This question's explanation and your answers so far.
表計算ソフトを用いて，二つの2進数α，βの和を求める。α，βの桁数は2桁に限定し，2の位，1の位の順に各桁の値がセルB3，セルC3とセルB4，セルC4に入力されている。ワークシートは2進数α＝10，β＝11の例である。結果は，4の位，2の位，1の位の順にセルA5，セルB5，セルC5に表示したい。セルB2は1の位からの繰り上がりの情報を格納するために用いられ，式IF(C3＋C4＝2，1，0)が入力されている。
| A | B | C | D | |
|---|---|---|---|---|
| 1 | 4の位 | 2の位 | 1の位 | |
| 2 | 0 | ：1の位からの繰上げ | ||
| 3 | 1 | 0 | ：2進数αの値 | |
| 4 | 1 | 1 | ：2進数βの値 | |
| 5 | 1 | 0 | 1 | ：結果の表示 |
セルB5に入力する式はどれか。
正解 B5 represents the 2's place of the result. Three things are added into the 2's-place digit: the carry from the 1's place B2, the 2's place of α B3, and the 2's place of β B4; their total SUM(B2:B4) takes a value from 0 to 3. The 2's-place digit to display is the remainder of this total divided by 2 (total 0→0, 1→1, 2（＝binary 10）→0, 3（＝binary 11）→1). Choice ア outputs the value itself as the digit when the total is less than 2 (0 or 1), outputs 0 when the total is 2, and otherwise (total 3) returns 1, so it gives the correct 2's-place digit in every case. In the example, SUM(B2:B4)＝0＋1＋1＝2, so ア returns 0, matching B5＝0 in the figure.
The inner branch is reversed, returning 1 when the total is 2 and 0 when the total is 3. But the 2's place in binary must be 0 when the total is 2（＝10）and 1 when the total is 3（＝11）, so the values are inverted. In the example the total is 2, so it outputs 1, which conflicts with the correct 0 (B5＝0 in the figure).
The outer condition is 「SUM(B2:B4)＞2」, so when the total is 0 or 1 it falls into the inner “otherwise” branch and always returns 1, wrongly outputting 1 even when the total is 0. Moreover, when the total is 3 it displays the total value 3 as is, which is not a single binary digit (0 or 1) and breaks down as a digit.
Because the outer condition is 「SUM(B2:B4)＞2」, when the total is 3 it displays 3 as is and does not become a single digit. Also it returns 0 for both total 0 and 1, so when the total is 1 the 2's place that should be 1 becomes 0. Even in the example (total 2) it returns 1, not matching the correct 0.