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]]>**Prime numbers**

2 is the smallest prime number

2 is the only prime even number

1 is not a prime

3, 5, 7 are the only three consecutive odd integers that are each a prime.

**Digits**

In the integer 5,432, 5 is the thousands digit, 4 is the hundreds digit, 3 is the tens digit, and 2 is the units digit.

In a multiple-digit number, there are 10 possible units digits (0 – 9). In the first digit, i.e. the digit farthest to the left, there are only nine possible digits. A zero as the first digit would not be valid. For instance, 0,453 is not a four-digit number. It’s – if anything – the three-digit number 453.

**Divisibility Rules**

3: Add up the digits in any number. If they are multiple of 3, then the number is divisible by 3. For example, 258 is divisible by 3 because 2 + 5 + 8 is 15. And 15 is a multiple of 3.

4: If the tens and units digits of integer X is a multiple of 4, then integer X is divisible by 4. For example, 364 is divisible by 4, because 64 is a multiple of 4.

5: If a number ends in 5 or 0, it is divisible by 5.

6: If the rules to divisibility to 2 and 3 are fulfilled, then a number is divisible by 6. One hundred and eleven is not divisible by 6. Two hundred and forty six, because it is divisible by both 2 and 3, is divisible by 6.

9: If the sum of the digits of a number is divisible by 9, then that number is divisible by 9. For example, adding up the digits in 4,536, we get 4+5+3+6 = 18. Because 18 is divisible by 9, the number 4,536 is also divisible by 9.

**Odds (so to speak) and Ends**

Zero is an even integer

Integers can be negative numbers (-3, -6, -17, etc.)

0! = 1

x^0 = 1

The problems below can be solved by referencing the above. Good luck!

**Practice Problems**

1. What is the smallest integer that can be divided by the product of a prime number and 7 while yielding a prime number?

(A) 7

(B) 14

(C) 24

(D) 28

(E) 35

2. The sum of five consecutive even integers is 20. What is the product of the median of the series and the smallest integer in the series?

(A) 12

(B) 10

(C) 6

(D) 4

(E) 0

3. Which of the following integers is NOT divisible by 3?

(A) 624

(B) 711

(C) 819

(D) 901

(E) 969

Answers:

1. D

2. E

3. D

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]]>Oftentimes, math problems will require that you know more than one concept. Sometimes, you’ll have to know the difference between an integer and a number in order to get the question right. Below, I’ve combined prime numbers with probability, two subjects already covered in this blog.

The question below isn’t easy and actually takes a little bit of work. Let’s see if you can crack it.

*What is the probability that the sum of two rolled dice will equal a prime number?*

*(A) **1/3*

*(B) **5/36*

*(C) **2/9*

*(D) **13/36*

*(E) **5/12*

First let’s list the prime numbers that pertain to this problem.

The pertinent prime numbers are 2, 3, 5, 7, and 11. Notice I stopped at 11. Why? Well, the greatest number you can roll on two six-sided dice is 12.

Next, we have to remember this is a probability question. Therefore, we have to divide the number of total outcomes by the number of desired outcomes. First, let’s find the number of desired outcomes. To do this, we have to make sure that each desired outcome conforms to the problem, i.e. how many different ways can we sum two dice to get a prime.

Let’s start with 2. There is only one way to roll a 2, and that is with a 1 and a 1. Therefore, we have one desired outcome.

What about the number 3? Well, we can get 2,1 and 1,2, or two possible ways.

Mind you, to do this problem you will have to make sure you write down each outcome. Do not try to do this in your head, for you’ll most likely get the problem wrong and induce dizziness.

Next we have the number 5, which we can get by rolling the following combinations:

1,4

4,1

2,3

3,2

We add these four possible outcomes to the prior three, giving us a total of 7.

Next, we look to see which numbers sum to 7 and find a total of six possibilities.

1,6

2,5

3,4

4,3

5,2

6,1

Our total is now 13.

Finally, don’t forget 11 as final prime number.

6,5

5,6

We add these two possibilities to the 13 possibilities, giving us a total of 15.

Finally, we want to find the total number of outcomes. Remember, we divided the total outcomes by the total number of desired outcomes, which we just found out was 15.

The total number of outcomes equals the number of different ways we can roll two six-sided dice. We find this number by multiplying 6 x 6. The logic is there are six sides to each die, so for each number on one die you can pair with six different numbers on the other die.

Therefore, the probability of rolling a prime number on two dice is 15/36, which reduces to 5/12 (E).

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]]>Here is a good problem to test your understanding of prime factors, not to add multiples. See if you can finish the problem in less than 2 minutes.

If P is the product of all of the positive multiples of 11 less than 100, then what is the sum of the distinct prime factors of P?

(A) 22

(B) 28

(C) 45

(D) 49

(E) 89

Approach: The first way to crack this problem is to figure out what the multiples of 11 are. Start with the lowest and begin writing them on your scratch paper. Doing so will help you map out the problem.

11, 22, 33… by the time you get to 33, you might want to stop and think what the range of the multiples is. Meaning, what is the highest multiple of 11 that is still less than 100. The answer is 99 = 11 X 9. Therefore there are 9 multiples of 11, starting with 11 and ending with 99.

The crux of the problem—at least in terms of saving time—is how to factor out the primes of P. Note that 11 x 22 x 33…x 99 is going to give you a really large number. How do we make it easy to look for the prime factors?

One way is to see which of the primes (2, 3, 5, 7, 11, 13 etc.) we can extract from 11 x 22 x 33…x 99. A quicker way is to factor out the 11 so that the multiples of 11, up until 99, can then be written as P = 11^{9} (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9).

Note that the problem asks us for the distinct primes, so we only need to pull one of each prime from P. The first prime to notice is 11. There were nine of these in P but because we only need one of each distinct prime, we don’t have to worry about the exponent.

P = 11 (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9).

Now we just need to add up 11 + 2 + 3 + 5 + 7 and we get 28. Answer (B).

These types of problems are time-consuming. Yet, there are always ways to make a problem less laborious. In the problem above, we looked for a pattern once we started writing out the numbers. Remember, if you are writing out all of the terms of the sequence that will cost you too much time. On other hand if you do not write out anything at all then you will not move forward on the problem. At the end of the day, the long way is better than the no way (unless of course you are running out of time.)

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