Ku saabsan adkaanta IBAN04
22

IBAN-ka Jarmalka wuxuu ka kooban yahay koodhka waddanka (DE), lambar hubin laba-god ah (sida waafaqsan ISO 7064 ), koodhka bangiga (8 god), iyo lambarka akoonka (oo ay ku jiraan lambarka akoonka hoose, 10 god, tiro maqan oo lagu buuxiyay eber hormood ah), taasoo ka dhigaysa 22 xaraf oo dheer. Si loo xisaabiyo lambarka jeegga, waxaa marka hore la sameeyaa taxane laga soo bilaabo waxa loogu yeero BBAN (koodhka bangiga iyo lambarka akoonka), koodhka waddanka ee lambarrada ah \(1314\) ee Jarmalka, iyo koodhka jeegga \(00\) .

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Tusaale ahaan, lambarka bangiga 21050170 iyo lambarka akoonka 12345678 waxay soo saaraan BBAN 210501700012345678. Kordhinta tan iyadoo la adeegsanayo lambarka waddanka iyo lambarka jeegga 00 waxay keenaysaa \(x = 210501700012345678131400\) , halkaas oo lambarka jeeggu hadda yahay: \(98 - (x \mod 97)\) . Qaybinta \(97\) ma aha mid shil ah. Maadaama ay tahay lambarka ugu weyn ee suurtogalka ah ee laba-god ah, waxay ogaanaysaa khaladaadka sida lambarrada la beddelay oo leh suurtogalnimada ugu sarreysa ee suurtogalka ah. Hadda waxaan soo bandhigi doonnaa indha-indheynta soo socota.:

1. Beddelida hal lambar oo IBAN ansax ah waxay keeni doontaa IBAN aan sax ahayn.
2. Beddelka laba lambar oo kala duwan oo IBAN ansax ah waxay keeni kartaa IBAN ansax ah.
3. Haddii laba boos oo kala duwan oo IBAN ansax ah la isku beddelo, IBAN aan sax ahayn ayaa la abuuray.
4. Haddii aad beddesho laba boos oo kala duwan oo IBAN ansax ah laba jeer, IBAN ansax ah ayaa ku dhalan kara.

U daa $$A = DE P_1 P_2 N_1 N_2 N_3 N_4 N_5 N_6 N_7 N_8 N_9 N_{10} N_{11} N_{12} N_{13} N_{14} N_{15} N_{16} N_{17} N_{18}$$ IBAN ansax ah.

Dabadeed $$A_B = N_1 N_2 N_3 N_4 N_5 N_6 N_7 N_8 N_9 N_{10} N_{11} N_{12} N_{13} N_{14} N_{15} N_{16} N_{17} N_{18} 131400$$ BBAN laxiriira (oo lagu dheereeyay lambarka summada dalka DE iyo lambarka jeegga \(00\) ).

1. Isbeddel hadda \(N_k\), waa \(A_B^* = A_B + l \cdot 10^{24-k}\) leh \(1 \leq k \leq 18\) iyo \((-1) \cdot N_k \leq l \leq 9-N_k \wedge l \neq 0\). Leh \( P = 98 - (A_B \mod 97) \) waase \(P^* = 98 - \left((A_B + l \cdot 10^{24-k}) \mod 97\right) \). Guud ahaan waxay khusaysaa \( a \equiv a' \mod m, b \equiv b' \mod m \): \(a + b \equiv a' + b' \mod m\). Leh \(A_B \equiv R_1 \mod 97\) iyo \(l \cdot 10^{24-k} \equiv R_2 \mod 97\) waa \( (A_B + l \cdot 10^{24-k}) \equiv R_1 + R_2 \mod 97 \). Laakiin hadda waa \( 0 < R_2 < 97 \) iyo sidaas \( P^* = 98 - (R_1+R_2) \neq 98 - R_1 = P \) sidaas darteed \( P_1 \neq P_1^* \vee P_2 \neq P_2^* \). Tani waxay ka tagaysaa hal isbeddel oo suurtagal ah oo lambar ah \( P \) ku \( P^* \neq P \). Halkan laakiin \( N_k \) waxba isma beddelin, jeegaggu waa la sameeyay \( P \neq P^* \).
2. Labada IBAN ee soo socda waa ansax:
   $$\begin{align} A_1 = DE89207300\boldsymbol{\color{red}01}0012345674 \\ A_2 = DE89207300\boldsymbol{\color{red}98}0012345674 \end{align}$$ Halkan, waxaan ka faa'iideysaneynaa xaqiiqda ah inaan kordhinay laba lambar oo isku xiga oo ku jira \(A_1\) \(97\) . Intaa waxaa dheer, IBAN ma aha oo kaliya mid rasmi ah, laakiin lambarrada bangiga ee hoose ee 20730001 iyo 20730098 ayaa dhab ahaantii jira.
3. Waxaan isku daynaa marka hore, \( N_{k_1} \) iyo \( N_{k_2} \) in la is dhaafsado. Marka hore waa \( P = 98 - (A_B \mod 97) \) sida \(P^* = 98 - \left((A_B + l \cdot 10^{24-k_1} - l \cdot 10^{24-k_2}) \mod 97\right) \) leh \(l = N_{k_2} - N_{k_1}\) iyo \(1 \leq k_1, k_2 \leq 18\). Hadda waa sababtoo ah
   
   $$\begin{array} {|c|c|} \hline k & R = 10^{24-k} \mod 97 \\ \hline 1 & 56 \\ \hline 2 & 25 \\ \hline 3 & 51 \\ \hline 4 & 73 \\ \hline 5 & 17 \\ \hline 6 & 89 \\ \hline 7 & 38 \\ \hline 8 & 62 \\ \hline 9 & 45 \\ \hline 10 & 53 \\ \hline 11 & 15 \\ \hline 12 & 50 \\ \hline 13 & 5 \\ \hline 14 & 49 \\ \hline 15 & 34 \\ \hline 16 & 81 \\ \hline 17 & 76 \\ \hline 18 & 27 \\ \hline \end{array}$$
   \( \forall k_1 \neq k_2 \in \left\{ 1, \ldots, 18 \right\} : R_{k_1} \neq R_{k_2}\). Waa sidaas oo kale \( P \neq P^* \). Markaa way u hadhsan tahay in taas la hubiyo \(P_n\) iyo \(N_k\) leh \( 1 \leq n \leq 2 \) iyo \( 1 \leq k \leq 18 \) ganacsi Waxaa laga yaabaa in \(P = 98 - (A_B \mod 97)), (R_1 = (A_B \mod 97)\), \(P^* = 98 - (A_B + (l \cdot 10^{24-k}) \mod 97)\), \(R_2 = (A_B + (l \cdot 10^{24-k}) \mod 97)\). Maadaama aanu \(A_B\) agagaarka \(l \cdot 10^{24-k}\) waa inaan bedelnaa \(P_1\) ama \(P_2\) agagaarka \(-l\), sidaas \(P\) agagaarka \(-10^m l\) leh \(m \in \{0,1\}\) beddel: Markaa waa \(P^* = 98 - R_2\) laakiin sidoo kale \(P^* = P - 10^m l = 98 - R_1 - 10^m l\), sidaas awgeed \(R_2 = R_1 + 10^m l,\) iyo sidaas
   $$((A_B \mod 97) + (l \cdot 10^{24-k} \mod 97)) \mod 97 = (A_B \mod 97) + 10^m l$$ Si kastaba ha ahaatee, isla'egtaani weligeed lama fulin, sida qoraalka soo socda uu muujinayo:
   
   

   See the Pen IBAN FORMULA CHECK by David Vielhuber (@vielhuber) on CodePen.

   Tani waxay ka tagaysaa kaliya beddelka suurtagalka ah \(P_1\) iyo \(P_2\). Halkan laakiin \( N_k \) waxba isma beddelin, jeegaggu waa la sameeyay \( P \neq P^* \).
4. Labada IBAN ee soo socda waa ansax:
   $$\begin{align*}A_1 = DE\boldsymbol{\color{red}8}\boldsymbol{\color{green}3}20220800\boldsymbol{\color{red}1}000000\boldsymbol{\color{green}0}00 \\ A_2 = DE\boldsymbol{\color{red}1}\boldsymbol{\color{green}0}20220800\boldsymbol{\color{red}8}000000\boldsymbol{\color{green}3}00\end{align*}$$ Halkan, sidoo kale, BIC 20220800 runtii waa jiraa.