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<h2>Quadratic primes</h2><div class="info" style="cursor:help;width:200px;margin-bottom:10px;"><h3>Problem 27</h3><span style="width:300px;color:#666;">Published on Friday, 27th September 2002, 06:00 pm; Solved by 38139</span></div>
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<p>Euler discovered the remarkable quadratic formula:</p>
<p style='text-align:center;'><i>n</i>&sup2; + <i>n</i> + 41</p>
<p>It turns out that the formula will produce 40 primes for the consecutive values <i>n</i> = 0 to 39. However, when <i>n</i> = 40, 40<sup>2</sup> + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when <i>n</i> = 41, 41&sup2; + 41 + 41 is clearly divisible by 41.</p>
<p>The incredible formula &nbsp;<i>n</i>&sup2; <img src='images/symbol_minus.gif' width='9' height='3' alt='&minus;' border='0' style='vertical-align:middle;' /> 79<i>n</i> + 1601 was discovered, which produces 80 primes for the consecutive values <i>n</i> = 0 to 79. The product of the coefficients, <img src='images/symbol_minus.gif' width='9' height='3' alt='&minus;' border='0' style='vertical-align:middle;' />79 and 1601, is <img src='images/symbol_minus.gif' width='9' height='3' alt='&minus;' border='0' style='vertical-align:middle;' />126479.</p>
<p>Considering quadratics of the form:</p>
<blockquote>
<i>n</i>&sup2; + <i>an</i> + <i>b</i>, where |<i>a</i>| <img src='images/symbol_lt.gif' width='10' height='10' alt='&lt;' border='0' style='vertical-align:middle;' /> 1000 and |<i>b</i>| <img src='images/symbol_lt.gif' width='10' height='10' alt='&lt;' border='0' style='vertical-align:middle;' /> 1000<br /><br />
<div class='info' style='text-align:left;'>where |<i>n</i>| is the modulus/absolute value of <i>n</i><br />e.g. |11| = 11 and |<img src='images/symbol_minus.gif' width='9' height='3' alt='&minus;' border='0' style='vertical-align:middle;' />4| = 4</div>
</blockquote>
<p>Find the product of the coefficients, <i>a</i> and <i>b</i>, for the quadratic expression that produces the maximum number of primes for consecutive values of <i>n</i>, starting with <i>n</i> = 0.</p>
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