This question's explanation and your answers so far.
表計算ソフトを用いて社員コード中のチェックディジットを検算する。社員コードは 3 けたの整数値で，最下位の 1 けたをチェックディジットとして利用しており，上位 2 けたの各けたの数を加算した値の 1 の位と同じ値が設定されている。セル B2 に社員コードからチェックディジットを算出する計算式を入力し，セル B2 をセル B3～B5 に複写するとき，セル B2 に入力する計算式のうち，適切なものはどれか。
| A | B | |
|---|---|---|
| 1 | 社員コード | チェックディジット |
| 2 | 370 | |
| 3 | 549 | |
| 4 | 538 | |
| 5 | 763 |
10 - INT(A2/100) + MOD(INT(A2/10), 10). The whole expression is not reduced to the ones place by an outer MOD(…, 10), so there is no guarantee the result stays within 0 to 9, and on top of that it contains a subtraction 10 - that has no justification. For A2＝370 it gives 10-3+7＝14, which is not a single-digit check digit (the correct value is 0). For 763 it gives 10-7+6＝9, which does not match the lowest digit 3. For 549 and 538 it happens to produce the same value as the correct answer, so be careful: testing only one row and choosing from that will lead you astray.
MOD(10 - INT(A2/100) + INT(A2/10), 10). Because the inner MOD(…, 10) that extracts the tens digit is missing, INT(A2/10) is not the tens digit but the value of the upper 2 digits themselves (37 for 370). The unnecessary subtraction 10 - is also still there. For A2＝370 it gives MOD(10-3+37, 10)＝MOD(44, 10)＝4, which does not match the correct value 0. For 763 it gives MOD(10-7+76, 10)＝9, which does not match the lowest digit 3.
正解 The check digit is defined as “the ones place of the value obtained by adding the digits in the upper 2 places (the hundreds and the tens).” In a spreadsheet, to pull each digit out of a 3-digit integer, use INT(A2/100) for the hundreds digit and MOD(INT(A2/10), 10) for the tens digit. INT(A2/10) by itself gives the value of the upper 2 digits (37 for 370), so the key point is to keep only the tens digit by using the inner MOD(…, 10). To then take the ones place out of the sum of these two, you wrap the whole thing in one more MOD(…, 10) on the outside. Therefore MOD(INT(A2/100) + MOD(INT(A2/10), 10), 10), which is ウ, is correct. Checking the actual values: 370 → 3+7＝10 → MOD(10, 10)＝0, 549 → 5+4＝9, 538 → 5+3＝8, 763 → 7+6＝13 → MOD(13, 10)＝3; all four match the lowest digit of the employee code. Because A2 is a relative reference, copying B2 to B3 through B5 shifts the reference automatically to A3, A4, and A5, so each row is calculated correctly.
INT((INT(A2/100) + INT(A2/10)) / 10). Taking INT after dividing by 10 is an operation that extracts not the ones place but the upper part left after discarding the ones place, which is the opposite of what is wanted here. When you want the ones place, use MOD(…, 10). In addition, the inner INT(A2/10) is again not the tens digit. For A2＝370 it gives INT((3+37)/10)＝4 and for 763 it gives INT((7+76)/10)＝8, neither of which matches the lowest digit (0 and 3).