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.<\/p>\n
Download HCF and LCM Questions for IIFT PDF<\/a><\/p>\n Get 50% Off on IIFT & Other MBA Test Series\u00a0<\/a><\/p>\n Download IIFT Previous Papers PDF<\/a><\/p>\n Practice IIFT Mock Tests<\/a><\/p>\n Question 1:\u00a0<\/b>How many pairs of natural numbers are there such that their LCM is 30 and HCF is not 5?<\/p>\n Question 2:\u00a0<\/b>The sum of two numbers is 2720, and their HCF is 17. How many ordered pairs of numbers (a, b) satisfy this condition?<\/p>\n Question 3:\u00a0<\/b>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?<\/p>\n a)\u00a00<\/p>\n b)\u00a072<\/p>\n c)\u00a064<\/p>\n d)\u00a080<\/p>\n IIFT Free Mock Test<\/a><\/p>\n Enroll for CAT\/MBA Courses<\/a><\/p>\n Question 4:\u00a0<\/b>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?<\/p>\n a)\u00a0128<\/p>\n b)\u00a0116<\/p>\n c)\u00a096<\/p>\n d)\u00a0112<\/p>\n e)\u00a082<\/p>\n Question 5:\u00a0<\/b>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?<\/p>\n a)\u00a00<\/p>\n b)\u00a072<\/p>\n c)\u00a064<\/p>\n d)\u00a080<\/p>\n Question 6:\u00a0<\/b>The LCM of $7^{1403}-1$ and $7^{1403}+1$ when divided by 10 leaves a remainder of<\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer:\u00a012<\/b><\/p>\n The possible pairs that satisfy the given criteria are: Hence there are 12 such pairs.<\/p>\n 2)\u00a0Answer:\u00a064<\/b><\/p>\n Let the two numbers be 17a and 17b, here a and b are co-prime to each other. Top 500+ Free Questions for IIFT<\/a><\/p>\n IIFT Previous year question\u00a0 answer PDF<\/a><\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the smaller number be 8x so that the larger number would be 8x+8, 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given, HCF of N and 315 is 1 i.e. N and 315 are co-primes. 5)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Let the smaller number be 8x so that the larger number would be 8x+8, 6)\u00a0Answer:\u00a04<\/b><\/p>\n $7^{1403}-1$ and $7^{1403}+1$ are two consecutive even numbers. Thus, their HCF has to be 2.
\n(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).<\/p>\n
\nSo given that 17a+17b = 2720
\n=> a+b = 2720\/17
\n=> a+b = 160
\n=> 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}$.
\n=> E(n) for 160 = $160*(1-\\frac{1}{2})(1-\\frac{1}{5})$ = 160*$\\frac{1}{2}$*$\\frac{4}{5} = 64$
\nSo the pair of numbers which satisfy the given condition = $\\frac{64}{2}$ = 32
\nBut 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<\/p>\n
\nNow 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.
\nHence the largest possible value of smaller number is 8*10 = 80, so the required product of digits is 0<\/p>\n
\n315 = $5 \\times 7 \\times 3^2$
\n105 = $ 5 \\times 7 \\times 3$
\nSince 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
\n= 315( 1 – 1\/7)(1 – 1\/5)(1 – 1\/3) – 105(1 – 1\/7)(1 – 1\/5)(1 – 1\/3)
\n= 48 x 3 – 48 = 96<\/p>\n
\nNow 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.
\nHence the largest possible value of smaller number is 8*10 = 80, so the required product of digits is 0<\/p>\n
\nNow, HCF X LCM = Product of numbers
\n=> 2 X LCM= $7^{2806}-1$ =>LCM=$\\frac{7^{2806}-1}{2}$
\nThe remainder when this number is divided by 10 is the unit digit of the number.
\nThe unit digit of $7^{2806}$ is 9. Thus the remainder will be (9-1)\/2=4.<\/p>\n