Integers - PSAT Math
Card 1 of 1267
8 people locked in a room take turns holding hands with each person only once. How many hand holdings take place?
8 people locked in a room take turns holding hands with each person only once. How many hand holdings take place?
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The first person holds 7 hands. The second holds six by virtue of already having help the first person’s hand. This continues until through all 8 people. 7+6+5+4+3+2+1=28.
The first person holds 7 hands. The second holds six by virtue of already having help the first person’s hand. This continues until through all 8 people. 7+6+5+4+3+2+1=28.
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If a series of license plates is to be produced that all have the same pattern of three letters followed by three numbers, roughly how many alphanumeric combinations are possible?
If a series of license plates is to be produced that all have the same pattern of three letters followed by three numbers, roughly how many alphanumeric combinations are possible?
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The total number of possible combinations of a series of items is the product of the total possibility for each of the items. Thus, for the letters, there are 26 possibilities for each of the 3 slots, and for the numbers, there are 10 possibilities for each of the 3 slots. The total number of combinations is then: 26 x 26 x 26 x 10 x 10 x 10 = 17,576,000 ≈ 18 million.
The total number of possible combinations of a series of items is the product of the total possibility for each of the items. Thus, for the letters, there are 26 possibilities for each of the 3 slots, and for the numbers, there are 10 possibilities for each of the 3 slots. The total number of combinations is then: 26 x 26 x 26 x 10 x 10 x 10 = 17,576,000 ≈ 18 million.
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If x and y are integers and at least one of them is even, which of the following MUST be true?
If x and y are integers and at least one of them is even, which of the following MUST be true?
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Since we are only told that "at least" one of the numbers is even, we could have one even and one odd integer OR we could have two even integers.
Even plus odd is odd, but even plus even is even, so x + y could be either even or odd.
Even times odd is even, and even times even is even, so xy must be even.
Since we are only told that "at least" one of the numbers is even, we could have one even and one odd integer OR we could have two even integers.
Even plus odd is odd, but even plus even is even, so x + y could be either even or odd.
Even times odd is even, and even times even is even, so xy must be even.
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If xy = 100 and x and y are distinct positive integers, what is the smallest possible value of x + y?
If xy = 100 and x and y are distinct positive integers, what is the smallest possible value of x + y?
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Consider the possible values for (x, y):
(1, 100)
(2, 50)
(4, 25)
(5, 20)
Note that (10, 10) is not possible since the two variables are to be distinct. The sums of the above pairs, respectively, are:
1 + 100 = 101
2 + 50 = 52
4 + 25 = 29
5 + 20 = 25, the smallest possible value.
Consider the possible values for (x, y):
(1, 100)
(2, 50)
(4, 25)
(5, 20)
Note that (10, 10) is not possible since the two variables are to be distinct. The sums of the above pairs, respectively, are:
1 + 100 = 101
2 + 50 = 52
4 + 25 = 29
5 + 20 = 25, the smallest possible value.
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m, 3m, 5m, ...
The first term in the above sequence is m, and each subsequent term is equal to 2m + the previous term. If m is an integer, then which of the following could NOT be the sum of the first four terms in this sequence?
m, 3m, 5m, ...
The first term in the above sequence is m, and each subsequent term is equal to 2m + the previous term. If m is an integer, then which of the following could NOT be the sum of the first four terms in this sequence?
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The fourth term of this sequence will be 5m + 2m = 7m. If we add up the first four terms, we get m + 3m + 5m + 7m = 4m + 12m = 16m. Since m is an integer, the sum of the first four terms, 16m, will have a factor of 16. Looking at the answer choices, 60 is the only answer where 16 is not a factor, so that is the correct choice.
The fourth term of this sequence will be 5m + 2m = 7m. If we add up the first four terms, we get m + 3m + 5m + 7m = 4m + 12m = 16m. Since m is an integer, the sum of the first four terms, 16m, will have a factor of 16. Looking at the answer choices, 60 is the only answer where 16 is not a factor, so that is the correct choice.
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How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
How many integers in the following infinite series are positive: 100, 91, 82, 73 . . . ?
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The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.
The difference between each number in the series is 9. You can substract nine 11 times from 100 to get 1: 100 – 9x11 = 1. Counting 100, there are 12 positive numbers in the series.
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A sequence of numbers is as follows:
What is the sum of the first seven numbers in the sequence?
A sequence of numbers is as follows:
What is the sum of the first seven numbers in the sequence?
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The pattern of the sequence is (x+1) * 2.
We have the first 5 terms, so we need terms 6 and 7:
(78+1) * 2 = 158
(158+1) * 2 = 318
3 + 8 + 18 +38 + 78 + 158 + 318 = 621
The pattern of the sequence is (x+1) * 2.
We have the first 5 terms, so we need terms 6 and 7:
(78+1) * 2 = 158
(158+1) * 2 = 318
3 + 8 + 18 +38 + 78 + 158 + 318 = 621
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What is the next number in the following series: 0, 3, 8, 15, 24 . . . ?
What is the next number in the following series: 0, 3, 8, 15, 24 . . . ?
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The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.
The series is defined by n2 – 1 starting at n = 1. The sixth number in the series then equal to 62 – 1 = 35.
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If n and m are both positive even integers, which of the following must be odd?
I. (n + 1)(m + 1)
II. nm + 1
III. nm + m
If n and m are both positive even integers, which of the following must be odd?
I. (n + 1)(m + 1)
II. nm + 1
III. nm + m
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Let us analyze I, II, and III one at a time.
Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.
Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.
We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.
Only choice I and II will always produce odd numbers.
The answer is I and II only.
Let us analyze I, II, and III one at a time.
Because n and m are both even, if we increase either by 1, the result will be an odd number. Thus, n + 1 and m + 1 are both odd. When two odd numbers are multiplied together, the result is always an odd number. Thus (n + 1)(m + 1) must be an odd number.
Because n and m are even, when we multiply two even numbers together, we always get an even number. Thus nm is even. However, when we then add one to an even number, the result will be an odd number. Thus, nm + 1 is odd.
We just established that nm is even. If we subtract an even number from an even number, the result is always even. Thus, nm – m is an even number.
Only choice I and II will always produce odd numbers.
The answer is I and II only.
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odd * odd * odd =
odd * odd * odd =
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The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
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When 
is evaluated, what number is in the ones digit?
When
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Let us say we have two numbers, 
and 
whose ones digits are 
and B, respectively. If we want to know the ones digits of the product of 
and 
, all we need to do is to look at the ones digit of the product of 
and 
. For example, if we multiply 137 and 219, then the ones digit will be the same as the ones digit of 
. Since the ones digit of 63 is 3, the ones digit of 137 x 219 will also be 3. In short, we really only need to worry about the ones digits of the numbers we multiply when we try to find the ones digit of their product.
We want to find the ones digit of 
. An exponent is essentially just a short hand for repeated multiplication. Let us look at the ones digit of the first few exponents of 2013.
- the ones digit is 3.
To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as 
.
- ones digit is 9.
Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.
- ones digit of 7.
To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.
- ones digit of 1.
To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.
- ones digit of 3.
Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.
The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):
3, 9, 7, 1, 3, 9, 7, 1,....
We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.
The answer is 3.
Let us say we have two numbers,
We want to find the ones digit of
To find the ones digit of the 2013 to the second power, we need to think of it as the product of 2013 and 2013. As discussed previously, if we want the ones digit of two numbers multiplied together, we just need to multiply their ones digits. Thus, if we multiply 2013 by 2013, then the ones digit will be the same as
Next, we want to find the ones digit of 2013 to the third power. In order to do this, we will multiply the square of 2013 by 2013. It does not matter that we do not know exactly what 2013 squared equals, beacuse we only need to worry about the ones digit, which is 9. In other words, 2013 to the third power will have a ones digit that is equal to the ones digit of the product of 9 (which was the ones digit of 2013 squared) and 3 (which is the ones digit of 2013). When we multiply 9 and 3, we get 27, so the ones digit of 2013 to the third power is 7.
To find the ones digit of 2013 to the fourth power, we only need to worry about multiplying the ones digit of 2013 to the third power (which is 7) by the ones digit of 2013. When we mulitply 7 and 3, we get 21, which means that the ones digit of 2013 to the fourth power is 1.
To find the ones digit of 2013 to the fifth power, we will multiply 1 by 3, which gives us 3.
Notice that we are back to a ones digit with 3. If we multiply this by 2013, we will end up with a ones digit of 9. In other words, the ones digits repeat every fourth power.
The value of the ones digits of the powers of 2013 is as follows (starting with 2013 to the first power):
3, 9, 7, 1, 3, 9, 7, 1,....
We essentially want to find the 2013th term of the sequence above. Notice that every fourth term is 1, i.e. the sequence repeats every four terms. If a terms position in the sequence is a multiple of 4, then the term will be 1. In short, the 4th, 8th, 12th, 16th terms, and so on, will be 1. Because 2012 is a multiple of 4, the 2012th term in the sequence will be 1. (We can determine if a number is a multiple of 4 by looking at its last two digits.) This means that that 2013th term will be 3. Thus, 2013 to the power of 2013 has a ones digit of 3.
The answer is 3.
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odd * odd * odd =
odd * odd * odd =
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The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
The even/odd number properties are good to know. If you forget them, however, it's easy to check with an example.
Odd * odd = odd. If you didn't remember that, a check such as 1 * 3 = 3 will give you the same answer. So if odd * odd = odd, (odd * odd) * odd = odd * odd = odd, just as 3 * 3 * 3 = 27, which is odd. This means we are looking for an answer choice that also produces an odd number. Let's go through them.
even * even = even (2 * 2 = 4)
even * odd = even (2 * 3 = 6)
odd * odd = odd (1 * 3 = 3) This is the correct answer! But just to double check, let's go through the last two.
even * even * even = even * even = even (2 * 2 * 2 = 8)
odd * odd * even = odd * even = even (1 * 3 * 2 = 6)
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If 
and 
are both odd integers, which of the following is not necessarily odd?
If
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With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables. However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.
Therefore, our best approach is to simply analyze each answer choice.
: Since 
is odd, 
is also odd, since and odd number multiplied by an odd number yields an odd product. Since 
is also odd, multiplying it by 
will again yield an odd product, so this expression is always odd.
: Since 
is odd, multiplying it by 2 will yield an even number. Subtracting this number from 
will also give an odd result, since an odd number minus an even number gives an odd number. Therefore, this answer is also always odd.
: Since both numbers are odd, their product will also always be odd.
: Since 
is odd, multiplying it by 2 will give an even number. Since 
is odd, subtracting it from our even number will give an odd number, since an even number minus and odd number is always odd. Therefore, this answer will always be odd.
: Since both numbers are odd, there sum will be even. However, dividing an even number by another even number (2 in our case) does not always produce an even or an odd number. For example, 5 and 7 are both odd. Their sum, 12, is even. Dividing by 2 gives 6, an even number. However, 5 and 9 are also both odd. Their sum, 14, is even, but dividing by 2 gives 7, an odd number. Therefore, this expression isn't necessarily always odd or always even, and is therefore our answer.
With many questions like this, it might be easier to plug in numbers rather than dealing with theoretical variables. However, given that this question asks for the expression that is not always even or odd but only not necessarily odd, the theoretical route might be our only choice.
Therefore, our best approach is to simply analyze each answer choice.
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Theodore has 
jelly beans. Portia has three times that amount. Harvey has five times as many as she does. What is the total count of jelly beans in the whole group?
Theodore has
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To find the answer to this question, calculate the total jelly beans for each person:
Portia: 
* <Theodore's count of jelly beans>, which is 
or 
Harvey: 
* <Portia's count of jelly beans>, which is 
or 
So, the total is:
(Do not forget that you need those original 
for Theodore!)
To find the answer to this question, calculate the total jelly beans for each person:
Portia:
Harvey:
So, the total is:
(Do not forget that you need those original
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If 
is an odd integer and 
is an even integer, which of the following must true of 
?
If
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An even number subtracted from an odd number will always produce an odd result.
None of the other answer choices are correct.
An even number subtracted from an odd number will always produce an odd result.
None of the other answer choices are correct.
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If x is the greatest prime factor of 42, and y is the greatest prime factor of 55, what is the value of xy?
If x is the greatest prime factor of 42, and y is the greatest prime factor of 55, what is the value of xy?
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Find the prime factors of 42: 7, 3, 2
Find the prime factors of 55: 5, 11
Product of the greatest factors: 7 and 11 = 77
Find the prime factors of 42: 7, 3, 2
Find the prime factors of 55: 5, 11
Product of the greatest factors: 7 and 11 = 77
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If p is a prime number greater than 1, how many positive factors does p4 have?
If p is a prime number greater than 1, how many positive factors does p4 have?
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3 is a prime number that is easy to work with, so we can plug that in for p. p4 = 81. The positive factors of 81 are 1, 3, 9, 27, and 81. Thus, the answer is five factors.
Let's look at another prime number to plug in for p. If we plug in 2, another easy prime number to work woth, we get p4 = 16. The positive factors of 16 are 1, 2, 4, 8 and 16.
Notice that when p = prime number greater than 1, the positive factors for p4 are 1, p, p2, p3 and p4. Multiplying pn by p only adds pn as a factor when p is prime, so you will have n factors plus 1 which is a factor, so p4 has 5 factors.
3 is a prime number that is easy to work with, so we can plug that in for p. p4 = 81. The positive factors of 81 are 1, 3, 9, 27, and 81. Thus, the answer is five factors.
Let's look at another prime number to plug in for p. If we plug in 2, another easy prime number to work woth, we get p4 = 16. The positive factors of 16 are 1, 2, 4, 8 and 16.
Notice that when p = prime number greater than 1, the positive factors for p4 are 1, p, p2, p3 and p4. Multiplying pn by p only adds pn as a factor when p is prime, so you will have n factors plus 1 which is a factor, so p4 has 5 factors.
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What is the sum of all prime factors of 152?
What is the sum of all prime factors of 152?
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Since 152 is divisible by 2, start by dividing 152 by 2 which gives you factors of 2 and 76. 2 is a prime factor (cannot be divisible by anything other than itself and 1) but 76 can still be divided by 2. Continue dividing until there are only prime factors. You should get prime factors of 2, 2, 2, and 19. 2+2+2+19 = 25.
Since 152 is divisible by 2, start by dividing 152 by 2 which gives you factors of 2 and 76. 2 is a prime factor (cannot be divisible by anything other than itself and 1) but 76 can still be divided by 2. Continue dividing until there are only prime factors. You should get prime factors of 2, 2, 2, and 19. 2+2+2+19 = 25.
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Let x, y, and z be three distinct positive integers. Which of the following values of y is possible,
if x3y2z = 1000
Let x, y, and z be three distinct positive integers. Which of the following values of y is possible,
if x3y2z = 1000
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If y = 2, then x3z = 250. This means that x could be equal to 1, and z could be equal to 250. This would make x, y, and z distinct positive integers. Thus, it is possible for y to equal 2.
If y = 5, then x3z = 40. This means x could be 1, and z could be 40. This would make x, y, and z distinct positive integers. Therefore, y can equal 5.
If y = 10, then x3z = 10. This means that the only value that x could be is 1. If x were equal to 2, for example, then z would be equal to 10/8, which is not an integer. The same would be true if x were equal to anything other than 1. Thus, x must equal 1, and z must equal 10. However, because y and z must be distinct, y cannot equal 10.
Thus, y could equal only 2 and 5.
If y = 2, then x3z = 250. This means that x could be equal to 1, and z could be equal to 250. This would make x, y, and z distinct positive integers. Thus, it is possible for y to equal 2.
If y = 5, then x3z = 40. This means x could be 1, and z could be 40. This would make x, y, and z distinct positive integers. Therefore, y can equal 5.
If y = 10, then x3z = 10. This means that the only value that x could be is 1. If x were equal to 2, for example, then z would be equal to 10/8, which is not an integer. The same would be true if x were equal to anything other than 1. Thus, x must equal 1, and z must equal 10. However, because y and z must be distinct, y cannot equal 10.
Thus, y could equal only 2 and 5.
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The product of three consecutive positive integers is 210. What is the sum of the integers?
The product of three consecutive positive integers is 210. What is the sum of the integers?
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Let x be the first integer. The next two integers would then be x+1 and x+2. We must find x such that:
x(x+1)(x+2) = 210.
To do this, we need to look at the factors of 210, and find a set of three factors that are also consecutive integers. First, let's find the prime factorization of 210, because then we can use that to determine all of the possible factors of 210.
210 = 10 x 21 = 2 x 5 x 21 = 2 x 5 x 3 x 7
Thus, the prime factorization of 210 is 2 x 3 x 5 x 7.
We can notice now that 5 and 7 are almost consecutive integers, and the only number missing between them is 6. If we multiply 2 and 3, we get 6. Thus 210 can be written as 5 x 6 x 7. This means that the consecutive integers are 5, 6, and 7, and their sum is 18.
Let x be the first integer. The next two integers would then be x+1 and x+2. We must find x such that:
x(x+1)(x+2) = 210.
To do this, we need to look at the factors of 210, and find a set of three factors that are also consecutive integers. First, let's find the prime factorization of 210, because then we can use that to determine all of the possible factors of 210.
210 = 10 x 21 = 2 x 5 x 21 = 2 x 5 x 3 x 7
Thus, the prime factorization of 210 is 2 x 3 x 5 x 7.
We can notice now that 5 and 7 are almost consecutive integers, and the only number missing between them is 6. If we multiply 2 and 3, we get 6. Thus 210 can be written as 5 x 6 x 7. This means that the consecutive integers are 5, 6, and 7, and their sum is 18.
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