Two circles have their centers 21 cm apart. The radii of the circles are 10 cm and 17 cm. Find the length (in cm) of the common chord of the two circles.

 if two circles having radii a and b with the distance between them as c,
Length of common chord= 2* Sqrt [ a^2 - ((a^2+c^2-b^2)/2c)^2]

Ans. 84

If the HCF of two number is 23 and the LCM is 4186, one of the two numbers is one of the following: 

a) 276 b) 300 c) 322 d) 345. Which one is it?

Ans. c) 322 

 Both number should be divisible by its HCF and LCM should be divisible by  both the numbers whose  HCF and LCM are to determine. 
 a) 276%23=0
      4186%276=46

b)300%23=1
  4186%300=286

c)322%23=0
 4186%322=0

d)345%23=0
 4186%345=46

How do I find two numbers when I know that their LCM is 36 and HCF is 15?

If the HCF of the two numbers say x and y is 15; the numbers can be represented as following:

x = 15a and y = 15b where a and b are positive integers

It is given that LCM of x and y is 36

Now LCM (x,y) * HCF (x,y) = x * y

Hence, 36 * 15 = 15a * 15b

540 = 225*a*b

or 2.4 = a*b

The product of two integers cannot be 2.4 hence such two numbers do not exist.

The sum of two numbers is 52 and their LCM is 168. What are the numbers?

As LCM of two numbers is 168 that two numbers must be factors of 168

now think about what are all factors of 168 that are less than 52(because sum of those two is 52)

factors=(2,3,4,6,8,7,8,12,14,24,28,42)

from above factors u have to select two numbers such that sum of two is 52

only possibility is 24+28

What is the difference between two numbers if their LCM is 630 and HCF is 9 and the sum of the two numbers is 153?

Given GCD of two numbers is 9 and LCM is 630.

Let a, b be two numbers
As we know,

LCM × HCF = The product of given numbers

a×b=9×630
Given a+b=153
By using formula
(a-b)^2=(a+b)^2-4×a×b
(a-b)^2=(153^2)-(4×9×630)
(a-b)^2=729
x-y=27
Ans.27

Another Solution:

Let's assume the numbers are 9*a and 9*b (as the hcf is 9).

As we know,

LCM × HCF = The product of given numbers

So, (9*a)*(9*b) = 630*9

So, a*b = 70
Again 9*(a + b)= 153

By solving those two equations, we get either a=10, b=7 or a=7 , b=10
So either way, the two numbers are 90 and 63.

So, $(\mathrm{90}-\mathrm{63})=27$ or the difference between the two numbers is 27.

Ans.27

The LCM of 1, 2, 3, ... 60 is N. What is the LCM of 1, 2, 3, ... 65?

Answer: 61*2*n

We are given LCM (1, 2, 3 ... 60) = n
Now, we need to find out LCM (1, 2, 3... 65)

= LCM (n, 61, 62, 63, 64, 65) 
{The first 60 numbers can be replaced with 'n' as we already know their LCM}

= LCM (n, 61,64) 

{62, 63, and 65 can be broken down into prime factors which are in turn factors of n} 

= 61*2*n  
{61 is a prime number. 64 is 2^6. 2^5 occurs in 32. An extra 2 is required to accommodate it}

= 122n

Explanation:

Lcm of first 60 numbers is n, so we need to consider 5 more numbers 61...65. Lets use prime factorization and try to figure out how inclusion of each of these numbers will affect lcm.
61 - It is a prime number and so it has not been included in n till now. So Lcm of 1...61 become 61*n.
62 - Prime factorization of 62 is 2*31 and both of these numbers are already included in n. So lcm doesn't change by inclusion of 62 and it remains 61*n.
63 - Prime factorization of 63 is 3*3*7 and these numbers are already there in n. In fact 3 is there three times (for 27) and 7 is there two times (for 49). So inclusion of 63 also doesn't change lcm and lcm remain 61*n for numbers 1....63.
64 - Factorization of 64 is 2^6. Now 2^5 is already there in n for 32. So we need to include only one more 2, and lcm of numbers 1..upto 64 becomes 61*2*n.
65 - 65 factorizes as 5*13. Both of these numbers are there in n already and by inclusion of 65, lcm doesn't change. So it remains 61*2*n.

How-many-pairs-of-2-numbers-are-there-whose-LCM-is-N

If N= a^m*b^n *c^o then ordered solution such that LCM = N is (2m+1)*(2n+1)*(2o+1)
Unordered will be (2m+1)*(2n+1)*(2o+1) /2 +0.5