Four Methods:Using the Middle NumberUsing the Highest NumberUsing the Lowest NumberUsing Other Amounts of Consecutive Numbers
Bet someone you can add five consecutive integers faster than they can. You can use this as a bar trick, with your friends, or (if you are a student) impress your teacher!
EditSteps
EditMethod One of Four:
Using the Middle Number
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1Mentally multiply the middle number by 5 ... finished!? That's all you do!. e.g. with your example 53 X 5 = 265. Here's how to multiply this mentally:
- First separate 53 into 50 and 3.
- Now 50 X 5 = 250.
- Also 3 X 5 = 15.
- Now add those separate answers. 250 + 15 = 265.
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2Learn how it works:
- Let the smallest number be (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 above method: 10x/2 = 5x
EditMethod Two of Four:
Using the Highest Number
EditMethod Three of Four:
Using the Lowest Number
EditMethod Four of Four:
Using Other Amounts of Consecutive Numbers
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1To add four consecutive numbers, multiply the highest by 4 and subtract 6.
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2To add six consecutive numbers, multiply the highest by 6 and subtract 15.
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3To add seven consecutive numbers, multiply the highest by 7 and subtract 21.
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4To add eight consecutive numbers, multiply the highest by 8 and subtract 28.
EditTips
- You can add any sequence of consecutive numbers, even or odd, regardless of the number of integers the sequence contains, by adding the first and the last number in the sequence, dividing it by two and multiplying the answer by the number of integers in the sequence. In algebra, we can say ((a+b)/2)*n, or rearrange it to n*(a+b)/2 to remove a set of parenthesis.
- Method 2 can be used for any odd amount of consecutive numbers, but instead of using "5x" you must use "(amount of consecutive numbers)x"
- e.g. in 6 + 7 + 8, seven is x.
- (3)7 = 21, and 6 + 7 + 8 = 21
EditAdvanced Usage
- They don't need to be consecutive numbers. They just need to be a sequential subset of any linear equation. ( The examples above use the linear equation of x = c + 1 * n )
- For example, let's use the linear equation x = 10 + 7y, therefore, {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 also do not need to be integers. Let's use the linear equation x = 1 + y / 20, therefore, {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 also do not need to be positive values. The set can contain negative numbers, positive numbers, or both.
- This method can be extended to be used (as above) for any ODD number of consecutive integers --> 5, 7, 13, 25, 99 consecutive as long as you can locate the median digit and then multiply it be the number of integers. (Example 12, 13, 14, 15, 16, 17, 18, 19, 20 = 144 = 16(median) x 9(integer count). This can be most impressive if coupled with the simple trick of multiplying by 11.
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