<p>What is the greatest four digit integer that meets the following three restrictions?
<p>Without guessing, is there any way to do this? I spent liek 5 minutes guessing and that’s a real time killer.</p>
<p>i glanced at it, got 8521.</p>
<p>if its correct, this is my reasoning. </p>
<p>since it must be divisible by 10, then there must be a 2 and 5 in there.</p>
<p>naturally, you want the highest integer as the thousands digit, so u put 9.
But if 9 is the starting one, then it will be 9 5 2 2, so that doesnt work.</p>
<p>Then we go down to 8, which will yield 8 5 2 1. that works, so i put that into the highest integer which is 8521.</p>
<p>Ah, you are correct, good solution. anyone else have alternates?</p>
<p>Well, start with the greatest four digit integer which is:
9999</p>
<p>Then go to #1 where all the digits have to be different. The greatest number that fulfills this requirement would be:
9876</p>
<p>After that, look at #2: the greatest number that has four unique digits with the last three equaling the first is:
9810</p>
<p>The last part is the trickiest. You have to have the product divisible by 10. The factors of 10 are 1, 2, 5, 10. Unfortunately, 10 is not a digit, so that and 1 don’t work. Zero cannot be a digit, so you have to have a 2 and a 5 in the number. </p>
<p>Let’s use the following equation as a model:</p>
<p>X = w + z + y</p>
<p>X has to be greater than w, z, and y. (XWZY/10) has to equal floor(XWZY/10) according to number 3 and XWZY =/= 0. </p>
<p>let’s make w=5 and z=2 since you want the largest number possible (x52y is greater than x25y). </p>
<p>X = 2 + 5 + Y which also takes care of the divisible by 10 thing. </p>
<p>Lets try X = 9</p>
<p>9 = 2 + 5 + 2 </p>
<p>Digits W and Y are the same, so that doesn’t work. We should now try X = 8</p>
<p>8 = 2 + 5 + 1 </p>
<p>XWYZ = 80 </p>
<p>80/10 = 8 (an integer)</p>
<p>8521</p>
<p>This is the greatest number that fulfills every requirement.</p>
<p>and thats why, kids, wedgedawg got a 2400 :)</p>