GCD and LCM

 

GCD and LCM

M.Velrajan

1. Which one of the following is not true?

a) GCD of two consecutive even numbers is 2

b) Any two consecutive odd numbers are relatively prime

c) Any two consecutive numbers are relatively prime

d) The product of any three consecutive numbers is not divisible by 3

(a), (b) and (c) are true. But (d) is not true.  


2. Find the LCM of 45, 4-81, 412 and 47  (a)412     (b)4        (c)42     (d)4-2

Since am divides an, if m < n, 45, 4-81 and 47 divide 412. Hence 412 is the LCM


3. Find the H. C.F of 4/9, 2/5, 6/8, 2/5    (a)1/180      (b)2/481      (c)2/350         (d)1/142

Any nonzero rational number is a factor of all the rational numbers.

Hence HCF can not be defined in the rational numbers


4. The GCD of 112, 192 and 336 is  a) 4    b) 8    c) 16    d) 32

The required GCD have to divide 192 - 112 = 80 and 336 - 192 = 144.

The required GCD = GCD of 80 and 144 = 16.

Aliter : The largest one in the answer choices is 32.

32 does not divide 112. So 32 is not the GCD

16 is the largest of 4, 8 and 16.

112 = 16 × 7, 192 = 16 × 12 and 336 = 16 × 21.

So 16 divides 112, 192 and 336. Hence 16 is the GCD.


5. The LCM of 16, 24 and 32 is  a) 96    b) 48    c) 32    d) 192

Since 16 divides 32, the required LCM = LCM of 24, 32.

24 = 3 × 8 and 32 = 4 × 8. LCM = 3 × 4 × 8 = 96


6. The GCD and LCM of two numbers are 12 and 216. If one of the number is 108 then

the other number is   a) 24    b) 36    c) 48    d) 84

LCM × GCD = product of the numbers.

12 × 216 = 108 × the otther number 

The other number = (12 × 216) / 108 = 24


7. If the LCM of two numbers is their product then a) they are relatively prime  

b) one of the number divides the other  c) they are even numbers  d) they are odd numbers

LCM × GCD = product of the numbers, so GCD = 1, (a).


8. Three numbers are in the ratio 3 : 4: 5 and their L.C.M is 2400. Their H.C.F is  

(a)40   (b)80  (c)120   (d)200

The numbers are 3k, 4k, 5k. Since 3, 4, 5 are relatively prime, 

LCM of the numbers = (3 × 4 × 5)k = 60k and HCF = k. 

Hence 60k = 2400 and HCF = k = 40


9. The greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case is

 (a)4                             (b)7                    (c)9                                (d)13

Since the number leaves the same remainder, it divides 91 - 43 = 48, 183 - 91 = 92 and 

183 - 43 = 140.

The number = gcd(48, 92, 140) = 4 (since 48 = 3 × 4 × 4, 92 = 4 × 23 and 140 = 4 × 5 × 7)


10. Find the greatest number which will  divide 2112 and 2792 leaving the remainder 4 in each case

(a)63                     (b)64                           (c)68                       (d)78

2112 - 4 = 2108, 2792 - 4 = 2788.

Required number = gcd of 2108, 2788.

The gcd has to divide 2788 - 2108 = 680. Among the answer choices only 68 divides 680.

Required number = 68


11. What is the greatest number that will divide 676 and 942 and will leave remainders

4 and 18 respectively?         (a) 64          (b) 46 (c) 84 (d) 42

676 - 4 = 672, 942 - 18 = 924

Required number = gcd of 672, 924

Among the answer choice 64 and 46 do not divide 672

The largest of 84 and 42 is 84. And 84 divides 672 and 942, so 84 is the answer  


12. The greatest four digit number divisible by 15, 25, 40 and 90 is  

(a)9600   (b)9000   (c)9400  (d)9200

In the answer choices 90 divides 9000 only.       (b).

Aliter : LCM {15, 25, 40, 90} = 1800.

Hence the largest four digit number divisible by the given numbers 

                                 = the largest four digit number that is a multiple of 1800 = 9000.

 

13. Find the greatest number of 4 digits which is divisible by 15, 25, 40 and 75

(a)9600  (b)3000  (c)9800  (d)8540

LCM {15, 25, 40, 75} = LCM(40, 75)   (since 15 and 25 divide 75)

                                     = 8 × 75 = 600.

The required number = the largest 4 digit number that is a multiple of 600 = 9600 

(note that 16 × 600 = 9600, but 17 × 600 = 10200)

Aliter : In the answer choices, 75 does not divide 9800 and 15, 25, 40 and 75 divide 9600. 

Ans. 9600


14. Find the least number which, upon being divided by 2, 3, 4, 5 and 6 leaves in each case a reminder of 1, 

but when divided by 7 leaves no remainder.    (a)  271        (b) 291      (c) 301     (d) 311

LCM(2, 3, 4, 5, 6) = LCM(4, 5, 6) = 2 × 5 × 6 = 60

60 + 1 = 61 is the least number which leaves remainder 1 upon being divided by 2, 3, 4, 5, 6. 

But 61 is not divisible by 7.

The required number is the least number which is of the form (multiple of 60) + 1 that is divisible

by 7. 

In the answer choices only 301 is of the form (multiple of 60) + 1, hence 301 is the answer

(or) 60 + 1, 120 + 1, 180 + 1, 240 + 1 are not divisible by 7 but 300 + 1 is divisible by 7.

The required number = 301 

  

15. The sum of two numbers is 528 and their H.C.F is 33. How many pairs of such numbers can be

found?   (a) 1                (b) 2                (c) 3 (d) 4

528 = 16 × 33

Hence the pairs of numbers with sum 528 and 33 as common divisor are

1 × 33, 15 × 33 (note 1 + 15 = 16);  2 × 33, 14 × 33 (note 2 + 14 = 16);

3 × 33, 13 × 33 (note 3 + 13 = 16); 4 × 33, 12 × 33 (note 4 + 12 = 16);

5 × 33, 11 × 33;  6 × 33, 10 × 33;  7 × 33, 9 × 33; 8 × 33, 8 × 33. 

Among these only for the pairs 1 × 33, 15 × 33;  3 × 33, 13 × 33;  5 × 33, 11 × 33; 7 × 33, 9 × 33

the gcd is 33, since 2 × 33 divides all the other pairs. 

So there are only 4 such pairs. Ans. (d)


16. The product of two numbers is 2646 and their H.C.F is 21. Find the number of possible pairs of

numbers.   (a) 1                 (b) 2 (c) 3 (d) 4

2646 = 126 × 21, 126 = 6 × 21. Hence HCF of 126, 21 is 21. 

The possible pairs of numbers are 126, 21 ; 

63, 42 (divide 126 by 2 and multiply 21 by 2 so that their product is 2646). 

Note that dividing 126 by 3 and multiplying 21 by 3, we get the same pair 42, 63.

So, there are only 2 possible pairs.


17. Six bells commence tolling together, afterwards they 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

LCM of 2, 4, 6, 8, 10, 12 = LCM of 8, 10, 12 = 8 × 5 × 3 = 120.

Hence they toll together for every 120 seconds = 2 minutes. 

In 30 minutes they toll together 30/2 = 15 times 


18. A, B, C start at the same time in the same direction to run around a rectangular garden. 

A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds starting at the same point.

After what time will they meet again at the starting point?      

(a)20 minutes 18 seconds      (b)40 minutes 20 seconds     (c)46 minutes 12 seconds    (d)30 minutes

LCM(252, 308, 198) = LCM(4 × 9 × 7, 4 × 7 × 11,  2 × 9 × 11) = 4 × 7 × 11 × 9 = 308 × 9 = 2772

So they will meet again at the starting point after 2772 seconds = 46 minutes 12 seconds

 

19. A, B,C, D start at the same time to run around a circular  garden in the same direction. 

A completes a round in 30 minutes, B in 60 minutes, C in 90 minutes and D in 105 minutes. 

After what time will they meet again at the starting point?  

(a)15 hours     (b)10 hours     (c)21 hours    (d)20 hours

LCM(30, 60, 90, 105) = 1260.

So they will meet again at the starting point after 1260 minutes = 21 hrs


20. The number of numbers – pairs lying between 40 and100 with their H. C.F. as 15 is

(a)3   (b)4   (c)5    (d)2

The smallest multiple of 15 greater than 40 is 45 = 3 × 15 and 

the largest multiple of 15 less than 100 is 90 = 6 × 15. 

Required pairs are 3 × 15, 4 × 15 (since 3 and 4 are relatively prime the HCF of the pair is 15);

3 × 15, 5 × 15 (since 3 and 5 are relatively prime the HCF of the pair is 15);

4 × 15, 5 × 15 (since 4 and 5 are relatively prime the HCF of the pair is 15);

and 5 × 15, 6 × 15 (since 5 and 6 are relatively prime the HCF of the pair is 15).  

Number of pairs = 4. 

Note that 3 × 15, 6 × 15 and 4 × 15, 6 × 15 are not included, since their HCF are 45 and 30

respectively.


21. The LCM and HCF of two numbers are 45 and 3 respectively, their sum is 24, what is their

difference?              (a)2                                 (b)4                   (c)6                             (d)8

xy = Product of two numbers = LCM × HCF = 45 × 3 = 15 × 9 = 135 and

x + y = 24

The possible values of the pairs x, y so that xy = 135 are 45, 3; 15, 9.

45 + 3 = 48 and 15 + 9 = 24

Hence the value of the pair x, y is 15, 9 and their difference = 6

Aliter : (x-y)2 = (x+y)2 - 4xy = 576 - 540 = 36

So x - y = 6


22. The H.C.F. of x3 +1 and x4 – 1 is      (a) x3 - 1          (b) x3 + 1      (c) x +1     (d) x – 1

a3 + b3 = (a + b)(a2  + b2 - ab) and a2 – b2 = (a + b)(a - b).

x3 + 1 = (x + 1)(x2 - x + 1) and x4 – 1 = (x2 + 1)(x2 - 1) = (x2 + 1)(x + 1)(x - 1)

(x2 + 1) and (x - 1) are  not divisors of (x2 - x + 1)

HCF = x + 1

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