LCM and HCF - Solved Examples

11.The total number of prime factors of the product (8)20×(15)24×(7)15$(8{)}^{20}\times (15{)}^{24}\times (7{)}^{15}$ is

a. 59

b. 98

c. 123

d. 4

Correct Option: D

Explanation:

The prime numbers are 2,3,5,17 in the expression.  The expression can be written as (23)20×(3×5)24×(17)15⇒260×324×524×1715$(2^3{)}^{{}^{20}}\times (3\times 5{)}^{24}\times (17{)}^{15}\Rightarrow 2^{60}\times 3^{24}\times 5^{24}\times {17}^{15}$

So number of prime factors are 4. i.e., 2, 3, 5, 17

12. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divisible by 3, then the first number is

a. 264

b. 132

c. Both a and b

d. 33

Correct Option: C

Explanation:

Let the numbers are ah, bh respectively.  Here h is HCF of two numbers.  (obviously a, b are coprimes i.e., HCF (a, b) = 1)

Given that HCF = h = 44 and LCM = abh = 264

Dividing LCM by HCF we get ab = 6.

ab can be written as 1 x 6, 2 x 3, 3 x 2, 6 x 1.

But given that the first number is divisible by 3. So only two options possible for A. 3 x 44, 6 x 44. So option C is correct

13. What least number must be subtracted from 1294 so that the remainder when divided 9, 11, 13 will leave in each case the same remainder 6 ?

a. 0

b. 1
c. 2
d. 3

Correct Option: B

Explanation:

LCM of 9,11,13 is 1287. Dividing 1294 with 1287, the remainder will be 7, to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.

14. The least number which is divisible by 12, 15, 20 and is a perfect square, is

a. 400

b. 900
c. 1600

d. 3600

Correct Option: b

Explanation: 

LCM = 5 × 3 × 2² = 60
To make this number as a perfect square, we have to multiply this number by 5 × 3

The number is 60 × 15= 900

15. The least perfect square number which is divisible by 3,4,5,6 and 8 is

a. 900

b. 1200

c. 2500

d. 3600

Correct Option: D

Explanation: 

LCM = 2×2×2×3×5=23×3×5$2\times 2\times 2\times 3\times 5=2^3\times 3\times 5$

But the least perfect square is = 23×3×5×(2×3×5)=24×32×52=3600$2^3\times 3\times 5\times (2\times 3\times 5)=2^4\times 3^2\times 5^2=3600$  as the perfect squares have their powers even.



16. Three piece of timber 42 m, 49 m and 63 m long have to be divided into planks of the same length. What is the greatest possible length of each plank?
a. 7 m
b. 14 m
c. 42 m
d. 63 m
Correct Option: A
Explanation:
The maximum possible length = (HCF of 42, 49, 63) = 7

17. The greatest number which can divide 1354, 1866, 2762 leaving the same remainder 10 in each case is :
a. 64
b. 124
c. 156
d. 260
Correct Option: A
Explanation:
The needed number = HCF of 1344, 1856 and 2752=64

18.  Find the value of LCM(1,2,3,4,.........100)LCM(51,52,53,......100)${\displaystyle \frac{LCM(1,2,3,4,.........100)}{LCM(51,52,53,......100)}}$

Solution:

LCM is defined as the product of all the prime numbers with maximum powers in the given numbers. There are 25 primes below 100 and we need to consider all the prime number where they take their maximum power.

LCM of (1, 2, 3, .........100) = 26×34×52×72×11×13.............97$2^6\times 3^4\times 5^2\times 7^2\times 11\times \mathrm{13.............97}$

so LCM of (51, 52, 53, ...........100) = 26×34×52×72×11×13.............97$2^6\times 3^4\times 5^2\times 7^2\times 11\times \mathrm{13.............97}$

For example 2 takes its maximum power 6 in 64. i.e., 26$2^6$.  Similarly 3 takes its maximum power 4 in 81. i.e., 34$3^4$.  so 

Now to find the LCM of (51, 52, 53......100) we need to consider the prime numbers up to 100 as 100 is the maximum number.  Again we can find the maximum power of 2 is 6 in 64. for 3 it is 4 in 81. for 5 it is 2 in 75 or 100...

LCM(1,2,3,4,.........100)LCM(51,52,53,......100)${\displaystyle \frac{LCM(1,2,3,4,.........100)}{LCM(51,52,53,......100)}}$ =  26×34×52×72×11×13.............9726×34×52×72×11×13.............97${\displaystyle \frac{2^6\times 3^4\times 5^2\times 7^2\times 11\times \mathrm{13.............97}}{2^6\times 3^4\times 5^2\times 7^2\times 11\times \mathrm{13.............97}}}$= 1

19. Find the number of combinations of (a, b, c) if LCM (a, b) = 1000, LCM (b, c) = 2000, LCM (c, a) = 2000

Sol: 

LCM (a, b) = 23×53$2^3\times 5^3$

LCM (b, c) = 24×53$2^4\times 5^3$

LCM (c, a) = 24×53$2^4\times 5^3$

Let a = 2p×5s$2^p\times 5^s$

b = 2q×5t$2^q\times 5^t$

c = 2r×5u$2^r\times 5^u$

Calculation for powers of 2:

Maximum power of 2 is 4 in LCM (c, a). So r is 4.  and either p or q will take 3 and other will take 0, 1, 2, 3.

r = 4;  p = 3;  q = 0, 1, 2, 3

r = 4;  p = 0, 1, 2;  q = 3,

Total combinations are (4,3,0), (4, 3, 1), (4, 3, 2), (4, 3, 3) and (4, 0, 3), (4, 1, 3), (4, 2, 3). 

Total options for powers of 2 = 7

Calculation for powers of 5:
Maximum power of 5 is 3.  So any 2 of s, t, u have maximum power 3.  and other will take 0, 1, 2, 3

a = 3, b = 3, c  = 0, 1, 2, 3

a = 0, 1, 2; b = 3;  c= 3, 

a = 3; b = 0, 1, 2; c = 3

Total options are 7 x 10 = 70

20. The letters A, B, C, D, E, F and G represent distinct digits chosen from (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) such that A*B*C = B*G*E = D*E*F, where ‘*’ means multiplication.  What does the letter G represent?

Assume A*B*C = B*G*E = D*E*F = X then A, B, C, D, E, F, G all divides X exactly.  That means, X is the LCM of all these digits.  LCM of (1, 2, 3, 4, 6, 8, 9) = 72
As A*B*C = B*G*E = D*E*F - - - -  (i), so each expression must posses those numbers whose integral multiple or sub-multiples is/are possessed by the other expressions. Thus 0, 5 & 7 are ruled out.

But 72 = 1 x 8 x 9 = 2 x 4 x 9 = 6 x 4 x 3

Here 4, 9 appeared two times so these values take B, E in some order so G = 2

We can't determine the remaining values uniquely.