HCM & LCM Quiz – 1

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.

Clearly, the numbers are (23 x 13) and (23 x 14).
Therefore,Larger number = (23 x 14) = 322.

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 mins, they will toll together 30/2 +1=16

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Let the numbers be 3x, 4x, and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
Therefore, The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Therefore,Number to be added = (60 - 37) = 23

L.C.M. of 252, 308 and 198 = 2772.
So, A, B, and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec

Video solution:

Given two numbers, finding HCF, LCM is very easy. Given some property about HCF and the numbers, finding the numbers is much tougher.
We need to find ordered pairs (x, y) such that xy * HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.
So h^{ab}(ab) = 1080 = 2^{3} * 3^{3} * 5
We need to write 1080 as a product of a perfect cube and another number.

Four cases:
1. h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1, 1080), (8, 135), (27, 40) and (5, 216). (Essentially we are finding co-prime a,b such that a*b = 1080).

2. h = 2, We need to find number of ways of writing (33) * (5) as a product of two co-prime numbers. This can be done in two ways - 1 and (33) * (5) , (33) and (5)
number of pairs = 2, number of ordered pairs = 4

3. h = 3, number of pairs = 2, number of ordered pairs = 4

4. h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.
The pairs are (1, 1080), (8, 135), (27, 40), (5, 216), (2, 270), (10, 54), (3, 120), (24, 15) and (6, 30).
The question is "How many pairs of integers (x, y) exist such that the product of x, y and HCF (x, y) = 1080?"
Hence the answer is "9"

x = 525 and y = 525 works best.

If the question states x, y have to be distinct, then the best solution would be x = 350, y = 700, HCF = 350.

So the HCF is 525.
The question is "What is the maximum value of the HCF between x and y?"
Hence the answer is "525"

Find HCF and divide total sweets by that!
All sweets need to packed and each box has the same variety.
This implies the number of sweets in each box should be HCF of different count of sweets
HCF of 32, 216, 136, 88, 184, 120 = 23 = 8
Minimum number of boxes = (32 + 216 + 136 + 88 + 184 + 120) / 8 = 97
The question is "What is the minimum number of boxes required to pack?"
Hence the answer is "97 boxes"