# Math Tips for Exams

I created and developed this article with the intention of giving others access to these fundamental tricks necessary for improving mental math speed and arithmetic skills.
Of course, if you pay for math homework and do not understand anything in math, my advice will help you prepare for the exam faster.
This Math Tips and Tricks can be helpful for all kinds of people. If you are a cashier and would like to become much faster at your job, our tricks can help you. If you are studying for a standardized exam and are struggling to get through the math section on time, these ideas and tips can help you improve your time. These tricks have been helping me since I discovered several of them in my 8th grade math class. This helped me do an outstanding job in all my math courses, the math sections on SAT and ACT, and for everyday use since I began discovering these tricks.

## Exam Tip # 1

I have seen this type of problem come up dozens of times before from math classes to the SAT’s. The problem gives you a range of numbers. It will provide you with at least 2 relatively small numbers and will ask you to figure out how many integers in that range are divisible by the provided numbers. The majority of people pull out their calculators to count or take a sheet and list each number. I will show you a much faster way to solve this problem.

### Example 1:

How many numbers greater than 1 and less than or equal to 1,000 are divisible by 4 and 9.
Step 1: Find out how many numbers between 1 and 1,000 are divisible by 4:
Considering 1,000 is included, we can just divide 1000 by 4 to get the solution.
1000 / 4 = 250
Step 2: Find out how many numbers between 1 and 1,000 are divisible by 9:
1000 / 9 = 111
Remember to always round down. We can’t round up because the next number 112 is greater then 1,000 when multiplied by 9 so it doesn’t fit.
Step 3: Multiply 9 and 4: 9 * 4 = 36
Step 4: Find out how many numbers between 1 and 1,000 are divisible by 36:
1000 / 36 = 27
Step 5: Add the solutions to steps 1 and 2, then subtract the solution from step 4.
250 + 111 – 27 = 334
Review: The correct answer is 334. For these types of problems, it may be necessary to use calculators when needed to deal with large numbers especially for exams. To explain the steps, we found out how many different numbers between 1 and 1000 are divisible by 4 and 9, but what we need to take into account for, which is step 4, is that every number divisible by 36 between 1 and 1000 is counted twice because it is both divisible by 4 and 9. So therefore we subtract 361, the original total, with 27 and we get 334 as our final answer.

### Example 2:

How many numbers greater than 1 and less than 8,000 are divisible by 3, 4, and 5.
This is as difficult as this problem can get basically. Once you understand how to do this, all these types of problems will be very easy.
Step 1: We must consider that we are really looking from 2 to 7,999 because we are not including the 1 or the 8,000. Our first step is to find out how many numbers are divisible by 3:
7,999 / 3 = 2,666
Step 2: Find out how many numbers are divisible by 4:
7,999 / 4 = 1,999
Step 3: Find out how many numbers are divisible by 5:
7,999 / 5 = 1,599
Step 4: For this problem, we are dealing with 3 numbers so there are more duplicates to consider. Every number divisible by ( 3 and 4 ) are counted twice, as well as numbers divisible by ( 3 and 5 ), ( 4 and 5 ), and ( 3, 4 and 5). So we must subtract all these duplicates.
3 * 4 = 12 ——————— 7,999 / 12 = 666
3 * 5 = 15 ——————— 7,999 / 15 = 533
4 * 5 = 20 ——————— 7,999 / 20 = 399
3 * 4 * 5 = 60 —————– 7,999 / 60 = 133
Step 5: Combine steps 1, 2 and 3, then subtract the sums from step 4:
2,666 + 1,999 + 1,599 = 6,264
6,264 – 666 – 533 – 399 – 133 = 4,533
Review: Hopefully this makes sense and is useful to everyone. I used to do this the long way by writing down each term and then adding up the number of terms until I discovered this method. Know this for the SAT, ACT, AMC, AIME or other exams in case you are asked.

## Math Tip # 2

The least common multiple is the smallest number that any two or more numbers are divisible by. For example, if you are given 8 and 6, the least common multiple is 24. To find the least common multiple of any two numbers, we simply factor the two numbers into prime numbers, then we combine the factors together and multiply them to come up with the solution. I’ll provide detailed explanations for each example.

### Example 1

Find the least common multiple of 48 and 36
Step 1: Factor the two numbers until you are left with only prime numbers.
48 = 24 * 2 = 12 * 2 * 2 = 6 * 2 * 2 * 2 = 3 * 2 * 2 * 2 * 2
36 = 18 * 2 = 9 * 2 * 2 = 3 * 3 * 2 * 2
Step 2: Rewrite using powers
48 = 3 * 2 ^ 4
36 = 3 ^2 * 2 ^2
Step 3: For both numbers provided, look at the factors and see if the two numbers have the same factor. If they do, take the one with the higher power. For this example, 48 and 36 both have 2′s as factors, but 48 has 2 ^ 4 whereas 36 has 2 ^ 2, so we take the 2 ^ 4 and multiply it by the other factors to find the least common multiple. Every unique factor must be used to find the least common multiple.
2 ^ 4 * 3 ^ 2 = 144
Therefore 144 is the least common multiple of 48 and 36

### Example 2

Find the least common multiple of 7, 210, and 25
Step 1: Factor the three numbers until you are left with only prime numbers.
7 = 7 * 1
25 = 5 * 5
210 = 70 * 3 = 7 * 10 * 3 = 7 * 2 * 5 * 3
Step 2: Rewrite using powers:
25 = 5 ^ 2
7 = 7 * 1
210 = 2 * 3 * 5 * 7
Step 3: We must take each unique factor from every number as well as the larger power for similar factors. For this problem, the unique factors are 2 and 3 since they only exist as factors for 210. The repeat factors are 7 and 5. We will take the 5 ^ 2 factor from 25 for the least common multiple because it is larger than the 5 factor for 210.
Therefore 7 * 3 * 2 * 5 ^ 2 = 1050

### Example 3

Find the least common multiple of 14, 49, 32 and 80
Step 1: Break down each term:
14 = 2 * 7
49 = 7 * 7 = 7 ^ 2
32 = 2 * 16 = 2 * 2 * 8 = 2 * 2 * 2 * 4 = 2 * 2 * 2 * 2 * 2 = 2 ^ 5
80 = 2 * 40 = 2 * 2 * 20 = 2 * 2 * 2 * 10 = 2 * 2 * 2 * 2 * 5 = 5 * 2 ^ 4
Step 2: Find each unique factor, then take the largest power for each factor:
Our factors are 2, 5 and 7. The largest factor of 2 is 2 ^5. The largest factor of 5 is just 5. The largest factor of 7 is 7 ^ 2
Step 3: Multiply the largest factors:
2 ^ 5 * 5 * 7 ^ 2 = 32 * 5 * 49 = 7,840
This means the least common multiple of 14, 49, 32 and 80 is 7,840
Hopefully this made sense. If not let me know by leaving a comment and i’ll try to explain it some more the best, I can.