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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?

 Jul 20, 2018
 #1
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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

 Jul 20, 2018
 #2
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Yes it is correct thank you very very much

 Jul 21, 2018

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