From the given conditions, we have $10(a-c)+b-d=11$, because $b-d \leq 9<11$ and $10(a-c) \leq 10 < 11$, we have $a-c=1$ and $b-d=1$. Now, using the condition $\overline{abcd} \vdots 3 \Rightarrow a+b+c+d \vdots 3 \Leftrightarrow c+d+1 \vdots 3$ and doing some caseworks, we're done!(hope so)

Nguyenhuyhoang wrote: From the given conditions, we have $10(a-c)+b-d=11$, because $b-d \leq 9<11$ and $10(a-c) \leq 10 < 11$, we have $a-c=1$ and $b-d=1$. Now, using the condition $\overline{abcd} \vdots 3 \Rightarrow a+b+c+d \vdots 3 \Leftrightarrow c+d+1 \vdots 3$ and doing some caseworks, we're done!(hope so) Not true since $ b - d$ is not necessarily positive Assuming $c$ can be 0, the question is asking for pairs of integers both $\le$ 99 with difference 11 and sum $\vdots$ 3 So X + X+11 $\vdots$ 3 with 0 $\le$ X $\le$ 88 Which means X + 1 $\vdots$ 3 so X = 2 , 5 , 8 , ... , 86 There are thus 29 integers which satisfy the conditions: 1302, 1605, 1908, ... , 9786

OK. You just need to take out of your list the numbers with $c=0$ since $\overline{cd}$ is a two-digit number.

Mualpha7 wrote: Find all four-digit numbers $\overline{abcd}$ such that they are multiples of $3$ and that $\overline{ab}-\overline{cd}=11$. ($\overline{abcd}$ is a four-digit number; $\overline{ab}$ is a two digit-number as $\overline{cd}$ is). if $\overline{ab}-\overline{cd}=11$, then $\frac{\overline{abcd}-\overline{cd}}{100}-\overline{cd}=11 \Longrightarrow \overline{abcd}-\overline{cd}-100\overline{cd}=1100 \Longrightarrow \overline{abcd}=1100 + 101\overline{cd}$ since $\overline{abcd}\equiv 0 \mod 3$, and $1100\not\equiv 0 \mod 3$, it follows that $101\overline{cd}\not\equiv 0 \mod 3$, so $\overline{cd}$ need to be of the form $3n+2$* * just notice the remainders of 1100 and 101 mod 3 as $101\overline{cd}\le 9999 - 1100=8899$, hence $\overline{cd} \in [2, 5, 8, ... , 83, 86]$, yielding $\overline{ab} \in [13, 16, 19, ... , 94, 97]$