Number Puzzle Help - Why does 5th grader get such hard homework assignment.

Nearly impossible or challenging? Maybe your son's teacher has high expectations and thinks that your son (and/or his whole class) can cope with it. What's the point of setting something that doesn't make your pupils think? I hope you aren't going to do his homework for him.

By the way, I was a teacher until I retired a month ago.

I don't know why Pete believes the largest 3-digit number is 900.

(In the final analysis, it turns out that he is right. The largest 3-digit number is 879. But how can we know that a priori?)

Since each digit must be unique, I believe the range of 3-digit numbers is 203 to 987, noting that the minimum 4-digit sum is 1023 and the maximum sum is 1640 (987+653). That is, 1 is always the most-significant digit of the 4-digit sum.

So, the max number of brute-force iterations is 616,225 (987-203+1 = 785, squared).

However, the inner loop can be conceptually y=MAX(1023-x0, 203) to 987. That reduces the number of brute-force iterations to 425,572.

I assume that 246+789 = 1035 is the same as 789+246 = 1035. So, I believe the number of solutions is 48, not 96.

The VBA procedure below finds the 48 solutions in less than 0.15 sec on my computer. (YMMV) I don't know if it qualifies as ``more efficient algorithm that works a little nicer and faster``.

If you want to count "duplicates", change #If 1 to #If 0. Ironically, it takes the same amount of time to find 96 solutions.

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That said, I believe an optimal algorithm requires at most 47,040 iterations. That is 8*7*7*6*5*4. 8*7 because the first digit of each 3-digit number is 2 to 9, and they are unique. 7*6*5*4 because the subsequent digits can include 0, and again, they are unique.

But I suspect that implementation is messier. (TBD)

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My brute-force implementation....

Errata.... Change Me.UsedRange.Clear to ActiveSheet.UsedRange.Clear if you put the procedure into a normal module instead of a worksheet module, as I did.

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A much more straight-forward approach, building on the previous analysis.

1. Generate all 3-digit numbers with unique digits between 203 and 987. 575 qualified numbers of 800 combinations.

2. Generate the sums of all pairs of 3-digit numbers between 1023 and 1640 (987+653), and check for uniqueness of all digits. 165,025 iterations.

The algorithm directly generates only the 48 unique pairs that qualify. No need to check for duplicate pairs.

It executes in about 0.05 sec on my computer. (YMMV)

Again, change Me.UsedRange.Clear to ActiveSheet.UsedRange.Clear if you put the procedure into a normal module instead of a worksheet module, as I did.

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Sorry for the incessant postings; and I might be beating a dead horse. But I had an epiphany that might be helpful to future readers who must deal with "fifth grade" math.

The problem might not be so intractable for young children who are taught "common core math". I don't know much that. But I believe it teaches students to look at the structure of numbers, not just the mechanics of arithmetic.

So the expected reasoning might go something like this....

1. The 4-digit number (z) must be 1023 to 1640 (897+653), because digits must be unique.
Thus, z is of the form 10gh, 12gh, 13gh, 14gh, 15gh or 16gh.

2. Also, #1 ==> ("implies") given x=abc and y=def ("is of the form of"),
a+d or a+d+1 is 10, 12, 13, 14, 15 or 16.
a+d+1 results from a carry from b+e.

3. Suppose z=10gh.
Then a+d is 2+8, 3+7, 4+6 if b+e<10 (no carry),
or a+d is 1+8, 2+7, 3+6, 4+5 if b+e>9 (carry)

4. Suppose z=10gh and x=3bc and y=7ef.
That leaves 2,4,5,6,8,9 for the remaining digits.
And #3 (b+e<10) ==> b+e is 2+4 or 4+5 if c+f<10 (no carry),
or b+e is 2+5 if c+f>9 (carry),
or b+e is 2+6 regardless of c+f (carry or not).

5. Suppose x=32c and y=76f and z=108h (c+f<10; no carry).
That leaves 4,5,9 for the remaining digits.
Thus c=4, f=5 and h=9 (c+f).

So, x=324, y=765 and z=1089.

And due to commutativity, we also have 3 other solutions, to wit:
x=325, y=764 and z=1089
x=365, y=724 and z=1089
x=364, y=725 and z=1089

Of course, I got lucky with my choices. (wink)

The solution might require more trial-and-error, iterating steps #3-5 for different choices of a+d and b+e, as well as for z (12gh, 13gh, etc).

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And to that end, it might be useful to note another structural relationship that we did not need for z=10gh.

If z<>10gh, one 3-digit number includes zero.

But zero cannot be the last digit, because xx0+yyy = zzzy. The last digit is not unique.

Thus, if z<>10gh, x=a0c.

And c+f>9 (carry) must be true, because otherwise, x0x+yyy = zzyz. The middle digit is not unique.

(Similarly for y. But since A+B = B+A (commutativity), it does not matter which 3-digit number we choose to include zero.)

Another brute-force takes 0.02 s

i + k = t and i > k, so i start at 502
1xxx reserve for total
234567890 available for the rest

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