Difficult Math Question #56 - Number Theory

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Difficult Math Question #56 - Number Theory

by 800guy » Fri Nov 17, 2006 10:35 am
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

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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
Same logic as one of the previous problem...
(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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OA

by 800guy » Sun Nov 19, 2006 2:57 pm
OA:

The first digit can be 2, 3, 5, or 7 (4 choices)
The second digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 choices)
The third digit can be 2, 3, 5, or 7 (4 choices)

4 * 4 * 10 = 160

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by BTGmoderatorRO » Sun Sep 03, 2017 6:48 am
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

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3- digit number

by Admin1 » Sun Sep 03, 2017 8:35 am
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.

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by ErikaPrepScholar » Tue Sep 12, 2017 12:42 pm
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.
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