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

A 4 digit no. is formed using the digits 0,2,4,6,8 without repeating any 1 of them. What is the sum of all such possible no.s?

1)519960 2)402096 3)133320 4)4321302

If you fix 8 as the last digit, you see that there are 4*3*2 ways to complete the number. Thus, 8 appears 24 times as the last digit. By the same logic, if we enumerate all possible numbers using these 5 digits, each number appears 24 times in each of the 4 positions. That is, the digit 8 contributes (24*8+240*8+2400*8+24000*8). In total, we have
(0+2+4+6+8)(24+240+2400+24000)=533280
as our total sum.
In case 4-digit numbers cannot start with 0, then we have over-counted. Now we have to subtract the amount by which we over-counted, which is found by answering: "What is the sum of all 3-digit numbers formed by using digits 2,4,6, and 8?" Now if 8 appears as the last digit, then there are 6 ways to complete the number, so 8 contributes (6*8+60*8+600*8). In total, we have
(2+4+6+8)(6+60+600)=13320.
Subtracting this from the above gives us 519960.

Which is the Smallest number that has 4 prime factors & 48 total factors?

Take first 4 prime numbers as
2^a*3^b*5^c*7^d
Give maximum powers to 2 and 3 to get the minimum number 
(a+1)(b+1)(c+1)(d+1)=48
4*3*2*2=48
a=3 b=2 c=1 d=1

8*9*5*7=2520 Ans.

A ball is dropped from a height of 96 feet and it rebounds 2/3 of the height it falls. If it continues to fall and rebound, find the total distance that the ball can travel before coming to rest? 

a.240 ft b 360 ft c.480 ft d.none of these

Here is a trick if ball rebounds for 2/3 or for any other fraction (p/q) like 3/4, 1/2 or 5/2 
then use this (p+q)*height.
so we get, according to this question,
(2+3)*96= 480 Feet.

Another Trick:

96(1+r)/(1-r)=96(5/3)/(1/3)=480

Power of 2

How do you find out if an unsigned integer is a power of 2?

The trick here is to know that an unsigned integer which is a power of 2 has only one of its bits as 1. So a simple solution would be to loop through the bits and count the number of 1s.

There is another solution which uses bit manipulation.

isPower = (x !=0 && !(x & (x-1)))

x & (x-1) will always give you a 0 if x is a power of 2. !(x & (x-1)) should give us what we want but there is one corner case. If x is 0, then the second term alone would return true when the answer should be false. We need the check to make sure x is not 0.

Reverse a String

First Solution

void reverseString(char* str) { int len, loop;
char temp; len = strlen(str); for(loop = 0; loop < (len / 2); loop++) { temp = str[loop];
str[loop] = str[len - loop - 1];
str[len - loop - 1] = temp; } }

Second Solution

The second solution is slightly better than the first as it does not need the char space. It uses bit manipulation (XOR in this case) to swap the chars in place.

void reverseStringBetter(char* str) { int loop1 = 0, loop2; loop2 = strlen(str) - 1; for (loop1 = 0; loop1 < loop2; loop1++, loop2--) { str[loop1] ^= str[loop2] ; str[loop2] ^= str[loop1] ; str[loop1] ^= str[loop2] ; } }