Do You Know Divisibility Rules?

Method 1 : Trial and error.

If the 6-digit number is divisible by both 3 and 11, then it is divisible by $3\times11=33$ .

Since there are only 3 options to choose, let's just try out each case.

Case 1 : $x = 7, y = 5$ , then the 6-digit number is $476750$ , a long division shows that it is not divisible by 33. So $x=7,y=5$ cannot be a solution.
Case 2 : $x = 8, y = 5$ , then the 6-digit number is $476850$ , a long division shows that it is divisible by 33. So $x=8,y=5$ can be a solution.
Case 3 : $x = 7, y = 4$ , then the 6-digit number is $476740$ , a long division shows that it is not divisible by 33. So $x=7,y=4$ cannot be a solution.

Thus the answer is $x=8,y=5$ .

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Method 2 : Apply Divisiblity rules .

If a number is divisible by 3, then the sum of digits of that number is divisible by 3 as well, $4+7+6+x+y+ 0= x+y +17$ is divisible by 3. Looking at the options tells us that either $x = 8,y=5$ is the only possible solution.

However, let's double check by confirming that it's divisible by 11 as well. If a number is divisible by 11, then the alternating sum of digits of that number is divisible by 11 as well, $4-7+6-x+y -0 = -x+y+3$ is divisible by 11. Substituting $x=8,y=5$ shows that it is indeed true.

Thus, among all these three choices, only $x=8,y=5$ is the possible solution.

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Bonus : Can you find the other two pairs of digits $(x,y)$ satisfying this constraint?

A solution can be found by the Chinese Remainder Theorem. In fact, ALL solutions for $x$ and $y$ can be found by this theorem.

$\begin{cases} 476000+x \equiv 0 \mod{3}\\ 476000+x \equiv 0 \mod{11} \end{cases} \implies \begin{cases}x\equiv 1 \mod{3}\\ x\equiv 3\mod{11} \end{cases}$

By the Extended Euclidean Algorithm:

$\begin{aligned} 11&=3\cdot 3 +2 & 2&=11-3\cdot 3\\ 3&=1\cdot 2 +1 & 1&=3-2\\ & & &=3-(11-3\cdot 3)\\ & & &=4\cdot 3-11 \end{aligned}$

Hence, $3^{-1}\equiv 4\mod 11$ . The Chinese Remainder Algorithm provides the following solution for $x$ :

$x\equiv 1+4\cdot 3\cdot (3-1)\equiv 25 \mod{33}$

Hence, $x\equiv 25+k\cdot 33\equiv 0 \mod{10}$ , where $k\in Z$ . It immediately follows that $k\cdot 33\equiv 5 \mod{10}$ . Again, by the Extended Euclidean Algorithm:

$\begin{aligned} 33&=3\cdot 10+3 & 3&=33-3\cdot 10\\ 10&=3\cdot 3+1 & 1&=10-3\cdot 3\\ & & &=10-3(33-3\cdot 10)\\ & & &=4\cdot 10-3\cdot 33 \end{aligned}$

Hence, $33^{-1}\equiv -3\equiv 7\mod{10}$ . Thus, $k\equiv 7\cdot 5\equiv 35\equiv 5\mod{10}$ . This can be written as $k=5+m\cdot 10$ , where $m\in Z$ .

$k$ values for the original question which do not alter the 4 given digits are $k=5, 15, 25$ .

Substituting this back into $x=25+k\cdot 33$ , we obtain $x=190, 520, 850$ . Substituting again for the original question provides $476190, 476520$ and $476850$ . These values fulfil the requirements of the question.

In terms of $(x,y)$ pairs, $(1,9)$ , $(5,2)$ and $(8,5)$ are solutions.

If it is divisible by 3, then the sum of all the digits should be divisible by 3. Only case two qualifies.