Cindy Smith asks:
From: Cindy Smith
To: Dr. Math
Subject: Factoring Tricks
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
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.
Dr. Rob answers:
From: Dr. Math
To: Cindy Smith
Subject: Re: Factoring Tricks
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
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
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:
On to 3, 7, 9, 11, 13, 17, and some higher numbers.
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.