, and 0.90. H.C.F. of 108, 36, and 90 is 18,
∴ H.C.F. of given numbers = 0.18.

12. প্রশ্নঃ The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

A. 74
B. 94
C. 184
D. 364

L.C.M. of 6, 9, 15, and 18 is 90.
Let the required number be 90x + 4, which is multiple of 7.
Least value of k for which (90x + 4) is divisible by 7 is x = 4.
∴ Required number = (90 x 4) + 4 = 364.

13. প্রশ্নঃ The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

A. 3
B. 13
C. 23
D. 33

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
∴ Number to be added = (60 – 37) = 23.

14. প্রশ্নঃ The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

A. 279
B. 283
C. 308
D. 318

Given that, H.C.F. of two numbers = 11
L.C.M. = 7700 & One number = 275
We know, 2nd number =
∴ Other number =

15. প্রশ্নঃ The ratio of two numbers is 3: 4 and their H.C.F. is 4. Their L.C.M. is:

A. 12
B. 16
C. 24
D. 48