 Quandaries and Queries Who is asking: Parent Level: Elementary Question: My daughter's 4th grade teacher sent home a weekly challenge asking her to make as many equations as possible to make the number 99999 using all of the numbers 0-9 but only once per equation. example 01234 + 98765 = 99999 she needs to make 150+ equations in order to get an A. Is there a formula to use? Can you show me the work? Hi Lisa, Here is a procedure to generate lots of equations: the key idea is to combine digits that add to 9 (e.g., 4+5=9), like you did in your example (e.g., 01234+98765=99999), in which case, there is never any carrying when you add. Thus, it doesn't matter what order you place the digits, as long as you keep the appropriate order in the second number. So, here are a few examples, which were generated by shifting the digits one place: 01234+98765=99999 40123+59876=99999 34012+65987=99999 23401+76598=99999 12340+87659=99999 Note that this procedure produces 5 possibilities. By being careful not to generate duplicates, we can repeat the procedure using a different arrangement of 0,1,2,3 and 4. Provided we change the order, this will produce distinct possibilities. For example: 01243+98756=99999 30124+69875=99999 43012+56987=99999 24301+75698=99999 12430+87569=99999 Again, this produces a list of 5 distinct possibilities. We can do the above procedure, and always produce 5 distinct possibilities, using the following arrangements of 0, 1,2, 3 and 4 as the starting number: 01234, 01243, 01324, 01342, 01423, 01432 02134, 02143, 02314, 02341, 02413, 02431 03124, 03142, 03214, 03241, 03412, 03421 04123, 04132, 04213, 04231, 04312, 04321 Each of these 24 possible starting numbers produces 5 distinct possibilities, so we have now generated a list of 24x5=120 distinct possibilities. It is easy to double this list by swapping digits between the two numbers. For example, swapping the first digits of 01234+98765=99999 produces 91234+08765=99999 Swapping the first digit for each of the 120 already generated produces 120 more distinct possibilities. So we now have 240, which is more than the required 150+ required, so I'll quit. But it should be mentioned that there are a heck of a lot more possibilities than 240. To find how many requires a careful counting procedure, which is probably beyond the ability of a grade 4 student to comprehend. Being careful not to generate duplicates there are (10x8x6x4x2)/2=1920 distinct possibilities. Paul Go to Math Central