Problems on LCM and HCF Part 1


Problems on LCM and HCF - 1


  1. Factors and Multiples:
  2. If 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.
  3. 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 the each one of the given numbers as the product of prime factors. The product of 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.
  4. 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 highest powers of all the factors.
    2. Division Method (short-cut): Arrange the given numbers in a rwo in any order. Divide by a number which divided exactly at least two of the given numbers and carry 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.
  5. Product of two numbers = Product of their H.C.F. and L.C.M.
  6. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
  7. H.C.F. and L.C.M. of Fractions:
        1. H.C.F. =H.C.F. of Numerators
    L.C.M. of Denominators
        2. L.C.M. =L.C.M. of Numerators
    H.C.F. of Denominators
  8. H.C.F. and L.C.M. of Decimal Fractions:
    In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without 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.
  9. Comparison of Fractions:
    Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.


1. 
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
Answer: Option A
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
     = H.C.F. of 48, 92 and 140 = 4.
2. 
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
Answer: Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
 Larger number = (23 x 14) = 322.
3. 
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
Answer: Option D
Explanation:
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 together30+ 1 = 16 times.
2
4. 
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
A.4
B.5
C.6
D.8
Answer: Option A
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
  = H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
5. 
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
A.9000
B.9400
C.9600
D.9800
Answer: Option C
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
 Required number (9999 - 399) = 9600.

6. 
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
Answer: Option C
Explanation:
Let the numbers be 37a and 37b.
Then, 37a x 37b = 4107
 ab = 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.
7. 
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
Answer: Option A
Explanation:
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, required H.C.F. = 40.
8. 
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
Answer: Option C
Explanation:
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.
9. 
The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
A.1
B.2
C.3
D.4
Answer: Option B
Explanation:
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
 ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
10. 
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
Answer: Option D
Explanation:
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
 Required number = (90 x 4) + 4   = 364.
11. 
Find the lowest common multiple of 24, 36 and 40.
A.120
B.240
C.360
D.480
Answer: Option C
Explanation:
 2 | 24  -  36  - 40
 --------------------
 2 | 12  -  18  - 20
 --------------------
 2 |  6  -   9  - 10
 -------------------
 3 |  3  -   9  -  5
 -------------------
   |  1  -   3  -  5
   
L.C.M.  = 2 x 2 x 2 x 3 x 3 x 5 = 360.   
12. 
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
Answer: Option C
Explanation:
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.
13. 
Reduce128352to its lowest terms.
238368
A.
3
4
B.
5
13
C.
7
13
D.
9
13
Answer: Option C
Explanation:
 128352) 238368 ( 1
         128352
         ---------------
         110016 ) 128352 ( 1
                  110016
                 ------------------  
                   18336 ) 110016 ( 6       
                           110016
                           -------
                                x
                           -------
 So, H.C.F. of 128352 and 238368 = 18336.
 
             128352     128352 ÷ 18336    7
 Therefore,  ------  =  -------------- =  --
             238368     238368 ÷ 18336    13                    
14. 
The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:
A.1677
B.1683
C.2523
D.3363
Answer: Option B
Explanation:
L.C.M. of 5, 6, 7, 8 = 840.
 Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
 Required number = (840 x 2 + 3) = 1683.
15. 
A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
A.26 minutes and 18 seconds
B.42 minutes and 36 seconds
C.45 minutes
D.46 minutes and 12 seconds
Answer: Option D
Explanation:
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
16. 
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
Answer: Option C
Explanation:
Other number =11 x 7700= 308.
275
17. 
What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?
A.196
B.630
C.1260
D.2520
Answer: Option B
Explanation:
 L.C.M. of 12, 18, 21 30                 2 | 12  -  18  -  21  -  30
                                         ----------------------------
   = 2 x 3 x 2 x 3 x 7 x 5 = 1260.       3 |  6  -   9  -  21  -  15
                                         ----------------------------
   Required number = (1260 ÷ 2)            |  2  -   3  -   7  -   5

                   = 630.
18. 
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
Answer: Option D
Explanation:
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
19. 
The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:
A.1008
B.1015
C.1022
D.1032
Answer: Option B
Explanation:
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
   = 1008 + 7
   = 1015
20. 
252 can be expressed as a product of primes as:
A.2 x 2 x 3 x 3 x 7
B.2 x 2 x 2 x 3 x 7
C.3 x 3 x 3 x 3 x 7
D.2 x 3 x 3 x 3 x 7
Answer: Option A
Explanation:
Clearly, 252 = 2 x 2 x 3 x 3 x 7.

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