<p>What is the greatest four digit integer that meets the following three restrictions?
1. All of the digits are different.
2. The greatest digit is the sum of the other three digits.
3. The product of the four digits is divisible by 10 and not equal to zero.</p>
<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>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>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>