How many 3-digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number?
A) 16
B) 80
C) 160
D) 180
E) 240
Difficult Math Question #56 - Number Theory
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Same logic as one of the previous problem...800guy wrote:How many 3-digit numerals begin with a digit that represents a prime and end with a digit that represents a prime number?
A) 16
B) 80
C) 160
D) 180
E) 240
(Thanks to rajs, he helped refresh that part of the memory...)
Number of prime digits : 2,3,5,7 = 4
(assuming repeatition of number is allowed)
4 * 10 * 4 = 160
Answer: C
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for the first digit, it can be 2, 3, 5 or 7 = which means 4 options
for the second digit, it can be 0,1,2,3,4,5,6,7,8 or 9 = 10 options
assuming the repetition of the first digit is accepted
then our third digit gives 2,3,5 or 7 = 4 options
so, 4*10*4 = 160
Hence option C is correct
for the second digit, it can be 0,1,2,3,4,5,6,7,8 or 9 = 10 options
assuming the repetition of the first digit is accepted
then our third digit gives 2,3,5 or 7 = 4 options
so, 4*10*4 = 160
Hence option C is correct
We have to build a 3 digit number, so we have 3 places: _ _ _ .
We need to write a 1 digit prime number in the first and the third place. The 1 digit prime numbers are: 2, 3, 5, 7. We have 4 different options.
For the second place we can write any of the numbers from 0 to 9, i.e. We have 10 different options.
The total of numbers we can build is the product of the options for each place.
4 * 10 * 4 = 160
The answer is C.
PD: the problem let us write the same number for the first place and the third place, this is why we have 4 options for the first and the third place.
We need to write a 1 digit prime number in the first and the third place. The 1 digit prime numbers are: 2, 3, 5, 7. We have 4 different options.
For the second place we can write any of the numbers from 0 to 9, i.e. We have 10 different options.
The total of numbers we can build is the product of the options for each place.
4 * 10 * 4 = 160
The answer is C.
PD: the problem let us write the same number for the first place and the third place, this is why we have 4 options for the first and the third place.
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This problem relies on the multiplication principle (also known as the rule of product). If there are m ways of doing X and n ways of doing Y, and if X and Y are independent events, there are xn ways of doing both X and Y. For example, if I have 3 shirts and 8 pants, and if I can match any of the shirts with any of the pants to make an outfit, I can make 3*8 = 24 outfits.
Here, we can select any single digit prime number for the hundreds digit: 2, 3, 5, or 7. So there are 4 options for the hundreds digit.
We can select any single digit integer for the tens digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. So there are 10 options for the tens digit.
Like the hundreds digit, we can select any single digit prime number for the ones digit: 2, 3, 5, or 7. So there are 4 options for the ones digit.
Applying the multiplication principle, there are 4*10*4 = 160 options for the hundreds, tens, and ones digits combined. The correct answer is C.
Here, we can select any single digit prime number for the hundreds digit: 2, 3, 5, or 7. So there are 4 options for the hundreds digit.
We can select any single digit integer for the tens digit: 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. So there are 10 options for the tens digit.
Like the hundreds digit, we can select any single digit prime number for the ones digit: 2, 3, 5, or 7. So there are 4 options for the ones digit.
Applying the multiplication principle, there are 4*10*4 = 160 options for the hundreds, tens, and ones digits combined. The correct answer is C.
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