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Showing posts with label algebra. Show all posts
Showing posts with label algebra. Show all posts

## Tuesday, April 22, 2008

This article will show how to do multiplication of two three digits numbers from right to left. Consider 234 × 456.

Step 1: Multiply the last digits. Write 2, carry 4.

Step 2: Cross multiply the one and the tenth, and add them together. We get 53. Add the 4 that we carry to 53. We get 57. Write 7, carry 5.

Step 3: Multiply all digits, like in the figure below. We get 55. Add the 5 that we carry to 55. We get 60. Write 0, carry 6.

Step 4: Finally, multiply the two left digits. We get 22. Add the 6 that we carry to 22. Write 2, carry 2.

Step 5: Finally, multiply the left digits. Add the 2 that we carry to 8. We get 10. Write both numbers.

## Thursday, April 17, 2008

### Easy Multiplication I

This technique of multiplication will do the multiplication from right to left. Let’s look at an example. 23 x 12 = 276

Step 1: Multiply the last digits.

Step 2: Cross multiply both sets of numbers and add them together.

Step 3: Finally, multiply the left digits.

## Wednesday, April 16, 2008

### Problem #1 of Project Euler

Add all the natural numbers below 1000 that are multiples of 3 or 5.

### Analysis:

Sequence of multiple of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ..., 999

Sequence of multiple of 5: 5, 10, 15, 20, 25, 30, ..., 995 So, the sum is 3 + 5 + 6 + 9 + 10 + 12 + 15 + 18 + ... + 999.

Take note that we have to remove the duplicate of 15, 30, 45, ..., 990

### Method # 1: using Python

All the natural numbers below 1000 that are multiples of 3 or 5, are divisible by either 3 or 5. Hence, we need to add the numbers that are only divisible by 3 or 5. That is very easy in Python. Using the % operator will give use the remainder of the division.

For example, 4%2 = 0, 4%3 = 1. Therefore, in order to ad the number that is divisible by 3 or 5, we have to check for the remainder. We will only add the numbers that give zero for the remainder.
total = 0
for i in range(1000):
if not (i % 3 and i % 5):
total += i

### Method # 2: using Arithmetic sequence

To add all natural numbers from 1 to 10.

Notice that the sum will always gives 11, if we write it this way. 11 occurs 10 times, so, 11 x 10 = 110. However, that is the sum of 1 to 10 twice. So, we divide 110 by 2, we get the answer, which is 55.

The formula: Sum = n (a1 + a2) / 2 where n is the number of occurrence of the sequence from a1 to a2.

Back to our problem, the sum of 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ..., 999. Adding the first and last numbers gives us 3 + 999 = 1002. 1002 occurs 333 times, so 1002 * 333 = 333666. Divide it by 2, we get 166883.

Repeat the same procedure for the second sequence of 5s. Then, add the answers. However, as I mentioned previously, remember that we also have to remove the duplicates of 15, 30, 45, ..., 990.

The formula would be like this:

sum of multiple of 3 + sum of multiple of 5 - sum of multiple of 15