-3
$\begingroup$

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, A.P., N. F. Taussig, Tim Raczkowski, Harish Chandra Rajpoot
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – A.P. Nov 1 '15 at 12:08
1
$\begingroup$

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.

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

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:

  • $57$
  • $75$
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.