Integers
Integers are whole numbers – that is, they’re not fractions – but they can be positive, negative, or zero. Integer problems often involve prime numbers, factorials, and sequences. Here are a few helpful facts about integers:
- Prime numbers are positive integers other than 1 which are divisible only by 1 and themselves. Examples: 2, 3, 5, 7, 11, 13, etc.
- Composite numbers are positive numbers other than 1 which are not prime. Examples: 4, 6, 9, 10, 12, 14, 15, etc.
- Every integer has a unique factorization into powers of primes. For example: 9 = 32, 12 = 22∙3, 45 = 32∙5, 30 = 2∙3∙5.
- A factorial (shown with an exclamation point !) is the product of all positive integers less than or equal to a given integer. For example: 10! = 10∙9∙8∙7∙6∙5∙4∙3∙2∙1
- The sum of the first n positive integers is n(n+1)/2.
- Any even number is divisible by 2. Any number ending in 0 is divisible by 2 and 5 (and therefore by 10). Any number whose digits add up to a multiple of 3 is divisible by 3. Any number whose digits add up to a multiple of 9 is divisible by 9. An even number whose digits add up to a multiple of 3 is divisible by 6.
On this page we provide examples of integer problems of the following types:
- 1 General Integer problem
- 1 Factorial problem
- 1 Sum of a Series of Integers problem
- 2 Representations of Integers problems
- 1 Exponents of Integers problem
To practice more of these types of problems, click here.
Example 1)
m is an integer greater than 150 but less than 300. When m is divided by 6, the result is the square of a prime number. What is m?
A. 289
B. 225
C. 294
D. 256
E. 186
Explanation: This problem isn’t as hard as it perhaps looks. First, we divide 150 and 300 by 6, and we get the range as 25 to 50. Now we have to find a perfect square that’s between 25 and 50. Well, not exactly. It says that m is greater than 150 and less than 300. So, the range is between 25 and 50, but it’s not inclusive of 25 or 50. Moreover, we’re looking for the square of a prime number. Let’s look at the first 4 prime numbers:
\(\begin{align}2^2 = 4\end{align}\)
\(\begin{align}3^2 = 9\end{align}\)
\(\begin{align}5^2 = 25\end{align}\)
\(\begin{align}7^2 = 49\end{align}\)
Obviously, the only prime number whose square is in the proper range is 7. Its square is 49, and m = 6∙49 = 294.
So C is the correct answer.
To practice more problems like this, click here.
Factorials Example 1)
50! ends with how many zeroes?
A. 22
B. 15
C. 10
D. 12
E. 9
Explanation: Let’s illustrate how to solve this problem by looking again at 10!
10! = 10∙9∙8∙7∙6∙5∙4∙3∙2∙1
A zero is added to the end of a number whenever it’s multiplied by 10, or in other words by 5 and 2. As you can see, every second number in the factorial is even, so there are plenty of 2’s. On the other hand, a number divisible by 5 comes up only once in 5 numbers. In 10! there are two numbers – 5 and 10 – that are divisible by 5. That means that 10! will have two zeroes at the end. So to calculate the number of zeroes at the end of 50!, we need to know how many numbers divisible by 5 are in the 50 numbers. Obviously, there are 10 of them. However, 2 of these – 25 and 50 – are divisible by 5 twice. That is,
\(\begin{align}25 = 5^2\end{align}\) and \(\begin{align}50 = 2 \cdot 5^2\end{align}\)
So for each of these two numbers there is an extra 0 at the end of 50! Altogether then, there are 12 zeroes at the end of 50!
So D is the correct answer.
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Sums of a Series of Integers Example 1)
The Smith family eats a lot of ice cream, but decides to save money by reducing this expenditure. The first month they spend $5 less than usual, the second month $10 less, the third month $15 less, and so forth. After a year, how much money have they saved?
A. $455
B. $325
C. $400
D. $390
E. $415
Explanation: This is the sum of multiples of 5 from 5 to 60. It is easily calculated by determining the sum of 1 to 12 and multiplying it by 5. Remember, the formula for the sum of the integers from 1 to n is:
n(n+1)/2
so
12×13/2 = 78
78 x 5 = 390
Therefore D is the correct answer.
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Representations of Integers Example 1)
A 24-hour clock’s hour hand is pointing to 17:00. Where will it be pointing 83 hours from now?
A. 3:00
B. 16:00
C. 12:00
D. 6:00
E. 4:00
Explanation: On a 24 hour clock, the passage of time can be represented by multiplying 24 by some number and adding a remainder. So 83 hours can be represented as:
83 = 3∙24 + 11
If we now add the current time, 17 hours, to that representation, we get
100 = 3∙24 + 28
However, the remainder must be less than 24, so we adjust as follows:
100 = 4∙24 + 4
And the remainder of 4 will be the hour to which the hour hand will be pointing. In other words, 4:00.
So E is the correct answer.
Representations of Integers Example 2)
In 38 college courses each uses 1 and only 1 textbook from a group of 12. Each of the 12 texts is used in at least 2 but no more than 4 of the courses. If n is the number of texts used in exactly 2 courses, what is the possible maximum and minimum range of n?
A. 0 ≤ n ≤ 12
B. 0 ≤ n ≤ 5
C. 2 ≤ n ≤ 7
D. 1 ≤ n ≤ 5
E. 0 ≤ n ≤ 6
Explanation: Let k be the number of texts used in 4 courses, m the number used in exactly 3, and n the number used in exactly 2 courses. Then, since each course only uses one text, we have 2 equations in 3 unknowns:
k + m + n = 12 Because there are 12 texts
4k + 3m + 2n = 38 Because there are 38 courses
We need to find the maximum and minimum range of n. Let’s start with the minimum range. Suppose n = 0. Then
k + m = 12 and 4k + 3m = 38
This gives us two equations with two unknowns.
m = 12 – k, so
4k + 3(12 – k) = 4k + 36 – 3k = k + 36 = 38, so
k = 2 and m = 10
That is, if k = 2 and m = 10, then there are 2 texts that are used in 4 courses each, and 10 texts used in 3 courses each. That makes 38 courses in all, so it’s possible for n to equal 0. That means that answers C and D are invalid, because they show a minimum value for n that is too large. So the only possible answers are A, B, and E.
The maximum value for n is more challenging. The 3 possibilities for the maximum given in answers A, B, and E, are 12, 6, and 5. We’ll start with the largest, and work our way down as needed. So let’s try 12. If n = 12, then all 12 texts are used in 2 courses each, and both k and m are zero. But
2·12 = 24
so we’re left with 14 courses of the 38 that have no texts. So n cannot be 12.
Now suppose n = 6. Then
k + m + 6 = 12, so k + m = 6 and
4k + 3m + 2·6 = 4k + 3m + 12 = 38, so 4k + 3m = 26
m = 6 – k so 4k + 3(6 – k) = 4k + 18 – 3k = k + 18 = 26, so k = 8 and m = -2.
Oops. m can’t be -2, so this solution doesn’t work, and n is not 6.
Now suppose n = 5. Then
k + m + 5 = 12, so k + m = 7 and
4k + 3m + 2·5 = 4k + 3m + 10 = 38, so 4k + 3m = 28
m = 7 – k so 4k + 3(7 – k) = 4k + 21 – 3k = k + 21 = 28, so k = 7 and m = 0.
This solution works because for k = 7, m = 0, and n = 5 we get
k + m + n = 12 and 4k + 3m + 2n = 4·7 + 3·0 + 2·5 = 28 + 10 = 38
Answer B is the correct answer.
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Exponents of Integers Example 1)
What is the units digit of \(\begin{align}24^{19} + 34^{20}\end{align}\)?
A. 4
B. 0
C. 9
D. 2
E. 8
Explanation: The units digit in a number is its rightmost digit. Therefore, we don’t have to pay attention to what happens anywhere else in the number, we just have to know what happens with the first digit. That is, we only care what units digit results when 4 is multiplied by itself many times. Notice then that
4 ∙ 4 = 16 and 4 ∙ 4 ∙ 4 = 64
So if 4 is a factor 2 times, then the units digit is 6, and if it’s a factor 3 times the units digit is 4 again. From this it’s easy to see that if a number ending in 4 is raised to an even power, then its units digit will be 6, and if it’s raised to an odd power, then its units digit is 4. Since 19 is odd and 20 is even, then the units digit of these two numbers is 4 and 6, respectively. Their sum, therefore, has a units digit of 0.
So answer B is correct.
For more problems like this to practice, click here.