Cindy Smith asks:

In reading a popular math book, I came across several arithmetic factoring tricks. Essentially, if the last digit of a number is zero, then the entire number is divisible by 10. If the last number is even, then the entire number is divisible by 2. If the last two digits are divisible by 4, then the whole number is. If the last three digits divide by 8, then the whole number does. If the last four digits divide by 16, then the whole number does, etc. If the last digit is 5, then the whole number divides by 5.

Now for the tricky ones.

If you add the digits in a number and the sum is divisible by 3, then the whole number is.

Similarly, if you add the digits in a number and the sum is divisible by 9, then the whole number is. For example, take the number 1233: 1 + 2 + 3 + 3 = 9. Therefore, the whole number is divisible by 9 and the quotient is 137. The number is also divisible by 3 and the quotient is 411. It works for extremely large numbers too (I checked on my calculator).

Now here's a really tricky trick. You add up the alternate digits of a number and then add up the other set of alternate digits. If the sums of the alternate digits equal each other then the whole number is divisible by 11. Also, if the difference of the alternate digits is 11 or a multiple of 11, then the whole number is divisible by 11. For example, 123,456,322. 1 + 3 + 5 + 3 + 2 = 14 and 2 + 4 + 6 + 2 = 14. Therefore, the whole number is divisible by 11 and the quotient is 11,223,302.

Also, if a number is divisible by both 3 and 2, then the whole number is divisible by 6.

The only single digit number for which there is no trick listed is 7.

I find these rules interesting and useful, especially when factoring large numbers in algebraic expressions. However, I'm not sure why all these rules work. Can you explain to me why these math tricks work? If I could understand why they work, I think it would improve my math skills. Thanks in advance for your help.

Cindy Smith
 

Dr. Rob answers:

Thank you for this long and very well-written question. I will try to write as clearly as you as I answer.

All these digital tests for divisibility are based on the fact that our system of numerals is written using the base of 10.

The digits in a string of digits making up a numeral are actually the coefficients of a polynomial with 10 substituted for the variable. For example,

    1233 = 1*10^3 + 2*10^2 + 3*10^1 + 3*10^0

which is gotten from the polynomial

    1*x^3 + 2*x^2 + 3*x^1 + 3*x^0 = x^3 + 2*x^2 + 3*x + 3

by substituting 10 for x. We can explain each of these tricks in terms of that fact.

Often the problem of divisibility can be broken up into two or more simpler problems. Whenever a positive integer can be written as a product of two numbers which have no common factor bigger than 1, testing divisibility by it can always be reduced to testing divisibility by the factors. Ultimately, this means that the only tests we need to deal with are powers of prime numbers.

  Numbers divisible by 10

Numerals ending in 0 represent numbers divisible by 10:

Since the last digit is zero, and all other terms in the polynomial form are divisible by 10, the number is divisible by 10. Similarly, if the number is divisible by 10, since all the terms except the last one are automatically divisible by 10 no matter what the coefficients or digits are, the number will be divisible by 10 only if the last digit is. Since all the digits are smaller than 10, the last digit has to be 0 to be a multiple of 10.

Actually, we can always test for divisibility by 10 by testing for divisibility by 2 and 5, since 10 = 2*5, and since 2 and 5 have no common factor bigger than 1.

  Numbers divisible by 2

Numerals ending in an even digit represent numbers divisible by 2

Same argument as above about all the terms except the last one being divisible by 2. The last digit is divisible by 2 (even) if and only if the whole number is.

  Numbers divisible by 4

Numerals ending with a two-digit multiple of 4 represent numbers divisible by 4:

Similar to the above, but since 10 is not a multiple of 4, but 10^2 is, we have to look at the last two digits instead of just the last digit. Notice that since 4 divides 20 = 2*10, we can subtract any multiple of 20 from the last two digits (or, in other words, we can subtract any multiple of 2 from the 10's digit) and keep the divisibility property.

  Numbers divisible by 8

Numerals ending with a three-digit multiple of 8 represent numbers divisible by 8:

Similar to 4, but now 10^2 is not a multiple of 8, but 10^3 is, so we have to look at the last three digits. Notice that since 8 divides both 40 = 4*10 and 200 = 2*100, we can subtract any multiple of 4 from the 10's digit and any multiple of 2 from the 100's digit and keep the divisibility property.

  Numbers divisible by 2^k

Numerals ending with a k-digit multiple of 2^k represent numbers divisible by 2^k:

Similar to 4 and 8, but now 10^(k-1) is not a multiple of 2^k, but 10^k is, so we have to look at the last k digits. As before, we can subtract any multiple of 2 from the 10^(k-1) digit, any multiple of 4 from the 10^(k-2) digit, and any multiple of 8 from the 10^(k-3) digit.

  Numbers divisible by 5

Numerals ending with 0 or 5 represent numbers divisible by 5:

Same argument as for 10 about all the terms except the last one being divisible by 5, since 10 = 5*2. The last digit is divisible by 5 if and only if the whole number is.

  Numbers divisible by 25

Numerals ending with 00, 25, 50, or 75 represent numbers divisible by 25:

Same argument as for 4 about all the terms except the last two being divisible by 25, since 10^2 = 25*4. The last two digits are divisible by 25 if and only if the whole number is. The only four combinations of last digits which are divisible by 25 are those mentioned above: 00, 25, 50, and 75.