Small long division

There are two possibilities.

$13\div3=4$ with remainder 1 or $33\div8=4$ with remainder 1.

Only the second version contains a number on the list, namely $\boxed8$

The number (we'll call y) must be such that y × 4 + 1 = number ending with 3 in the ones place. The possible numbers that work are 3 and 8. The number given is 8, thus, the only number on the list that works is 8.

8 × 4 + 1 = 33

A systematic algebraic approach: Let $k$ be the divisor, $m$ be the 10's place digit of the dividend, and $n$ the 10's place digit of the number below the dividend. $k, m, \text{ and } n$ are all digits in $\{1, 2, \ldots, 9\}$ . Then from the structure of the division problem:

$(1) \; 10m + 3 - (10n+2) = 1 \; \Rightarrow \; 10m - 10n = 0 \; \Rightarrow \; \underline{m = n}$

$(2) \; 10 m + 3 = 4 k + 1 \; \Rightarrow \; 5 m + 1 = 2 k \; \Rightarrow \; m$ must be odd and $\underline{k = \frac{1}{2}(5 m + 1)}$ for $m$ odd.

So, trying $m \in \{1, 3, 5, 7, 9\}$ gives only two possible single-digit $(m,k)$ solutions: $(1,3)$ and $(3,8)$ . Only $\boxed{8}$ is a solution option.

Only thing we have to focus is that 4 and 2

We have to find a sure number that is in the options and can fill any of the boxes, so the only thing we can calculate is a multiple of 4 whose one's digit 2

Only possibilities : 4 x 3 = 12

4 x 8 = 32

Only 8 is available in options, hence answer is 8

It is possible to solve this problem knowing only the two and the four. By following the long division procedure, we know that 4 times a single digit number must give a double digit number with two as its last digit. The second single-digit factor must be one of the options provided, so we have either: 4 x 5= 20, 4 x 6= 24, 4 x 7= 28, and 4 x 8= 32. Since 8 is the only factor that gives a product ending with 2, it must be the answer.

The first omitted digit, the divisor, has to be larger than the omitted number of the dividend, since the quotient is one digit. Also, since there is only a remainder of 1, the second two omitted numbers have to equal each other. There is only one number on the list which meets these requirements - 8, since 33 ÷ 8= 4, with a remainder of 1.

The two digit multiples of 4 with two ones are 12 and 32. You cannot go past 32 because then the divisor will not be a single digit anymore. Since 3 and 4 do not appear as choices, the answer must be 33/8.

it should be,

4x+1=..3

or, 4x+1=33.............[without 3, no numbers work here]

or. 4x= 32

or,x=8

Let be $n$ the divisor the division, $n\times4$ it must finish in $2$ . $\therefore$ $8\times4=32$ finish in $2$

Of the answer choices, only one of them (8) offers a product of 4 with 2 as its first digit.