HCF LCM

If a number N when divided by a, b or c leaves the same remainder x in each case, then the number will be of the form k(LCM of a,b,c) + x, where k = 0,1,2, etc.

Thus, the number will be of the form k (LCM of 6,7) + 2

-> Required number = 42K +2

Smallest such three-digit number = (42*3) +2 = 128

If a number N when divided by a, b or c leaves remainders p, q, r respectively, where (a-p) = (b-q) = (c-r) = x, then the number will be of the form k(LCM of a,b,c) – x, where k = 1,2,3 etc.

Thus, the number will be of the form k(LCM of 8,7) -6

-> Required number = 56k – 6

Now, largest 5-digit number = 99999, which may be written as 56(1785)+39

Now, 56(1785) = 99999-39 = 99960

-> 56k-6 = 99960-6= 99954

From the given information, N = 9k₁ +4

Also, N = 7k₂ + 3

Now, 9k₁ +4 when divided by 7, leaves a remainder of 3

-> 9k₁ +4 -3 or 9k₁ +1 is perfectly divisible by 7. Take different values of k₁ to check for which minimum value of k₁, will 9k₁ +1 be perfectly divisible by 7; we get k₁ =3 as the first such value, and the corresponding value of 9k₁ +1 will be 28

-> N= (9k₁ +1) +3 = 28+3 =31

The next higher number is obtained by adding LCM of 9 & 7 to 31. Therefore, any number of the form k(LCM of 9, 7)+31, will satisfy the given condition. Thus, the next number will be 63+31 = 94. This is also the highest two-digit such number.

The largest number which divides the numbers a,b,c giving remainders of p,q,r respectively will be the HCF of any 2 of the 3 numbers: (a-p), (b-q) and (c-r)

-> The largest number which divides the numbers 110 and 99 giving remainders of 2 and 3 respectively will be the HCF of (110-2) and (99-3), or HCF of 108 and 96.

Now, 108 = 2² * 3³  and 96 = 2⁵ * 3 -> HCF (108,96) = 2² * 3 = 12

The number of students in each room will be the HCF of 650, 910, and 1170.

650 = 2 * 5² * 13, 910 = 2 * 5 * 7 * 13, 1170 = 2 * 3² * 5 *13

-> HCF = 2 * 5 *13 = 130

-> Number of rooms = (650/130) + (910/130) + (1170/130) = 21

Given, N₁+N₂ = 544

Since HCF is 32, each number is a multiple of 32.

Let N₁ = 32x and N₂ = 32y, where x and y are natural numbers.

-> 32x+32y = 544 -> x+y = 17

Now, pairs of co-primes with sum 17 are (1, 16), (2, 15), (3, 14). (4, 13), (5, 12), (6, 11), (7, 10), (8, 9)

-> Required numbers are: (32*1, 32*16), (32*2, 32*15), (32*3, 32*14), (32*4, 32*13), (32*5, 32*12), (32*6, 32*11), (32*7, 32*10), (32*8, 32*9)

The largest such number will be HCF of (290-242) & (242-194)

-> HCF of 48 & 48

Therefore, 48 only.

If a number N is divided by a, b, c and leaves remainders p, q, r respectively, where (a-p) = (b-q) = (c-r) = x, then N = k(LCM of a, b, c) – x where k is a whole number

-> N = k(LCM of 2, 3, 4, 5, 6)-1 = 60k-1

When k = 1, N = 59; when k = 2, N = 119

Therefore, only 1 number i.e. 59, is between 0 & 100.

Let the minimum number of gifts be N. When the gifts are distributed among children, the difference between the divisor and the remainder in each case is the same i.e. 1.

Now, if a number N on being divided by a, b and c, leaves remainders p, q and r respectively, where (a-p) = (b-q) = (c-r) = x, then N = k(LCM of a, b, c) – x

-> N = k(LCM of 6, 7, 8, 9) – 1

-> N = 504k-1

Put k = 1 -> N = 504(1) -1 = 503

The product of LCM & HCF of two numbers is equal to the product of the two numbers.

-> LCM*HCF = 4641*21 = N₁*N₂, where N₁ & N₂ are the two numbers

Now, numbers can be expressed as multiples of their HCF.

Let N₁ = 21x and N₂ = 21 y, where x and y are positive integers

-> 4641*21 = 21x*21y

Now, 4641 = 13*17*21 (resolving 4641 into prime factors)

-> 13*17*21*21= 21x*21y

Comparing LHS and RHS of the above equation, one out of N₁ and N₂ can be written as 21*13, and the other one can be written as 21*17

-> The numbers are 21*13 = 273, and 21*17 = 357