How to find out if a number is prime - ISEE Upper Level: Quantitative Reasoning
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and 
are prime integers. 
and 
.
Which of the following sets gives all possible values of 
?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of 
. 97 is the only prime number between 90 and 100, so 
.
If 
, then 
.
If 
, then 
.
If 
, then 
.
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of
If
If
If
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and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of . 97 is the only prime number between 90 and 100, so .
If , then .
If , then .
If , then .
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of
If
If
If
← Didn't Know|Knew It →
and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of . 97 is the only prime number between 90 and 100, so .
If , then .
If , then .
If , then .
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of
If
If
If
← Didn't Know|Knew It →
and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of . 97 is the only prime number between 90 and 100, so .
If , then .
If , then .
If , then .
There are three prime numbers between 40 and 50 - 41, 43, and 47 - so these are the possible values of
If
If
If
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and are prime integers. 
and 
.
Which of the following sets gives all possible values of 
?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of 
. There are two prime numbers between 30 and 40 - 31 and 37 - so these are the possible values of 
.
Therefore, we find 
for six scenarios:
The possible values of 
are given by the set
.
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of
Therefore, we find
The possible values of
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Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since 
, we can remove 187; there are no multiples of 13, and we need go no further. The primes are
.
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
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Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 3 as their last digit are
.
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
,
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
The numbers between 100 and 200 that feature 3 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
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Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since , we can remove 187; there are no multiples of 13, and we need go no further. The primes are
.
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
← Didn't Know|Knew It →
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 3 as their last digit are
.
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
,
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
The numbers between 100 and 200 that feature 3 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
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and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of . There are two prime numbers between 30 and 40 - 31 and 37 - so these are the possible values of .
Therefore, we find for six scenarios:
The possible values of are given by the set
.
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of
Therefore, we find
The possible values of
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Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since , we can remove 187; there are no multiples of 13, and we need go no further. The primes are
.
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
← Didn't Know|Knew It →
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 3 as their last digit are
.
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
,
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
The numbers between 100 and 200 that feature 3 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
← Didn't Know|Knew It →
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 7 as their middle digit
(B) The number of primes between 100 and 200 that feature 7 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since , we can remove 187; there are no multiples of 13, and we need go no further. The primes are
.
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
The numbers between 100 and 200 that feature 7 as their middle digit are
We can immediately weed out the multiples of 2 and 5 by their last digit (0, 2, 4, 5, 6, or 8):
171 and 177 have digit sums 9 and 15, so both are multiples of 3 and can also be weeded out. This leaves
By attempting to divide by 7, 11 and 13 (no others are necessary, since the square of 17 exceeds 200), we can see both of these numbers are prime. The set in (A) has these two elements.
A similar process can be used to find the primes from the set of numbers between 100 and 200 that feature 7 as their last digit, which is
This time, since all numbers end in 7, we cannot weed out any multiples of 2 or 5. We can weed out 117, 147, and 177 as multiples of 3, since their digit sums are 9, 12, and 15, respectively. This leaves
Since
The set in (B) has six elements and is the set of greater cardinality. (B) is greater.
← Didn't Know|Knew It →
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Which is the greater quantity?
(A) The number of primes between 100 and 200 that feature 3 as their last digit
(B) The number of primes between 100 and 200 that feature 9 as their last digit
Tap to reveal answer
The numbers between 100 and 200 that feature 3 as their last digit are
.
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
,
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
The numbers between 100 and 200 that feature 3 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 3. Also, 123, 153, and 183 have digit sums 6, 9, and 12, respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers:
No prime factorization can be found for the other integers, so the set of numbers given in (A) is the set
a set with five elements.
A similar process can be used to identify the primes ending in 9; the numbers between 100 and 200 that feature 9 as their last digit are
There are no multiples of 2 or 5 here, as no multiple of 2 or 5 ends in 9. Also, 129, 159, and 189 have digit sums 12, 15, and 18 respectively. This immediately identifies them as multiples of 3, which can be removed to leave:
Of the remaining numbers,
No prime factorization can be found for the other integers, so the set of numbers given in (B) is the set
which also has five elements.
The sets described in (A) and (B) have the same cardinality and therefore, the quantities are equal.
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and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of . There are two prime numbers between 30 and 40 - 31 and 37 - so these are the possible values of .
Therefore, we find for six scenarios:
The possible values of are given by the set
.
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of
Therefore, we find
The possible values of
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and are prime integers. and .
Which of the following sets gives all possible values of ?
Which of the following sets gives all possible values of
Tap to reveal answer
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of . There are two prime numbers between 30 and 40 - 31 and 37 - so these are the possible values of .
Therefore, we find for six scenarios:
The possible values of are given by the set
.
There are three prime numbers between 70 and 80 - 71, 73, and 79 - so these are the possible values of
Therefore, we find
The possible values of
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Which is the greater quantity?
(a) The number of primes between 20 and 50
(b) The number of primes between 10 and 40
Which is the greater quantity?
(a) The number of primes between 20 and 50
(b) The number of primes between 10 and 40
Tap to reveal answer
The primes between 20 and 40 are included in both sets, so all we need to do is to compare the number of primes between 40 and 50 with the number of primes between 10 and 20.
(a) The primes between 40 and 50 are 41, 43, and 47 - three.
(b) The primes between 10 and 20 are 11, 13, 17, and 19 - four.
Since the number of primes between 10 and 20 outnumbers those between 40 and 50, the number of primes between 10 and 40 outnumber those between 20 and 50. Therefore, (b) is the greater.
The primes between 20 and 40 are included in both sets, so all we need to do is to compare the number of primes between 40 and 50 with the number of primes between 10 and 20.
(a) The primes between 40 and 50 are 41, 43, and 47 - three.
(b) The primes between 10 and 20 are 11, 13, 17, and 19 - four.
Since the number of primes between 10 and 20 outnumbers those between 40 and 50, the number of primes between 10 and 40 outnumber those between 20 and 50. Therefore, (b) is the greater.
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Which is the greater quantity?
(a) The number of prime numbers between 70 and 110
(b) The number of prime numbers between 80 and 120
Which is the greater quantity?
(a) The number of prime numbers between 70 and 110
(b) The number of prime numbers between 80 and 120
Tap to reveal answer
The primes between 80 and 110 are included in both sets, so all we need to do is to compare the number of primes between 70 and 80 and the number of primes between 110 and 120.
(a) The primes between 70 and 80 are 71, 73, and 79 - three primes
(b) The only prime between 110 and 120 is 113.
(a) is the greater quantity
The primes between 80 and 110 are included in both sets, so all we need to do is to compare the number of primes between 70 and 80 and the number of primes between 110 and 120.
(a) The primes between 70 and 80 are 71, 73, and 79 - three primes
(b) The only prime between 110 and 120 is 113.
(a) is the greater quantity
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Which is the greater quantity?
(a) The number of prime numbers between 1 and 20 inclusive
(b) The number of composite numbers between 1 and 20 inclusive
Which is the greater quantity?
(a) The number of prime numbers between 1 and 20 inclusive
(b) The number of composite numbers between 1 and 20 inclusive
Tap to reveal answer
The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, 19 - eight total. Since 1 is neither prime nor composite, this leaves 11 composite numbers. (b) is the greater quantity.
The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, 19 - eight total. Since 1 is neither prime nor composite, this leaves 11 composite numbers. (b) is the greater quantity.
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Which is the greater quantity?
(a) The number of prime numbers between 1 and 20 inclusive
(b) The number of composite numbers between 21 and 30 inclusive
Which is the greater quantity?
(a) The number of prime numbers between 1 and 20 inclusive
(b) The number of composite numbers between 21 and 30 inclusive
Tap to reveal answer
(a) The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, 19 - eight total.
(b) The prime numbers between 21 and 30 inclusive are 23 and 29 - two prime numbers out of ten integers. This leaves eight composite numbers.
(a) and (b) are therefore equal.
(a) The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, 19 - eight total.
(b) The prime numbers between 21 and 30 inclusive are 23 and 29 - two prime numbers out of ten integers. This leaves eight composite numbers.
(a) and (b) are therefore equal.
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