# Two digit products into Primes [closed]

How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9$\times$8=72, 72=7$\times$2=14, 14=1$\times$4=4. Consider only 4 prime no.s (2,3,5,7)

I would like to know, Is there any way we can approach this. Answer = 18. and Possibilities are 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75

## closed as off-topic by Travis, A.P., N. F. Taussig, Tim Raczkowski, Harish Chandra RajpootNov 2 '15 at 0:49

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There are $18$ such numbers. Here the PARI/GP - program and the output :

? q=0;for(m=10,99,n=m;x=digits(n);while(length(x)>1,n=prod(j=1,length(x),x[j]);x
=digits(n));if(isprime(x)==[1],q=q+1;print(q,"   ",m,"  ",x)))
1   12  [2]
2   13  [3]
3   15  [5]
4   17  [7]
5   21  [2]
6   26  [2]
7   31  [3]
8   34  [2]
9   35  [5]
10   37  [2]
11   43  [2]
12   51  [5]
13   53  [5]
14   57  [5]
15   62  [2]
16   71  [7]
17   73  [2]
18   75  [5]
?


You can also get this result by hand :

The final result must be one of the numbers $2,3,5,7$

So, the second last number must be one of $12,13,15,17,21,31,51,71$

From these numbers, only $12,15$ and $21$ can be represented by a product of two one-digit numbers. The numbers $26,62,34,43,35,53,37,73$ are added to the set.

Finally, only $35$ can be represented by a product of two one-digit numbers, so $57$ and $75$ are added to the set. Those numbers are no more representable in the desired way, so the set is complete.

• Is there anyway manually can we solve this problem without programming? – Ramakrishna S Nov 1 '15 at 12:29

First iteration:

• $2\leftarrow12,21$
• $3\leftarrow13,31$
• $5\leftarrow15,51$
• $7\leftarrow17,71$

Second iteration:

• $12\leftarrow26,34,43,62$
• $13$
• $15\leftarrow35,53$
• $17$
• $21\leftarrow37,73$
• $31$
• $51$
• $71$

Third iteration:

• $26$
• $34$
• $35\leftarrow57,75$
• $37$
• $43$
• $53$
• $62$
• $73$

Fourth iteration:

• $57$
• $75$