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Please look at the following observation made by trial and error.

Let us take some $2$-digit numbers like $12, 15, 24,\dots$

$12 = 1 * 2 = 2 \implies 2|12 $ ($2$ divides $12$)

$15 = 1 * 5 = 5 \implies 5|15$

for $3$ digit numbers: $315 = 3 * 15 = 45 \implies 45|315$

for $4$-digit numbers: $1352 = 13 * 52 = 676 \implies 676|1352$

Like this kind of applicability is applicable only for some rare such numbers. I would like to know from this observation the following:

1) In what cases the $2$-digit, $3$-digit cases we can write? Is there any formal method to justify this observation?

2) Is there any particular formula or method to generate such pairs?

Thanks in advance!

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In your third line, did you mean to say (2 divides 12)? – Ross Millikan May 13 '13 at 18:45
@RossMillikan!yes 2 divides 12. – Naroza_lary_sai May 13 '13 at 18:48

I will use concatenation to represent concatenation of digits and $\cdot$ for multiplication. For the two digit numbers, if the number is $ab$, you are asking when $a \cdot b | 10a+b$ We must then have that $b | 10\cdot a$ and $a|b$, which is when $a=b, b=2\cdot a,$ or $b=5\cdot a$
For three digit numbers, do you insist on breaking off the first digit (like your example) so $a\cdot bc|abc$ or would you also be interested in $ab\cdot c|abc$? For the first, we want $a\cdot bc | 100\cdot a +bc$ Again, we must have $bc | 100 \cdot a$ and $a |bc$ There are a number of possibilities, which you can catalog.

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! first of all thank you so much for your promt answer. If I have n-digit number and with all the possible . (multiplication of individual digits), how to generate such pairs, which will satisfies my criteria? Take n-digit number and then generalize. – Naroza_lary_sai May 13 '13 at 18:48
@Naroza_lary_sai: I think the pattern is pretty clear. Say we have a five digit number and want to split it $2+3$ Then we want $ab\cdot cde|1000\cdot ab+cde$ which requires $ab|cde$ and $cde|1000\cdot ab$ As $1000$ doesn't have many prime factors it restricts the choices for $\frac {cde}{ab}$ severely. – Ross Millikan May 13 '13 at 19:52
! when we taking know digits, it is clear. If you want to generalize for the size of n-digits, it may be tough. I think, you can. Please generalize for n-digits. – Naroza_lary_sai May 13 '13 at 19:55
@foss Millikan! Sir, I think it may be difficulty to know prime factors for n. Still, I think we may generalize by taking the two cases. one is n-digit = even number of digits and other case is n = odd number of digits. Could you please think in this way or your own way. – Naroza_lary_sai May 13 '13 at 20:00
@Naroza_lary_sai: We don't need the prime factors of $n$, we need the prime factors of $10^n$. These are easy to find-they are $2^n \cdot 5^n$ – Ross Millikan May 13 '13 at 20:21


I check all the numbers below $100000,$ here are the solutions: $\{1,2,3,4,5,6,7,8,9,11,12,15,24,36,111,112,115,128,132,135,144,175,212,216,224,312,315,384,432,612,624,672,735,816,1111,1112,1113,1115,1116,1131,1176,1184,1197,1212,1296,1311,1332,1344,1416,1575,1715,2112,2144,2232,2916,3111,3132,3168,3171,3276,3312,3915,4112,4224,4416,6144,6624,6912,7112,7119,8112,8832,9315,9612,11111,11112,11115,11133,11172,11212,11232,11313,11328,11331,11424,11664,11711,11872,12112,12128,12132,12216,12224,12288,12312,12432,12712,12768,13113,13131,13212,13248,13272,13311,13824,13932,14112,14144,16128,16224,16416,16632,17115,17136,18112,18432,18816,19116,21112,21132,21184,21216,21248,21312,21672,21728,22112,22144,22176,23112,23328,23424,24112,24192,24912,26112,26136,26712,27132,27216,28416,31113,31131,31212,31311,31488,32112,32172,32616,32832,33111,33264,34272,34992,35175,37212,38112,41112,41232,41616,42112,42192,42336,42432,42624,42912,43632,48128,51975,61344,61824,62112,64128,66312,71232,71316,72128,73332,77175,77616,81216,81312,82112,82944,83232,86112,89712,91728,92232,93312,93744\}$

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! how you got such all set of numbers. Is there any method to get... – Naroza_lary_sai May 14 '13 at 5:15
@Naroza_lary_sai I do it with mathematica 9.0 – Next May 14 '13 at 6:39
! could you generalize for 10^n-digits, instead of examples. Please generalize for 10^n digits for any n. – Naroza_lary_sai May 14 '13 at 6:40
I don't understand what's your mean,do you want all the solutions by a general formula? – Next May 14 '13 at 6:50
! yes! or how to find such all solutions mathematically by proving. not b y using any software etc – Naroza_lary_sai May 14 '13 at 7:02

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