Solution: Australian Mathematics Competition 2023 – Junior, Problem 29

Let's discuss a problem from AMC Junior 2023 that focuses on divisibility. From this solution, we will learn how to write two-digit numbers in terms of tenth place and 1's place.

Problem

A two-digit number has the property, that when it is divided by the sum of its digits the result is \(4\) with remainder \(3\). What is the sum of all two-digit numbers with this property?(AMC Junior 2023 Problem 29)

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Solution

Let's start this problem by representing the two-digit number as $10x + y$, where x represents the tenth-place digit and y represents the one's place digit.

In the next step, we have to divide this number by the sum of its digits: \(\frac {10x + y}{x + y}\).

The answer will have to be 4 : \(\frac {10x + y}{x + y} = 4\).

And we will also get a remainder: \( 10x + y = 4 (x + y) + 3\).

\(\Rightarrow 10 x + y = 4x + 4y +3\).

\(\Rightarrow 6x - 3y = 3\).

\(\Rightarrow 2x - y = 1\).

\(\Rightarrow2x - 1 = y\).

Making a table :

If we only have to consider two-digit numbers we will only consider till \(9\) for \(y\).
Thus the list of numbers are \(11, 23, 35, 47, 59\).

Now let's divide the numbers by the sum of their digits and find the remainder :

\(\frac{11}{2}and ans =5, rem =1\), \(\frac {23}{5} and ans = 4, rem =3\), \(\frac {35}{8} and ans = 4, rem = 3\), \(\frac {47}{11} and ans = 4, rem = 3\), \(\frac {59}{14} and ans = 4, rem = 3\).

As \(11\) doesn't satisfy the condition of getting the remainder as 3 and the answer as 4, after dividing the number by the sum of its digits, we will ignore 11.

Thus the sum of the remaining two digit numbers is = \( 23 + 35 +47 +59 = 164\).

So the answer to this problem is 164.