We all learn in elementary school that a number is divisible by 2 if the last digit is even. A number is divisible by 3 if the sum of the digits is divisible by 3. A number is divisible by 5 if its last digit is 0 or 5. Etc.
But imagine we were born with 8 fingers on each hand and did arithmetic in base 16, also called hexadecimal. Or suppose schools decided to teach children hexadecimal instead of decimal to give them a head start in computer science. We would count 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, …
What would divisibility rules look like? Here are the simplest rules.
n divisible by 2 if its last digit is even.
n is divisible by 3 if its digit sum is divisible by 3.
n is divisible by 4 if its last digit is 0, 4, 8, or C.
n is divisible by 5 if its digit sum is divisible by 5.
n is divisible by 6 if it is divisible by 2 and 3.
n is divisible by 8 if it ends in 0 or 8.
n is divisible by A if it is divisible by 2 and 5.
n is divisible by F if its digit sum is divisible by F.
n is divisible by 10 (i.e. sixteen) if its last digit is 0.
In this list “digit” means hexadecimal digit.
There are also rules for checking divisibility by 7 and 11 (i.e. seventeen) in hexadecimal analogous to the rules for divisibility by 7 and 11 in decimal.
To test divisibility by 7, split off the last digit from the rest of the number. Triple it and subtract from what’s left of the original number. Repeat this process until you get something you recognize as being divisible by 7 or not.
For example, suppose you start with F61. Split this into F6 and 1. Subtract 3 from F6 to get F3. Now split F3 into F and 3. Subtract 9 from F to get 6. F61 is not divisible by 7 because 6 is not divisible by 7. The explanation for how this works is completely analogous to the corresponding rule in base 10.
You can test for divisibility by 11 in base 16 (i.e. by seventeen) just as you’d test divisibility by 11 in base 10 (or any other base). Add up the digits in an odd position and subtract the sum of the digits in an even position. For example, suppose you start with 135B8. The digits in odd position sum to 1 + 5 + 8. The digits in even positions sum to 3 + B. Both sums are equal, so there difference is 0, which is divisible by 11, so 135B8 is divisible by 11.
There’s nothing special about base 16 or base 10. You could come up with analogous divisibility rules in any base.
Select 01-Select All Given a City table, whose fields are described
as +-------------+----------+ |
Field | Type | +-------------+----------+ |
ID | int(11) | |
Name | char(35) | |
CountryCode | char(3) | |
District | char(20) | |
Population | int(11) | +-------------+----------+ write a query that will fetch all columns for every row in the
table. My Solution SELECT*FROM city; --------------------------------------------------------------------------------- 02-Select by ID Given a City table, whose fields are described
Measuring Time Logic Puzzle
You are given with two ropes with variable width. However if we start burning both the ropes, they will burn at exactly same time i.e. an hour. The ropes are non-homogeneous in nature. You are asked to measure 45 minutes by using these two ropes.
How can you do it?
Please note that you can’t break the rope in half as it is being clearly stated that the ropes are non-homogeneous in nature. Answer & ExplanationSolution:
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.
Thus you have successfully calculated 30+15 = 45 minutes …