## HCF and LCM Questions for IIFT PDF

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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

**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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