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Solution to EST MODUS IN REBUS

Oct 28th, 2006 by tony

This is going to be quite a long post, since the solution is quite involved. Here is the original problem.

This is the diagram in question:

Est modus In rebus Math Puzzle

First, we note that there are both alphabets and asterisks in the equation. We see that there are 12 different letters in total, so while the same letters always represent the same digit, each digit may be represented by more than one letter. For example, while “E”s may always represent “3″, other letters, such as “M”, may also represent “3″.

To simplify our analysis, let’s call the eight digit multiplier (ESTMODUS) X and the other (INREBUS) Y. Since there are seven partial products (the seven lines of asterisks in the middle), we can deduce that there are no “0″ in Y. We also note that the last two digits of X, Y, and their product are the same, namely “US”. Thus, the square of “US” must also end in “US”; for example, 25^2=625 and 76^2=5776. It turns out that these are the only two possibilities (among two digit numbers, which we are looking for here), and so “US” must be 25 or 76. To prove this point, let the two digit number “US” be z, then the last few digits of the product X \times Y must be z^2. But the last two digits of z^2 is just z, therefore z^2-z=z\left(z-1\right) must be divisible by 100. From this, we can see that either z is divisible by 4 and z-1 is divisible by 25, or z is divisible by 25 and z-1 is divisible by 4. In the first possibility, z=76; in the other possibility, z=25. Q.E.D.

In order to determine whether our z is 76 or 25, we call the first six digits in X (other than the last two digits) x, call the first five digits in Y (other than the last two) y, then clearly

X=\left(100 \times x\right) + z
Y=\left(100 \times y\right) + z

Since only the third partial product out of the seven contain 8 digits (the others have 9), we conclude that the unit digit in y, which is the B, is the smallest among INREBUS, and that B \times E < 10. If B \times E \ge 10, this third partial product would not only contain eight digits, since E is the highest digit in X. Now we look at the fourth partial product (which is the product of the "E" in y and ESTMODUS). From the fact that this partial product has nine digits, we know that E>2, because if E \not> 2, then this partial product can at most contain eight digits. From this, we deduce that B can only be 1, or 2. If B>3, we will contradict our previously established fact that B \times E < 10.

Because X \times Y has the same last seven digits as the multiplier Y, XY-Y=Y(X-1) must be divisible by 10^7. Thus, if z=76, then it is obvious that Y is not divisible by 5 but divisible by 2^7=128, X-1 is not divisible by 2 but divisible by 5^7=78125. Because Y=100y+76, X-1=100x-1+76=100x+75, therefore Y \div 4 =25y+19 must be divisible by 128 \div 4 = 32, and (X-1) \div 25=4x+3 must be divisible by 78125 \div 25=3125. Let 4x+3=3125t. Since 3125 \equiv 1 \pmod 4, then t \equiv 3 \pmod 4 (because 4x+3 \equiv 3 \pmod 4). Let u and v be integers such that 3125=4u+1 and t=4v+3. Then
4x+3=\left(4u+1\right)\left(4v+3\right)=4\left(4uv+v+3u\right)+3.
Therefore, x=3125v+2343.

We mentioned before that 25y+19 is divisible by 32, and thus divisible by 4. But 25y+19=25(y-1)+44, so y-1 is divisible by 4. We also shown before that E>2 (second last digit of y) and B can only be 1, 2, or 3. In order to satisfy the condition y-1 is divisible by 4, B cannot be 2, and E can only be 4, 6, or 8. It follows that X is one of the following possibilities:
4 6 * * * * 7 6
6 6 * * * * 7 6
8 6 * * * * 7 6
Considering our result from above, x=3125v+2343, from the first possibility above, we have x is either 461718, 464843, or 467968; from the second possibility above, we have x is 661718, 664843, or 667968; from the third possibility above, we have x is 861718, 864843, or 867968. From these possibilities, we note that due to the fact that each of these have multiple copies of many of the digits, we won’t get all ten digits in X and Y. Also, there isn’t a “0″ anywhere in X, which is not reasonable. From these clues, we can claim that z=76 is not a viable choice, and turn to look at the possibility of z=25.

When z=25, Y is divisible by 5^7=78125 and X-1 is divisible by 2^7=128. Since Y=100y+25, X-1=100x-1+25=100x+24, then 4y+1 is divisible by 3125, 25x+6 is divisible by 32. x must be even and it is trivial to show shat x \equiv 10 \pmod {32}. From the results above, 4y+1=3125w, but 3125 \equiv 1 \pmod 4. In order for 3125w \equiv 1 \pmod 4 to be true, w \equiv 1 \pmod 4. Let w=4u+1, we have

y=3125u+\dfrac{3125-1}{4}=3125u+781

Because we determined the last digit of y (B) to be 1 or 2, u can’t be odd, and consequently we get B=1. Thus, y \equiv 781 \pmod {6250}. Since Y does not contain “0″, and its highest digit “I” cannot be “1″, and that there cannot be two set of identical digits in Y, we find that Y is 6328125, 6953125, or 9453125 through trial. For the last two possibilities, E=3, which means that the second partial product cannot contain nine digits. Thus, we can determine that Y=6328125. But what is X? From our value for Y, we see that X should be 8 5 * * * * 2 5 where each * is one of 0, 4, 7, or 9. Because x \equiv 10 \pmod {32}, D must be even (0 or 4); the second last digit of x (O) must be odd (7 or 9). From trial and error, we see that out of the eight possibilities, we must have T=0, M=7, O=9, and D=4. Therefore, X=85079425, and we are done!

To conclude, given that the last two digits of the multiplier are the same, and the last seven digits of the product is the same as that of one of the multipliers, this problem didn’t seem too difficult in the beginning. But the solution involves detailed analysis and modular arithmetic, making the decryption much more complex. You really need a lot of patience to solve this thing!

Posted in Solutions | 2 Comments

2 Responses to “Solution to EST MODUS IN REBUS”

  1. on 28 Oct 2006 at 6:22 pm   Percy

    Yikes!

  2. on 22 Sep 2008 at 2:06 am   Mary

    Wow!

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