AMC 2017 junior question

Question

the reverse of the number $129$ is $921$ and these add to $1050$ which is divisible by $30$. How many three-digit numbers have the property that, when added to their reverse, the sum is divisible by $30$?

Answer 1

Let the number be $100a+10b+c$.

Since the sum is divisible by $30$ , It should also be divisible by $10$ and $3$.

Taking $a=1$ , we get $c=9$ and since the number is divisible by $3$ , we can take $b = 2,5,8$.

Taking $a=2$ , we get $c=8$ and since the number is divisible by $3$ , we can take $b = 2 , 5,8$

$\vdots$

Can you similarly conclude for all cases ?

Answer 2

Any $3$ digit number can be expressed in the form $x=100+10+$, so reversing we number we obtain $y=100+10+$.

Now lets add these numbers together to obtain $$x+y=100(a+c)+20b+(c+a)$$. Now we can take the $\mod30$ (multiples of $30$) on both sides to obtain that

$$x+y≡10(a+c)+20b+(c+a)\mod30≡11(a+c)+20b \mod30$$

So when is $11(a+c)+20b≡0 \mod30?$ Well, since $20b$ gives a multiple of $20$, we only need to check $11(a+c)+20≡0\mod30$ so $11(a+c)≡10\mod30$, $11(a+c)+40≡0\mod30, 11(a+c)≡20\mod30$ and when $b=0$ so $11(a+c)≡0\mod30$. (Since $11(a+c)+60≡0\mod30$ gives back $11(a+c)≡0\mod30$ and so on).

Moreover since we have a three digit number $a$ is bigger than $0$ but less than $10$. So we only need to check a few cases.

Answer 3

Write the number as $abc $ and use the divisibility rules for 10 and 3. From the rule for 10, we know that the last digit of $abc + cba$ is 0, hence $a+c=10$. Thus the options are $a=1$ & $c=9$, $a=2$ & $c=8$, etc.

From the rule for 3, namely that the digit sum must be divisible by 3, we find that $2a+2b+2c=20+2b $ is divisible by 3. By hand it's easy to check that $b=2,5,8$ are the only options that work.

So there are 9 choices for $a $ and $c $ and 3 for $b $, giving 27 total.

Answer 4

So if $n=\overline{abc}$ then $$30 \mid \overline{abc}+\overline{cba} = 101a+20b+101c$$

Now since $$10\mid 101a+20b+101c -(100a+20b+100c) =a+c$$ and $2\leq a+c\leq 18$ we have $$a+c=10$$ so we are left with $$3\mid 202+2b$$

so $b\in\{1,4,7\}$. So we have $3$ possibilities for $b$ and $9$ posibilities for pair $(a,c)$. So there is $27$ such numbers.