+0  
 
+1
45
2
avatar+28 

Dr. Math's four-digit house number ABCD contains no zeroes and can be split into two different two-digit primes ``AB'' and ``CD'' where the digits A, B, C and D are not necessarily distinct. If each of the two-digit primes is less than 40, how many such house numbers are possible?

Memes4Life132  Jul 20, 2018
 #1
avatar+87564 
+1

The two digit primes < 40 are

11, 13, 17, 19, 23, 29, 31, 37

 

Any two of these eight can be selected  =  C(8,2)  = 28

 

And each of these can be arranged in two ways

 

So...the total number possible house  numbers  is  2 * 28  =  56

 

 

cool cool cool

CPhill  Jul 20, 2018
 #2
avatar
0

Yes it is correct thank you very very much

Guest Jul 21, 2018

17 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.