Bet with your friends that you are the fastest to add up five consecutive numbers. Use it as a fun joke with friends or (if you go to school) do it to amaze your teacher!
Steps
Method 1 of 4: Using the number in the center
Step 1. Mentally multiply the number in the center by 5
.. done!? It's all here! For example, 53 X
Step 5. = 265. Here's how to do it mentally:
- First separate 53 into 50 and 3.
- Now 50 X 5 = 250.
- And 3 X 5 = 15.
- Now add the two results together. 250 + 15 = 265.
Step 2. Learn how to:
- Let's say the smallest number is (x - 2). Then the other 4 are (x - 1), (x), (x + 1) and (x + 2).
- The sum: (x - 2) + (x - 1) + (x) + (x + 1) + (x + 2) = 5x
- Using the method above: 10x / 2 = 5x
Method 2 of 4: Using the greater number
Step 1. Choose 5 consecutive numbers
Step 2. Multiply the larger number by 5
Step 3. Subtract 10
- Ex. 11, 12, 13, 14, 15
- 15 x 5 = 75
- 75 - 10 = 65
Method 3 of 4: Using the lowest number
Step 1. Choose 5 consecutive numbers
Step 2. Multiply the minor number by 5
Step 3. Add 10
- Ex. 11, 12, 13, 14, 15
- 11 x 5 = 55
- 55 + 10 = 65
Method 4 of 4: Using a number of consecutive numbers other than 5
Step 1. To add four consecutive numbers, multiply the highest by 4 and subtract 6
Step 2. To add six consecutive numbers, multiply the highest by 6 and subtract 15
Step 3. To add seven consecutive numbers, multiply the highest by 7 and subtract 21
Step 4. To add eight consecutive numbers, multiply the highest by 8 and subtract 28
Advice
- You can add up any sequence of consecutive numbers, even or odd, no matter how many integers there are in the sequence. You just have to add the first and last number in the sequence, divide by two and multiply the result by the number of integers in the sequence. In algebra, we can say ((a + b) / 2) * n, or, removing the brackets, n * (a + b) / 2.
- The second method can be used for any quantity shots of consecutive numbers, but instead of using "5x", you must use "(quantity of consecutive numbers) x"
- ex. in 6 + 7 + 8, seven is x.
- (3) 7 = 21, and 6 + 7 + 8 = 21
- They don't have to be consecutive numbers. They must be just one sequential subset of "any" linear equation. (The examples above use the linear equation x = c + 1 * n)
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For example, we use the linear equation x = 10 + 7y, hence, {xϵN | 17, 24, 31, 38, 45, …}
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- So if we use: 17, 24, 31, 38, 45
- 31 x 10 = 310 and 310/2 = 155
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They don't have to be whole numbers. * For example, we use the linear equation x = 1 + y / 20, hence, {xϵN | 1, 05 1, 1 1, 15 1, 2 1, 25 …}
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- So if we use: 1, 05 1, 1 1, 15 1, 2 1, 25
- 1, 15 x 10 = 11, 5 and 11, 5/2 = 5, 75
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- They don't even have to be positive values. The group can contain negative, positive, or both numbers.
- This method can be used (as above) for an ODD number of consecutive integers 5, 7, 13, 25, 99, just being able to identify the median digit and multiply it by the number of integers. (Example 12, 13, 14, 15, 16, 17, 18, 19, 20 = 144 = 16 (median) x 9 (amount of integers). This can be even more impressive when combined with the simple trick of multiplying by 11.
