Mathematics

Under this topic I will not just show tricks, as the notion of tricks in mathematics is probably the most damning to the spread of mathematical knowledge. I will instead explain how these, so called, tricks work using logic and reason. So this topic is not for the savvy hackers or mathematicians out there.

Divisibility theorems

When it comes to divisibility there exists some neat theorems to test a certain number’s different divisibilities, or factors. Following are those theorems and their proof. Some of the proofs are more trivial than others, such as the proof for divisibility by 1, 2, 5 and 10. From here on out it’s assumed that every number, when working with divisibility, is an integer.

What is the smallest number that is divisible by 1 through 10? The answer is 2520. You can try the different divisibility theorems below on it.

Divisibility by 1

This theorem is quite easy to remember. Every integer is divisible by 1.

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Divisibility by 2

It’s common knowledge that every even number (numbers ending with an even number) is divisible by 2. This is because even numbers are multiples of 2. In short, if a number ends with either 0, 2, 4, 6 or 8 it is divisible by 2.

Say we have a four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. If d is divisible by 2 we can represent it as an even number 2n, like so:

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We can now see that the number is divisible by 2 if, and only if, the last digit is divisible by 2. And so the theorem is proven.

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Divisibility by 3

The theorem goes that if the sum of all digits in a number is divisible by 3, the whole number is divisible by 3, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. this number can be represented as 1000a + 100b + 10c + d. Now let’s break out one a, b and c from the first 3 terms, and factor out 3 like so:

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We can now see that the first term is divisible by 3, and the second term is divisible by 3 if, and only if, the sum of a + b + c + d is divisible by 3. And so the theorem is proven.

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We have shown that the procedure above will hold for all cases. The procedure is also recursive. If the sum is to hard to test divisibility for, the procedure can be repeated, until a smaller sum is reveald. This is rarely necessary as the divisibility of the sum often is easy to determined.

Divisibility by 4

The theorem goes that if the last two digits of a number is divisible by 4, the whole number is divisible by 4, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. Now let’s factor out 4 from the first two terms, like so:

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We can now see that the first term is divisible by 4, so the whole number is divisible by 4 if, and only if, the second term is divisible by 4. And so the theorem is proven.

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Divisibility by 5

The theorem goes that if the last digits of a number is divisible by 5, the whole number is divisible by 5, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. Now let’s factor out 5 from the first three terms, like so:

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We can now see that the first term is divisible by 5, so the whole number is divisible by 5 if, and only if, the second term is divisible by 5. And so the theorem is proven. As the only one digit numbers that are divisible by 5 are 0 and 5, another way of putting it is – if last digit is 0 or 5, the number is divisible by 5.

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Divisibility by 6

This theorem is a combination of the theorem for divisibility by 2 and divisibility by 3.

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Note: Be careful when combining theroems like this. Don’t be fooled and try to combine divisibility theorems for 2 and 4 to get the theorem for 8, as numbers divisible by 4 always are divisible by 2. The number 4 is for instance both divisible by 2 and 4, but not by 8. Make sure the theorems you combine doesn’t share a factor, like 4 and 2 sharing the factor 2. It’s of course possible to use theoerems 4 and 2 together, but it’s not certain in all cases that the two theorems prove divisibility by 8.

Divisibility by 7

Probably one of the most useful theorems is the theorem of divisibility by 7, as it is recursive (just like the theorems of divisibility by 3 & 9). The theorem states that if the difference between the last digit multiplied by 2 and the remaining digits in a number is divisible by 7, the whole number is divisible by 7. I’ll show the procedure with an example below:

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Neat! So how and why does it work? For simplicity’s sake we use a two digit number ab, represented as 10a + b. The theorem says that if (A) 7 | a - 2b then (B) 7 | 10a + b. Let’s prove it!

In order to prove the theorem, we must prove both A and B. So let’s start with A. If we have a - 2b, and it’s divisible by 7, we know that 7 must be a factor of the expression. We can now create an equation:

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Multiply the whole equation with 10, and add one extra b:

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Now add 20b to each side of the equation, and try to factor out 7:

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We can now see that the right side of the equation is divisible by 7, and our left side says 10a + b. Neat, we now know that the two digit number ab is divisible by 7. Now we must show that Bimplies A. That is if a - 2b is divisible by 7, then 10a + b is divisible by 7. Let’s prove B.

Just as for A, we know that 7 must be a factor of the expression. We can now create another equation:

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Subtract 21b from the whole equation, and factorize:

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We can now see that the right side of the equation is divisible by 7, and on our left side 10 is not divisible by 7 so the expression inside the parenthesis must be. But isn’t that expression a - 2b. Neat, we now have the proof for the theorem and can conclude that indeed:

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We have shown that the procedure above will hold for all cases.

Divisibility by 8

The theorem is quite similar to the theorem for divisibility by 4. The theorem goes that if the last three digits of a number is divisible by 8, the whole number is divisible by 8, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. Now let’s factor out 8 from the first term, like so:

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We can now see that the first term is divisible by 8, so the whole number is divisible by 8 if, and only if, the second term is divisible by 8. And so the theorem is proven.

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Divisibility by 9

Much like the theorem for divisibility by 3, the theorem goes that if the sum of all digits in a number is divisible by 9, the whole number is divisible by 9, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. Now let’s break out one a, b and c from the first 3 terms, and factor out 9 like so:

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We can now see that the first term is divisible by 9, and the second term is divisible by 9 if, and only if, the sum of a + b + c + d is divisible by 9. And so the theorem is proven.

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We have shown that the procedure above will hold for all cases. The procedure is also recursive. If the sum is to hard to test divisibility for, the procedure can be repeated, until a smaller sum is reveald. This is rarely necessary as the divisibility of the sum often is easy to determined.

Divisibility by 10

The theorem goes that if the last digits of a number is divisible by 10, the whole number is divisible by 10, i.e.

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Why is this proposition true?

Let’s use the four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. Now let’s factor out 10 from the first three terms, like so:

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We can now see that the first term is divisible by 10, so the whole number is divisible by 10 if, and only if, the second term is divisible by 10. And so the theorem is proven. As the only one digit number that is divisible by 10 is 0, another way of putting it is – if last digit is 0, the number is divisible by 10.

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Divisibility by 11

The theorem goes that a number is divisible by 11 if, and only if, the alternate sum of its digits is divisible by 11, like so:

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Neat! So how and why does it work? For simplicity’s sake we use a four digit number abcd, where a represents the number of thousands, b the number of hundreds, c the number of tens and d the number of ones. This number can be represented as 1000a + 100b + 10c + d. This expression can also be represented in another way by manipulating the terms. We give and take in an alternating fashion, like so:

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Now we factorize the expression, like so:

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We can see that the first term in the expression is divisible by 11. This means that if, and only if, the sum of the other terms is divisible by 11 the whole expression is divisible by 11, and so the theorem is proven.

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We have shown that the procedure above will hold for all cases, as the number can be extended with infinite digits and still follow the same pattern.

Fractions

The fraction flip when dividing

Many of you have probably been taught the trick of flipping the right fraction in a division to instead use simpler multiplication. This is how it works:

We start of with the division:

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From there we can reconstruct the two fractions as the dividend over the divisor with a horizontal line, for simplicity’s sake, like so:

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After that the "trick" can begin. First we multiply both the dividend and the divisor with the inverse of the divisor. As long as we treat the dividend and the divisor the same way, this is fine. Be aware of PEMDAS though! If any of the dividend or the divisor would have been an addition or subtraction you would have to multiply both terms by x, either by x(a + b) or xa + xb.

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As you can see the right fraction has now "flipped", and not by magic, but with logic and reason. As division by 1 is equal to the dividend, we can then solve the expression, like so:

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As the final cherry on top, we can prove the procedure by dividing 1 by 2 as we know this should result in one half.

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