I’ve started to develop a reputation as something of a brain-teaser pro, given some of the beastly brain teasers we’ve featured on the blog over the last few months.

And, as such, I’ve started to receive brain teasers from friends and fellow puzzlers, challenging me to unravel them AND explain my methods to the PuzzleNation audience.

I’ve never been one to shirk a challenge, so here we go! This puzzle is entitled Mystery Number, and a little googling after solving it reveals it most likely came from this *Business Insider* link. (Although their solution is slightly flawed.)

Enjoy!

There is a ten-digit mystery number (not starting with zero) represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?

1. A + B + C + D + E = a multiple of 6.

2. F + G + H + I + J = a multiple of 5.

3. A + C + E + G + I = a multiple of 9.

4. B + D + F + H + J = a multiple of 2.

5. AB = a multiple of 3.

6. CD = a multiple of 4.

7. EF = a multiple of 7.

8. GH = a multiple of 8.

9. IJ = a multiple of 10.

10. FE, HC, and JA are all prime numbers.

(And to clarify here for clues 5 through 9, AB is a two-digit number reading out, NOT A times B.)

[Image courtesy of Wikipedia.]

Now, anyone who has solved Kakuro or Cross Sums puzzles will have a leg up on other solvers, because they’re accustomed to dealing with multiple digits adding up to certain sums without repeating numbers. If they see three boxes (which would essentially be A + B + C) and a total of 24, they know that A, B, and C will be 7, 8, and 9 in some order.

[For those unfamiliar with Cross Sums or Kakuro solving, feel free to refer to this solving aid from our friends at Penny/Dell Puzzles, which includes a terrific listing of possible number-combinations that will definitely prove useful with this brain teaser.]

And since the digits 0 through 9 add up to 45, that provides a valuable starting hint for clues 1 and 2 (in which all 10 digits appear exactly once). A multiple of 6 (6, 12, 18, 24, 30, 36, 42) plus a multiple of 5 (5, 10, 15, 20, 25, 30, 35, 40, 45) will equal 45. And there’s only one combination that works.

So A + B + C + D + E must equal 30, and F + G + H + I + J must equal 15.

The same logic applies to clues 3 and 4 (in which all 10 digits appear exactly once). A multiple of 9 (9, 18, 27, 36, 45) plus a multiple of 2 (2, 4, 6, 8, 10, etc.) will equal 45. And there’s only one combination that works.

So A + C + E + G + I must equal 27, and B + D + F + H + J must equal 18.

And now, we jump to clue 9. Since IJ is a multiple of 10, and all multiples of 10 end in 0, we know J = 0.

This tells us something about JA in clue 10. J is 0, which means A can only be 2, 3, 5, or 7.

There may a quicker, more deductive manner of solving this puzzle, but I couldn’t come up with it. I went for a brute force, attrition-style solve.

So I wrote out all of the possibilities for clues 5 through 9, and began crossing them off according to what I already knew. Here’s what we start with:

AB = 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99

CD = 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96

EF = 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98

GH = 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

Now, we can remove any double numbers like 33 because we know each letter represents a different number.

AB = 12, 15, 18, 21, 24, 27, 30, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96

CD = 12, 16, 20, 24, 28, 32, 36, 40, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 92, 96

EF = 14, 21, 28, 35, 42, 49, 56, 63, 70, 84, 91, 98

GH = 16, 24, 32, 40, 48, 56, 64, 72, 80, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

[Sorry guys, you’re out.]

And we know that J = 0, so we can remove any numbers that end in zero for AB, CD, EF, and GH.

AB = 12, 15, 18, 21, 24, 27, 36, 39, 42, 45, 48, 51, 54, 57, 63, 69, 72, 75, 78, 81, 84, 87, 93, 96

CD = 12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96

EF = 14, 21, 28, 35, 42, 49, 56, 63, 84, 91, 98

GH = 16, 24, 32, 48, 56, 64, 72, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

And for AB, we know that A can only be 2, 3, 5, or 7, so we can delete any numbers that don’t start with one of those four digits.

AB = 21, 24, 27, 36, 39, 51, 54, 57, 72, 75, 78

CD = 12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96

EF = 14, 21, 28, 35, 42, 49, 56, 63, 84, 91, 98

GH = 16, 24, 32, 48, 56, 64, 72, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

Hmmm, that’s still a LOT of options. What else do we know?

Well, we know from clue 10 that FE and HC are prime numbers. So they can’t be even numbers OR end in a 5. So we can eliminate any options from CD and EF that begin with an even number or a 5.

AB = 21, 24, 27, 36, 39, 51, 54, 57, 72, 75, 78

CD = 12, 16, 32, 36, 72, 76, 92, 96

EF = 14, 35, 91, 98

GH = 16, 24, 32, 48, 56, 64, 72, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

Alright, now we need to look at those big addition formulas again. Specifically, we need to look at B + D + F + H + J = 18.

We know J = 0, so the formula becomes B + D + F + H = 18. Now, take a look at our lists of multiples for AB, CD, EF, and GH. Look at the second digit for each. There’s a little nugget of information hiding inside there.

Every D and H digit is an even number. Which means that B and F must either both also be even, or both be odd in order to make an even number and add up to 18.

But, wait, if they were both even, then they would use all of our even numbers, and some combination of B, D, F and H would be 2 + 4 + 6 + 8, which equals 20. That can’t be right!

So let’s delete any even numbered options from AB and EF.

AB = 21, 27, 39, 51, 57, 75

CD = 12, 16, 32, 36, 72, 76, 92, 96

EF = 35, 91

GH = 16, 24, 32, 48, 56, 64, 72, 96

IJ = 10, 20, 30, 40, 50, 60, 70, 80, 90

Okay, we’ve whittled down EF to 2 possibilities: 35 and 91. [Here is where the *Business Insider* solution goes awry, because they never eliminate one of these two options.]

Clue 10 tells us that FE is a prime number, but that doesn’t help, because both 53 and 19 are prime. So now what?

Let’s return to those starting formulas.

We know that A + B + C + D + E = 30, and our handy-dandy number-combination listing tells us there are six possible ways that five digits can add up to 30: 1-5-7-8-9; 2-4-7-8-9; 2-5-6-8-9; 3-4-6-8-9; 3-5-6-7-9; and 4-5-6-7-8.

Look at the possibilities for A, B, C, D, and E according to our work thus far:

AB = 21, 27, 39, 51, 57, 75

CD = 12, 16, 32, 36, 72, 76, 92, 96

EF = 35, 91

There’s not a single 8 in any of those pairings! And five of our six possible answers for A + B + C + D + E = 30 include an 8 as one of the five digits.

Therefore, 3-5-6-7-9 and A-B-C-D-E match up in some order.

EF is either 35 or 91, but with both 3 and 5 counted among the letters in A-B-C-D-E, EF cannot be 35, so EF is 91. Let’s eliminate any option for AB, CD, GH, or IJ that include 9 or 1.

AB = 27, 57, 75

CD = 32, 36, 72, 76

EF = 91

GH = 24, 32, 48, 56, 64, 72

IJ = 20, 30, 40, 50, 60, 70, 80

Because E = 9, that leaves 3, 5, 6, and 7 as the only possible digits available for A, B, C, and D. So let’s eliminate any combinations that use numbers other than those four.

AB = 57, 75

CD = 36, 76

EF = 91

GH = 24, 32, 48, 56, 64, 72

IJ = 20, 30, 40, 50, 60, 70, 80

We can also eliminate any combinations for GH and IJ that include those four numbers.

AB = 57, 75

CD = 36, 76

EF = 91

GH = 24, 48

IJ = 20, 40, 80

Since our only possibilities for AB use 5 and 7 in some order, CD cannot be 76, so it must be 36.

AB = 57, 75

CD = 36

EF = 91

GH = 24, 48

IJ = 20, 40, 80

So, here are our options at this point:

AB = 57, 75

CD = 36

EF = 91

GH = 24, 48

IJ = 20, 40, 80

All possible solutions for GH include the number 4, so we can delete 40 as a possibility for IJ.

AB = 57, 75

CD = 36

EF = 91

GH = 24, 48

IJ = 20, 80

Let’s look at those formulas one more time. We know A + C + E + G + I = 27.

We also know C = 3 and E = 9, so A + G + I = 15. And the only combination of available digits that allows for that is 5, 2, and 8, meaning AB = 57, GH = 24, and IJ = 80.

So ABCDEFGHIJ = 5736912480.

I don’t think I’ve tackled a puzzle this tough since the seesaw brain teaser!

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Wow this is really cool but how did you know J=0, couldn’t I also be 0?

No, I couldn’t be zero because the two-digit number represented by IJ was a multiple of 10. All multiples of 10 — 20, 30, 40, etc. — end in 0, so J has to be the zero. If I is the zero, then you have 01, 02, 03, etc. as options, and none of those are multiple of ten.

Also, each number 0-9 is only used once, so IJ couldn’t be 00.

Oh ok got it, thanks

No worries, happy to explain!