This question's explanation and your answers so far.
大文字の英字から成る文字列の暗号化を考える。暗号化の手順と例は次のとおりである。この手順で暗号化した結果が“EGE”であるとき，元の文字列はどれか。
| 暗号化の手順 | 例 “FAX”の暗号化<br>処理前 | 例 “FAX”の暗号化<br>処理後 | |
|---|---|---|---|
| 1 | 表から英字を文字番号に変換する。 | FAX | 5, 0, 23 |
| 2 | 1文字目に1，2文字目に2，n文字目にnを加算する。 | 5, 0, 23 | 6, 2, 26 |
| 3 | 26で割った余りを新たな文字番号とする。 | 6, 2, 26 | 6, 2, 0 |
| 4 | 表から文字番号を英字に変換する。 | 6, 2, 0 | GCA |
| 英字 | A | B | C | D | E | F | G | H | I | J | K | L | M |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 文字番号 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 英字 | N | O | P | Q | R | S | T | U | V | W | X | Y | Z |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 文字番号 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 |
Encrypting BED: B,E,D=1,4,3 with 1,2,3 added gives 2,6,6, and the remainders modulo 26 are also 2,6,6, i.e., C,G,G=“CGG”, which is not “EGE”.
正解 Decryption means retracing the encryption steps in reverse. First, convert the ciphertext “EGE” into character numbers: E=4, G=6, E=4. In encryption step 2, n was added to the n-th character, so in decryption we do the reverse and subtract n from the n-th character (if the result is negative, add 26). 1st character: 4-1=3 → D; 2nd character: 6-2=4 → E; 3rd character: 4-3=1 → B. Therefore the original string is “DEB”, so option イ is correct. To be sure, verify by forward calculation: adding 1, 2, 3 to DEB=3,4,1 gives 4,6,4, and the remainders when divided by 26 are also 4,6,4, i.e., E,G,E=“EGE”, which matches the given ciphertext.
Encrypting FIH: F,I,H=5,8,7 with 1,2,3 added gives 6,10,10, and the remainders modulo 26 are also 6,10,10, i.e., G,K,K=“GKK”, which is not “EGE”.
Encrypting HIF: H,I,F=7,8,5 with 1,2,3 added gives 8,10,8, and the remainders modulo 26 are also 8,10,8, i.e., I,K,I=“IKI”, which is not “EGE”.