Your performance on a math test, whether it is a quiz or the SAT, depends on three things. The first two things are obvious, you must know the material well and you must be well-rested.  The third thing is less obvious and often seemingly impossible to control - time management. Here are some tricks to help speed your progress during a test. After each trick, I will provide a sample problem using the trick.

In the age of the calculator, why do we care about divisibility? Because they can turn problems into "sight" problems or at least decrease the number of steps. 

For example: Solve for x, 3(4x - 9) = 6x + 51

I use a divisibility rule and note that 6 and 51 can be divided evenly by 3.

This gets rid of the need for distributing and decreases the number of steps.

                         4x - 9 = 2x + 17

                               2x = 26

                                x  = 13

Divisibility Rules are about sums. For example, 635 is divisible by 5 because 5 goes into 10 evenly, therefore 5 goes into 630. As 635 = 630 + 5 and 5 goes into each part of the sum, 5 goes into the sum. Of course, most people just remember that if a number ends with a 5 or a 0, it is divisible by 5. Similarly, if a number ends with an even number, the number is divisible by 2. For brevity's sake, I will sometimes use the symbol "|" to denote "divides evenly"

Here are the common divisibility rules:

2|n when the last digit of n is an even number

3|n when the digits of n sum up to a number which is divisible by 3. Example: Let's use the Battle of Hastings 1066, that's a nice random number. 

1066 = 1(1000) + 0(100) + 6(10) + 6

         = 1(999 + 1) + 0(99 + 1) + 6(9 + 1) + 6 

         = 1(999) + 0(99) + 6(9) + 1 + 0 + 6 + 6

 3 divides the sum 1(999) + 0(99) + 6(9), so we need to check the sum 1 + 0 + 6 + 6 = 13. 3 does not divide 13 evenly, 3 does not divide 1066.

4|n Since 4|100, we merely need to check the last two digits. Another random number, the date congress approved the Declaration of Independence, 7/4/1776.

7041776, check 76. 4x19 = 76, therefore 4 divides 7041776.

5|n when the numbers ends with 5 or a 0.

6|n when the number is even (2 divides it) and 3 divides it. 7041776 is even, but 7+0+4+1+7+7+6= 32 which is not divisible by 3. Therefore, 6 does not divide 7041776

7... there is no super-easy way to check 7, but if you cannot use a calculator there is a relatively quick way, abbreviated division.

  7041776

-7000000

     41776

   -35000

      6776

    - 6300

        476

      -420

         56    

What do you know! 7 does divide 7041776. BTW, it took me much longer to type and format this process than it took to do it.

8|n when 8 goes into the last 3 digits. But who is going to memorize all the multiples of 8 up to a 1000? There is a trick... If 8 divides the hundreds digit, the 8 must divide the last two digits. If 8 does not divide the hundreds digit, then 4 must divide the number (4 divides the last 2 digits) and 8 must NOT divide the last two digits.

Does 8 divide 7041776. We know that 8 does divide 7041000, so we only need to check 776. 8 does not divide 7, 8 does not divide 76, and 4 does divide 76. 8 divides 7041776.

9|n when 9 divides the sum of the digits. It's the same process as for 3's. For example, the year of the Great Depression, 1929. 1 + 9 + 2 + 9 = 21. 9 does not divide 21, but 3 does! 3|1929, 9 does not divide 1929.

10|n when the n ends in 0.

11|n when difference between consecutive digits is divisible by 11. This works because 10 = 11 -1 and 100=99+1. Therefore we can express even powers of ten as the sum of multiples of 99 and the digit. We can express odd powers of ten as the difference of multiples of eleven and the digit. A lot of words... let's do an example, say the year that gave us Woodstock.

1969 = 1(1000) + 9(100) + 6(10) + 9 

         = 1(1001 - 1) + 9(99 + 1) + 6(11 - 1) + 9

         = 1(1001) + 9(99) + 6(11) -1 + 9  - 6 + 9

11 divides the sum 1(1001) + 9(99) + 6(11), now we check -1 + 9 - 6 + 9 = 11. 11 divides 1969

Do we have to pay attention to the number of digits? No. We'll get a negative, but 11 divides the negative so it still works.

1 - 9 + 6 - 9 = -11.

Sample Problem - In searching for the cause of the Great Famine of in 1809 in Freedonia, biologist have discovered a new species of locust with a synchronized life cycle of 9 years. The combined events of an unusual drought and the mating phase of the locust destroyed the grain crop that year. Which of the following years will NOT have a recurrence of the mating phase of the locust?

a) 1836

b) 1971

c) 1999

d) 2016

Answer - By sight, we determine that 9 divides 1809. Therefore, every year in which the mating phase occurs will be divisible by 9. 1999 is not divisible 9, so during that year there will not be a locust incursion.

Divisibility tests are great to check one number, but what if you want to know all the prime numbers that divide a specific number? Even if you use a calculator, there could be a lot of random checks. Also, how many numbers do you check?

Let's look at 1776. We quickly note that 8 is a factor. 8 x 222 = 1776. Now it's easier, 2 x 111 = 222. 3 divides 111, 3 x 111 = 37. 24 x 3 x 37 = 1776. The prime factors of 1776 are 2, 3, 37. Note that we have only found the primes. Finding ALL the factors is more complicated and is left to a future blog. For those who desire a challenge, I will tell you that there are 20 factors of 1776.

By using previously determined factors we were able to whittle down the problem into manageable size. However, that won't always be the case.

What about 1777? It fails all divisibility tests listed. Okay, check 13, 17, and 19 with a calculator. Nope. Check 23, 29, and 31 with a calculator. Nope. How many numbers do we check. Do we have 1746 left to check? Or, do we have all the primes between 31 and 1777 left to go. That is still a lot of numbers. No worries, there is a trick which sets a limit on the how many primes that you have to check. Find the square root of the number and round to the nearest integer. Any number that is greater than the square root must have a factor that is less than the square root.

sqrt(1777) = 42.1544... Therefore we check all the primes up to 42. Already checked the primes up to 31, now we check 37 and 41. Nope! 1777 is prime.

sample problem - What is the sum of the prime factors of 253? sqrt(253) = 15.9... So we only have to check primes up to 13. Visual inspection tells us that 2, 3 and 5 are not factors, but 11 is a factor! 

11x23 = 253. Remember the problem; we need the sum. 11 + 23 = 34.

Squares are great for estimating. If you followed my advice from a previous blog you will have memorized the squares up to 20. How does that help? Squares give a range of numbers that an answer will fall between. For example, 14 x 16 will be between 196 & 256. Say 225? The real answer? 224. Not bad.

Sample Problem - Which of the following is equivalent to (4x - 13)(7x + 12)?

A) 28x2 - 43x + 84

B) 28x2 - 43x  - 84

C) 28x2 - 43x - 156

D) 11x2 - 43x - 156

This is a "sight" problem; you can solve it by visual inspection. First, all the answer choices have - 43x so we know to focus on the x2 term and the constant. 4 x 7 = 28, eliminate D. 122 = 144, therefore 12x13 must be larger than 144. Eliminate A & B. This leaves C as the only viable choice, without using a calculator or even a pencil!

Multiplying or dividing by 5: When you multiply by 5, divide by 2 and move the decimal to the right. 5 x 36 = 18 x 10 = 180. 5 x 13 = 6.5 x 10 = 65. When you divide by 5, multiply by 2 and move the decimal to the left. 56/5 = 112/10 = 11.2

Sample Problem - What is the sum of the interior angles of a heptagon? Using the formula 180(n-2), 180(7-2) = 180(5) = 90x10 = 900.

Estimating Square Roots. Find the perfect square less than the number and the perfect square larger that the number. Average the square roots of the perfect squares. I.E. add .5 to the lower square. If the number is closer to the smaller perfect square then add less that .5. If your number is closer to the larger perfect square then add more than .5. There is a more accurate way to approximate square roots, but it is cumbersome. However, this way is quick and will work for most multiple choice problems.

Sample Problem - One diagonal of a rhombus is 12 centimeters and the other is 32 centimeters. How many centimeters is a side of the rhombus? Round the the nearest hundredths. Hint: All the sides of a rhombus are congruent and the diagonals are perpendicular.

A) 8.54

B) 17.09

C) 17.72

D) 34.17

E) 35.78

At first glance, this problem looks like you absolutely need a calculator. Not so. Not if you remember your squares. Here goes... Since the diagonals of the rhombus are perpendicular, they form four congruent right triangles with sides 6 and 16. Therefore the sides are the hypotenuses of the triangles. Using the Pythagorean Theorem, s2 = 62 + 162 = 36 + 256 = 292. 289<292<324 means that 172<s2<182 which means that 17<s<18. As 292 is much closer to 289, choose the option less than 17.5, B) 17.09.

Magic 9's: Multiply by 9, multiply by 10 and subtract the number. 9x35 = 350-35 = 315.

Decimals with 9's. 1/9 = .1111.... 2/9=.2222... 3/9=.3333.... you get the idea. 

Sample Problem - Fred skis 28 meters in 9 seconds. How fast does he ski? 28/9 = (27+1)/9 = 3 + .1111... = 3.1111... He skis about 3.1 foot/sec.

These tricks will not take the place of knowledge of content nor will they help if you are not well-rested. However, they will free time up to use on how to solve problems as opposed to squandering it on the arithmetic. That will give you an edge.

Have fun on the road to numeracy!

Lorelei S.