Riddler Classic for 8 June 2018

Spoiler for last week’s fivethirtyeight.com Riddler Classic:

From Ben Gundry via Eric Emmet, find and replace with a twist:

Riddler Nation has been enlisted by the Pentagon to perform crucial (and arithmetical) intelligence gathering. Our mission: decode two equations. In each of them, every different letter stands for a different digit. But there is a minor problem in both equations.

In the first equation, letters accidentally were smudged on their clandestine journey to a safe room within Riddler Headquarters and are now unreadable. (These are represented with dashes below.) But we know that all 10 digits, 0 through 9, appear in the equation.

What digits belong to what letters, and what are the dashes?

In the second equation, our mathematical spies have said that one of the letters in the equation is wrong. But they can’t remember which one. Which is it?

First puzzle:

The smudged letter business seems to be an obvious red herring; all 10 digits are used, but 6 letters appear and 4 dashes, so each dash is a different letter and might as well be replaced by four different letters, say: EXMREEK + EHKREKK = AKBHCXDE. Right away we know A = 1.

2E > 9 and 2K\mod 10 = E, so E = 6\textrm{ or } 8. But 2E\mod 10 is K = X-1 or X = K-1. So E = 6 with K = 3 and X = 2. D = 9.

Now 2+H\ge 9, so H = 7 or H = 8.

1+2R\mod 10 = C. The only possibility still available is R = 7 and C = 5 with H = 8. The two digits left are 0 and 4 and those are indeed what B and M work out to respectively. The sum is 6247663+6837633=13085296.

Second puzzle:

We’ll assume everything’s okay until we hit a contradiction. By exactly the same logic as before E = 1, Y= 6, D = 3. In column 7, D+Y=9 so T=9\textrm{ or }0. Since in column 5 2B\mod{10} cannot be 1, there must be a carry into that column, meaning T=9.

But in column 9, K+D\mod{10}=T so K=6 which is not available.

If one of the letters column 5 is wrong then there could be no carry into that column, in which case we could have T = 0, and then K=7. But from column 4 we’d still need B = 6\textrm{ or }7 neither of which is possible. So the problem lies elsewhere.

Maybe a letter in column 9 is wrong. Then R=0, and B=5. H=7. If there’s a carry into column 8 then M+7=P, which doesn’t work, so there’s no carry and M+6=P means M=2, P=8. We’re left with K=4 which gives the incorrect sum 695513243+673596633=1369109896; if the first two numbers are right the sum should be 1369109876. Or if the first number is right and the sum is right the second number should be 673596653. Or if the second number is right and the sum is right the first number should be 695513263. We can’t determine which letter is incorrect.

This doesn’t rule out other possibilities, such as that one of the letters in column 7 or column 10 is wrong.

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Riddlers for 4 May 2018

Spoilers for last week’s fivethirtyeight.com Riddler Express and Classic:

Express

From Charlie Drinnan, find the letters’ numbers:

If A, B, C, D and E are all unique digits, what values would work with the following equation?

ABC,CDE × 4 = EDC,CBA

No problem. A must be even, and 4A < 10, so A=2. Then 4AB < 100 so B < 5, and BA\mod 4 = 0 so B = 1 and E = 8, or B = 3 and E = 9. But 4E \mod 10 = A = 2 so E = 8.

Now 4D+3 \mod 10 = B = 1 so 4D \mod 10 = 8. D\ne 2 so D = 7.

Finally 4C+3 \mod 10 = C or 3C\mod 10 = 7, so C = 9.

219,978 \times 4= 879,912.

Classic

This is fairly mechanical although I tripped up several times on the way.

Let n_i be the number of coconuts found by the ith pirate, and n_8 be the number of coconuts found in the morning. n_8 is divisible by 7 so write n_8 = 7m_8.

But n_8 is what was left when the seventh pirate discarded one coconut out of n_7 and hid one seventh of the rest, so n_7 = (7/6)n_8+1 = (7/6)(7m_8)+1. For that to be an integer m_8 must be a multiple of 6, so write m_8 = 6m_7. Then n_7 = 7^2m_7+1.

Now n_6 = (7/6)n_7+1 = (7/6)(7^2m_7+1)+1. For integer value, 7^2m_7\mod 6 = m_7\mod 6 = 5, so write m_7 = 6m_6+5. Then n_6 = (7/6)(7^2(6m_6+5)+1)+1 = 7^3m_6+288.

You see how this goes.

n_5 = (7/6)n_6+1 = (7/6)(7^3m_6+288)+1. For integer value, m_6\mod 6 = 0, so write m_6 = 6m_5. Then n_5 = (7/6)(7^3(6m_5)+288))+1 = 7^4m_5+337.

n_4 = (7/6)n_5+1 = (7/6)(7^4m_5+337)+1. For integer value, m_5\mod 6 = 5, so write m_5 = 6m_4+5. Then n_4 = (7/6)(7^4(6m_4+5)+337)+1 = 7^5m_4+14400.

n_3 = (7/6)n_4+1 = (7/6)(7^5m_4+14400)+1. For integer value, m_4\mod 6 = 0, so write m_4 = 6m_3. Then n_3 = (7/6)(7^5(6m_3)+14400)+1 = 7^6m_3+16801.

n_2 = (7/6)n_3+1 = (7/6)(7^6m_3+16801)+1. For integer value, m_6\mod 6 = 5, so write m_3 = 6m_2+5. Then n_2 = (7/6)(7^6(6m_2+5)+16801)+1 = 7^7m_2+705888.

n_1 = (7/6)n_2+1 = (7/6)(7^7m_2+705888)+1. For integer value, m_2\mod 6 = 0, so write m_2 = 6m_1. Then n_1 = (7/6)(7^7(6m_1)+705888)+1 = 7^8m_1+823537. The smallest value corresponds to m_1 = 0 so n_1^{min} = 823537.

These pirates were either very sleepy, or incredibly quick at counting coconuts.

Riddler Classic for 23 March 2018 (part 2)

We’ve seen a generalization of the problem to multiplicative factors other than 2 and bases other than 10. A third generalization would be to shift more than one digit.

Take a positive integer j in base b and move its last m digits to the front. What is the smallest j such that when you do this, the result k is exactly p times j (where p is in the range [2,b-1 ] )?

Let j= b^m \times y+x, where b^{m-1}\le x< b^m . So the digits of x are the final m digits of j and the initial m digits of k, and the digits of y are the remaining digits of both. Let n be the number of digits in y. Then we require

p(b^my+x)= b^nx+y.

Solving, (pb^m-1)y=(b^n-p)x or

y= \frac{b^n -p}{pb^m-1}x .

Now let g=\gcd(pb^m-1,x) and let w=x/g, v=(pb^m-1)/g. Then

y= \frac{b^n-p}{v}w

and since v and w are relatively prime, we must have

b^n=p\mod v.

So we must calculate pb^m-1 and then, for each possible value of x, obtain v and look for the least n that satisfies the above relation. (The minimal value of x is in fact not b^{m-1} but pb^{m-1}, since there can be no carry to an additional digit when j is multiplied by p.) Then we can compute y, and verify that it does indeed have n digits. The smallest such y and associated x solve the problem.

If (pb^m-1) is prime, then v = (pb^m-1) for all possible x and the smallest k results when x=pb^{m-1}. But if (pb^m-1) has a common factor with some x, then v is some factor of (pb^m-1). Generally n, the number of digits in y, is on the order of v, so when (pb^m-1) is prime the solution tends to be many orders of magnitude larger than when it is not.

For instance, consider p=3 and b=10. For m=2, (pb^m-1)=299=13\times 23. In that case some values of x have a common factor with (pb^m-1). E.g., x=46 gives

y= \frac{10^n-3}{13}w

where

10^n=3\mod 13.

The solution is n=4, leading to

j=153846, k=3j=461538.

But for m=1, (pb^m-1)=29 is prime. The number of digits in y then turns out to be 27, and

j=1034482758620689655172413793.

And for m=3, (pb^m-1)=2999 is prime too. We end up with the 1499 digit solution

j = 100033344448149383127709236412137379126375458486162054018006\\ 0020006668889629876625541847282427475825275091697232410803601200\\ 4001333777925975325108369456485495165055018339446482160720240080\\ 0266755585195065021673891297099033011003667889296432144048016005\\ 3351117039013004334778259419806602200733577859286428809603201067\\ 0223407802600866955651883961320440146715571857285761920640213404\\ 4681560520173391130376792264088029343114371457152384128042680893\\ 6312104034678226075358452817605868622874291430476825608536178726\\ 2420806935645215071690563521173724574858286095365121707235745248\\ 4161387129043014338112704234744914971657219073024341447149049683\\ 2277425808602867622540846948982994331443814604868289429809936645\\ 5485161720573524508169389796598866288762920973657885961987329109\\ 7032344114704901633877959319773257752584194731577192397465821940\\ 6468822940980326775591863954651550516838946315438479493164388129\\ 3764588196065355118372790930310103367789263087695898632877625875\\ 2917639213071023674558186062020673557852617539179726575525175058\\ 3527842614204734911637212404134711570523507835945315105035011670\\ 5568522840946982327442480826942314104701567189063021007002334111\\ 3704568189396465488496165388462820940313437812604201400466822274\\ 0913637879293097699233077692564188062687562520840280093364454818\\ 2727575858619539846615538512837612537512504168056018672890963654\\ 5515171723907969323107702567522507502500833611203734578192730910\\ 3034344781593864621540513504501500500166722240746915638546182060\\ 6868956318772924308102700900300.

If you’d asked me to guess, before I worked any of this out, the order of magnitude of the smallest positive integer which is exactly tripled when you move the last three digits to the front, I’m pretty sure I would have seriously underestimated it.

Riddler Classic for 23 March 2018

Spoiler for last week’s fivethirtyeight.com Riddler Classic:

From Joseph Converse, a puzzle of digital manipulation:

Imagine taking a number and moving its last digit to the front. For example, 1,234 would become 4,123. What is the smallest positive integer such that when you do this, the result is exactly double the original number? (For bonus points, solve this one without a computer.)

One way

Write the number (call it m) as 10y+x, where x<10; for instance, for 1,234, y=123 and x=4. Let n be the number of digits in y; then the number resulting from moving the last digit to the front is 10^nx+y. We want 2(10y+x)=10^nx+y.

Solving, we get y = (10^n-2)x/19. Now, 19 is prime and x<10 so isn’t a multiple of 19, and therefore (10^n-2) must be, that is, 10^n\mod 19 = 2.

Going ahead and doing the long division ― that is, dividing 100, 1000, 10000… by 19 until a remainder of 2 is found ― we get n=17; (10^{17}-2)/19 = 5263157894736842. That has 16 digits, so x=1 won’t work; x=2 will give us a 17-digit y=10526315789473684, though. So the smallest number that works is m = 105263157894736842 = 210526315789473684/2.

Another way

We require 2m to have the same number of digits as m, and that means the first digit of 2m can’t be 1. So let’s try making it 2. Then that’s also the last digit of m.

Now, the last digit of 2m is 4, which is also the second to last digit of m. And then the second to last digit of 2m is 8, which must be the third to last digit of m. Then the third to last digit of 2m is 6, with a carry. That means the fourth to last digit of m is 6 and the fourth to last digit of 2m is 3, with a carry.

And so on, until we reach a place where the digit of m is 1 and 2m is 2 without a carry. Stop there. Follow that procedure and you get n = 105263157894736842 as before.

Except we don’t know that this is the smallest answer. We need to do the same with last digit of m running from 3 to 9. In fact (and as you’d expect having seen the other method) this is the smallest.

I’m rather surprised the second way didn’t occur to me sooner; it seems a good deal more obvious in retrospect. Then again, had I done it first I think I might’ve looked at the size of the result and been unconfident I hadn’t overlooked something that might give a smaller answer. The first method confirms it and makes it more clear, I think, that this really is the minimum. It also provides the insight that the answer is not just a mysterious sequence of digits but is connected with the decimal representation of 1/19.

Triple?

Instead of moving the digit to get multiplication by 2, what about by 3? Or by k generally? Adapting either method, one finds moving the last digit of 1034482758620689655172413793 to the front gives triple that number. For quadrupling we get a much smaller answer: 102564 = 410256/4. For quintupling we encounter for the first time the necessity of checking all possible last digits, not just the minimum one: Using 5 for the last digit we find 102040816326530612244897959183673469387755, but with 7 we get 142857! This is related to the fact that if you try the first method, you need y = (10^n-5)x/49, and 49 is not prime, so when x=7 we need 10^n \mod 7 = 5 rather than 10^n \mod 49 = 5.

Here’s a table for k = 1 through 9. I am glad the Riddler puzzle didn’t ask for 6 (by hand)… I did these with a spreadsheet.

1 1
2 105263157894736842
3 1034482758620689655172413793
4 102564
5 142857
6 1016949152542372881355932203389830508474576271186440677966
7 1014492753623188405797
8 1012658227848
9 10112359550561797752808988764044943820224719

All your base

Another generalization of the original puzzle is to do it in base b instead of base 10. There’s no solution in binary, as you can figure out by trying to adapt either method above, or just by noting that doubling a number in binary means appending a 0 to the right end, so you can’t double a number and get a number with the same number of binary digits (bits). In base 3, starting with final digit 2 for m, the final digit for 2m is 1 with a carry of 1; 1 for the next digit of m gives 0 for the next digit of 2m with a carry; 0 for m gives 1 for 2m; 1 for m gives 2 for 2m and we’re done: 1012_3 = 2101_3/2. In decimal, that’s 32 and 64. (In higher bases we need to check last digits from 2 to b-1.) Alternatively, we want y = (3^n-2)x/5 and so require 3^n \mod 5 = 2. n = 3 works and then y = 5x. So with x = 2, y = 10 = 101_3 and m = 1012_3.

In all bases from 3 through 10:

3 1012
4 102
5 13
6 1031345242
7 103524563142
8 25
9 10467842
10 105263157894736842

Edit to add:

And if this Python script is bug free, then

\mathrm{10179b72334c73ad1496e2dbd2c4365203047e46699e76b293dd5c8b5986ca406090d8cd44}\\  \mathrm{ded7537cbba27b41da580c131c2ab89cebea70a88550783c5b119263956824ae8e5e1621aa_{15}}

is the smallest base 15 number which is multiplied by (decimal) 10 when the last digit is moved to the front. At 148 digits, it’s the longest such number for all bases up to 16 and all factors up to (base-1).