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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