1
$\begingroup$

So my mom- a teacher - called me and my brother and gave us each a paper and pencil.

She said: Think of three distinct single digit positive integers (0 to 9) and write down as many numbers (number combinations) you can make with those three without adding any math operations (like ^ for example).No repeating same digits in your numbers. So I want to see only single, double and triple digit numbers.Remember just numbers, no math operations.

We got busy. My brother got 26 different numbers. I got more! I was creative.

What did we both get?

Note: If you choose say a,b and c you cannot write aa or aab etc

$\endgroup$
1
  • $\begingroup$ This sounds like it should just be a polynomial but then it hits you with the 26. Very nice! $\endgroup$
    – xyldke
    Oct 5, 2022 at 14:38

1 Answer 1

2
$\begingroup$

At my first stab at this, I can get to 48 78. I avoided using superscript, assuming that that would violate your math rule.

We choose the digits:

5, 6, 7

Which can net us the following numbers:

756, 576, 765, 567, 675, 657

Flip the 6 around to get
759, 579, 795, 597, 975, 957

Tilt the 7 ever-so-slightly to get
156, 516, 165, 561, 615, 651

Flip the 6 and tilt the 7 to get
159, 519, 195, 591, 915, 951

Write the 5 very strongly on the back of the paper to show through and get
726, 276, 762, 267, 672, 627

Then the nonsense with the other two digits again:
129, 219, 192, 291, 912, 921
729, 279, 792, 297, 972, 927
126, 216, 162, 261, 612, 621

Then let's throw in the 1 and 2 digit numbers to the count (thanks @WeatherVane) for a bonus 30:

1, 2, 5, 6, 7, 9
12, 15, 16, 19
21, 26, 27, 29
51, 56, 57, 59
61, 62, 65, 67
72, 75, 76, 79
91, 92, 95, 97

Let's try to find 26 numbers. The regular permutations of non-zero digits gives 15 distinct numbers, which is 11 too few:

5, 6, 7, 56, 57, 67, 65, 75, 76, 567, 576, 657, 675, 756, 765

But then if we

arbitrarily choose the "flip the 6" from before, we get:
9, 59, 79, 95, 97, 597, 579, 759, 795, 957, 975

which gets us up to 26. Let's say that's where the overlap is, and that the OP found more by using one or more of the above tricks that I listed.

$\endgroup$
7
  • 1
    $\begingroup$ Not to mention 1 and 2 digit numbers, which the description allows. $\endgroup$ Oct 5, 2022 at 14:00
  • $\begingroup$ rot13(Ner qrpvzny cbvagf nyybjrq, be qb gubfr nyfb pbhag nf "zngu bcrengvbaf")? $\endgroup$
    – Ed Murphy
    Oct 5, 2022 at 14:34
  • $\begingroup$ Another option might be to rot13(jevgr lbhe qvtvgf yvxr n frira frtzrag qvfcynl naq pbzovar n fvk naq avar gb trg na rvtug) $\endgroup$
    – xyldke
    Oct 5, 2022 at 14:37
  • $\begingroup$ "What did we both get?" What 26 numbers did the brother get? Without any fudging, I can get 15 numbers (the regular perms excluding $0$ digit because $012$ and $12$ are not "different numbers"). $\endgroup$ Oct 5, 2022 at 19:11
  • $\begingroup$ You can convert an 8 to a double zero. $\endgroup$
    – Florian F
    Oct 5, 2022 at 20:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.