HCF and LCM Questions for IIFT PDF

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HCF and LCM Questions for IIFT PDF
HCF and LCM Questions for IIFT PDF

HCF and LCM Questions for IIFT PDF

Download important IIFT HCF and LCM Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice HCF and LCM questions and answers for IIFT exam.

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Question 1: How many pairs of natural numbers are there such that their LCM is 30 and HCF is not 5?

Question 2: The sum of two numbers is 2720, and their HCF is 17. How many ordered pairs of numbers (a, b) satisfy this condition?

Question 3: The HCF of two numbers is 8 and the difference between the two numbers is also 8. The LCM of the two numbers is a 3 digit number. What is the product of the digits of the largest possible value of the smaller number?

a) 0

b) 72

c) 64

d) 80

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Question 4: The value of a natural number N lies between 105 and 315. If the HCF of N and 315 is 1, what is the number of possible values of N lying between 105 and 315?

a) 128

b) 116

c) 96

d) 112

e) 82

Question 5: The HCF of two numbers is 8 and the difference between the two numbers is also 8. The LCM of the two numbers is a 3 digit number. What is the product of the digits of the largest possible value of the smaller number?

a) 0

b) 72

c) 64

d) 80

Question 6: The LCM of $7^{1403}-1$ and $7^{1403}+1$ when divided by 10 leaves a remainder of

Answers & Solutions:

1) Answer: 12

The possible pairs that satisfy the given criteria are:
(1,30), (2,30), (3,30), (6,30), (10,30), (15,30), (30,30), (2,15), (6,15), (3,10), (6,10) and (5,6).

Hence there are 12 such pairs.

2) Answer: 64

Let the two numbers be 17a and 17b, here a and b are co-prime to each other.
So given that 17a+17b = 2720
=> a+b = 2720/17
=> a+b = 160
=> So we essentially have to find the number of ways 160 can be written as sum of 2 co-prime natural numbers. This is given by $ \frac{E(n)}{2}$.
=> E(n) for 160 = $160*(1-\frac{1}{2})(1-\frac{1}{5})$ = 160*$\frac{1}{2}$*$\frac{4}{5} = 64$
So the pair of numbers which satisfy the given condition = $\frac{64}{2}$ = 32
But we have been asked about the ordered pairs so (a,b) and (b,a) are different pairs. Hence the required number of ways = 2*32 = 64

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3) Answer (A)

Let the smaller number be 8x so that the larger number would be 8x+8,
Now x and x+1 will be co-prime to each other, so LCM of these two numbers would be 8*x*(x+1) = > LCM is a 3 digit number, so the maximum possible value of x is 10.
Hence the largest possible value of smaller number is 8*10 = 80, so the required product of digits is 0

4) Answer (C)

Given, HCF of N and 315 is 1 i.e. N and 315 are co-primes.
315 = $5 \times 7 \times 3^2$
105 = $ 5 \times 7 \times 3$
Since both 315 and 105 have the same prime factors N would also be coprime to 105. Thus, the value of N = All co-primes of 315 – All co-primes of 105
= 315( 1 – 1/7)(1 – 1/5)(1 – 1/3) – 105(1 – 1/7)(1 – 1/5)(1 – 1/3)
= 48 x 3 – 48 = 96

5) Answer (A)

Let the smaller number be 8x so that the larger number would be 8x+8,
Now x and x+1 will be co-prime to each other, so LCM of these two numbers would be 8*x*(x+1) = > LCM is a 3 digit number, so the maximum possible value of x is 10.
Hence the largest possible value of smaller number is 8*10 = 80, so the required product of digits is 0

6) Answer: 4

$7^{1403}-1$ and $7^{1403}+1$ are two consecutive even numbers. Thus, their HCF has to be 2.
Now, HCF X LCM = Product of numbers
=> 2 X LCM= $7^{2806}-1$ =>LCM=$\frac{7^{2806}-1}{2}$
The remainder when this number is divided by 10 is the unit digit of the number.
The unit digit of $7^{2806}$ is 9. Thus the remainder will be (9-1)/2=4.

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