Many thanks to Sue Liu, who is in S6 at Madras College in St
Andrews, for this beautifully neat solution. 

<p>To find all 3-digit numbers abc (in base 10) such that<br />





 
<br />
<table width="100%"><tr><td align="center">
    <math xmlns="http://www.w3.org/1998/Math/MathML">
    <mstyle displaystyle="true"><mrow><mrow><mi>a</mi><mo>+</mo>
<msup><mrow><mi>b</mi></mrow><mrow><mn>2</mn></mrow>
</msup>
<mo>+</mo>
<msup><mrow><mi>c</mi></mrow><mrow><mn>3</mn></mrow>
</msup>
</mrow>
</mtd><mtd columnalign="center">
<mrow><mo>=</mo></mrow>
</mtd><mtd columnalign="center">
<mn>100</mn><mi>a</mi><mo>+</mo><mn>10</mn><mi>b</mi><mo>+</mo><mi>c</mi></mrow>
    </mstyle></math>
</td></tr></table>
<br /></p>

<p>For any three consecutive integers, one of them is divisible by
3. Since 3 divides 99, it follows that 3 divides b(10-b). Since 3
is a prime, this limits the possible choices of b:<br />
<br />
<strong>Either</strong></p>

<table width="300" border="1">
<tr>
<td>b=0, 10-b=10,</td>
<td>&#x00A0;</td>
<td>b(10-b) = 0</td>
</tr>

<tr>
<td>b=3, 10-b=7,</td>
<td>or b=7, 10-b=3,</td>
<td>b(10-b) = 21</td>
</tr>

<tr>
<td>b=6, 10-b=4,</td>
<td>or b=4, 10-b=6,</td>
<td>b(10-b) = 24</td>
</tr>

<tr>
<td>b=9, 10-b=1,</td>
<td>or b=1, 10-b=9,</td>
<td>b(10-b) = 9</td>
</tr>
</table>

<p>Hence the possible values of b(10-b) are 0, 9, 21, 24.</p>

<p>We now have to find a multiple of 99 which when subtracted from
a product of 3 consecutive natural numbers gives 0, 9, 21 or
24.<br />
Since a is at least 1, c(c+1)(c-1) is at least 99, so c is at least
5.</p>

<table border="1" width="300">
<tr>
<th>c</th>
<th>c(c+1)(c-1)</th>
<th>a</th>
<th>99a</th>
<th>c(c+1)(c-1)-99a</th>
</tr>

<tr align="center">
<td>5</td>
<td>120</td>
<td>1</td>
<td>99</td>
<td>21</td>
</tr>

<tr align="center">
<td>6</td>
<td>198</td>
<td>2</td>
<td>198</td>
<td>12</td>
</tr>

<tr align="center">
<td>7</td>
<td>297</td>
<td>3</td>
<td>297</td>
<td>39</td>
</tr>

<tr align="center">
<td>8</td>
<td>495</td>
<td>5</td>
<td>495</td>
<td>9</td>
</tr>

<tr align="center">
<td>9</td>
<td>693</td>
<td>7</td>
<td>693</td>
<td>27</td>
</tr>
</table>

<p>(Since 0, 9, 21, 24 &#x003C; 99, only the multiple of 99 which is
closest to (c+1)c(c-1) needs to be checked.)</p>

<p>From the table we can see that the following are the
possibilities for a, b and c:</p>

<ul>
<li>
<p>a=1, b=3 or b=7, c=5</p>
</li>

<li>
<p>a=5, b=1 or b=9, c=8</p>
</li>
</ul>

<p>giving the solutions 135, 175, 518, 598.</p>

<p>Vassil Vassilev, of Lawnswood High School in Leeds, also sent in
a good solution. He worked systematically through the possible
values of c, using the rearrangement c(c<sup>2</sup>-1) = b(10-b) +
99a to establish the range of possible values for a, and then
testing to find values of b which fitted.</p>