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Category: | Math |

Submitted By: | javaguru |

Fun: | (2.17) |

Difficulty: | (3.4) |

What is the smallest integer such that if you rotate the number to the left you get a number that is exactly one and a half times the original number?

(To rotate the number left, take the first digit off the front and append it to the end of the number. 2591 rotated to the left is 5912.)

x 1.5 =

1,764,705,882,352,941

Here's how you find the number:

You needed to find a repeating fraction such that the number of repeating digits is one less than the number. This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e., every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).

Now you need to find any occurrences where the nth digit is followed by the 1.5nth digit. 'n' will obviously be even.

The repeating digits in 1/7 and their corresponding 'n' where the digits are the first digit after the decimal point are:

142857 - repeating digits

132645 - n

The ordered pairs of adjacent values of n are (1,3), (3,2), (2,6), (6,4), (4,5) and (5,1). None of these has the second number equal to one and a half times the first number.

The next prime number p with a repeating fraction for 1/p containing p-1 repeating digits is 17. The repeating digits for 1/17 are:

.0588235294117647

The easiest way to assign 'n' for each digit is to list all of the digit pairs in order:

XX - n (digit)

05 - 1 (1)

11 - 2 (11)

17 - 3 (12)

23 - 4 (5)

29 - 5 (8)

35 - 6 (6)

41 - 7 (10)

47 - 8 (15)

52 - 9 (7)

58 - 10 (2)

64 - 11 (14)

70 - 12 (16)

76 - 13 (13)

82 - 14 (4)

88 - 15 (3)

94 - 16 (9)

Checking the even-numbered values of n in the above table reveals that ALL even values of n less than 2/3 of 17 are followed by the 1.5n digit. The values of n are and their digit places are:

2,3 (11,12)

4,6 (5,6)

6,9 (6,7)

8,12 (15,16)

10,15 (2,3)

So the five smallest numbers with this property are:

1176470588235294

2352941176470588

3529411764705882

4705882352941176

5882352941176470

Notice that 2352941176470588 can be rotated TWICE, yielding 1.5 times the number each time since the 4->6->9 n digits are in order.

Show Hint | Hide Answer |

Posted by ronhoward | 02/17/09 |

Wow, that's amazing! :o I had the right idea--I tired the 7ths first, but I just didn't take it far enough. Excellent teaser, keep 'em coming! :D | |

Posted by tamjp | 02/19/09 |

:o I always feel moderately intelligent and capable... Until I do teasers in the Math Category! :oops: :wink: | |

Posted by precious1026 | 02/22/09 |

:evil: This Teaser is absurd. In mathematics before you can calculate any number, solve a problem, or figure percentages, probability or calculus, there are given numbers. You cannot just think up a number and expect others to guess what you re thinking. You are unfair to give such a useless problem and go through half a page to solve something that only you know. I don't like this teaser. Why cheat yourself and everyone who looks forward to learning something new. :evil: :evil: 8) | |

Posted by javaguru | 02/22/09 |

If it is so hard, why did you not rate it as difficult? The idea for this came from another math teaser named "50% Off". I figured out how to solve that teaser, so I know it is possible to figure it out without having created the teaser. 8) | |

Posted by piratechicken92 | 02/22/09 |

WAAAAAAAAAAAAAAAAAAAAAAAYYYYYYYYYYYYYYYYYYY TOO HARD | |

Posted by KarateGirl098 | 02/23/09 |

Hard, but a great teaser! :D | |

Posted by javaguru | 02/23/09 |

Thanks! :D BTW, I got the name of the teaser this was based on wrong. The name is "50% Larger". | |

Posted by dancingd3 | 02/24/09 |

cool teaser, i didnt even try tho :oops: it was just gonna b 2 hard, i didnt even kno where 2 start! im just a sixth grade math student. honors math tho! haha :D | |

Posted by PiRsquared | 02/25/09 |

Wow! That is really cool, but really difficult. I like how the hint gave the suggestion about the repeating digits--I feel a little stupid that I didn't catch that part of the hint. :oops: Probably wouldn't have figured it out even if I'd caught the hint, but wow, just wow. :o Great teaser tho'!!! :D | |

Posted by stil | 03/03/09 |

I like math curiosities, but wish they didn't need to be classified as brain teasers to appear on this site. 7058823529411764 does not possess the basic properties required. One-and-one-half times this number requires 17 digits. (Curiously with the same trailing 15 digits sandwiched between a 1 and a 6.) Because the answer wasn't precisely confined to natural numbers, there are decimal answers. 0.0 is too tricked out to give serious consideration, but the decimal answer 1.176470588235294 is easily built a digit at a time. 150% of 1.0 is 1.5. 150% of 2.0 is 3.0. 150% of 1.10 is 1.65. 150% of 1.20 is 1.80. 150% of 1.160 is 1.740. 150% of 1.170 is 1.755. 150% of 1.180 is 1.770. 150% of 1.1750 is 1.7625. 150% of 1.1760 is 1.7640. 150% of 1.1770 is 1.7655. 150% of 1.17640 is 1.76460. ... 150% of 1.176470588235294 is 1.764705882352941. ... | |

Posted by javaguru | 03/03/09 |

I submitted a correction for the wrong value the moment the teaser was approved. Someday maybe the editors will get around to reviewing the correction. The correct value is 5882352941176470, which you can see if you follow the explanation. You decimal solutions don't satisfy the requirements of the teaser. By removing the number from the front and rotating the number you move the decimal place, and thus don't achieve the goal. I found this teaser by working it exactly as stated in the solution. I got the idea from a couple of other teaser that involved rotating numbers using fractions of both 7 and 13. All of these teasers I solved by discovering the relationships required in this teaser, so I'd counter your claim that this is simply a math curiosity. My "12-Digit Oddities" teaser is more of a math curiosity, although you can discover the common feature of the numbers as easily as many of the more esoteric group and series teasers. To each his own. 8) | |

Posted by stil | 03/03/09 |

javaguru- You've inferred an attack I did not imply. It definitely is not "simply a math curiosity." It is a spectacular curiosity of very high degree! It is excessive (?) :roll: experience that makes me question its status as brain teaser. Until a couple of decades ago, I only saw such direct questions classified as puzzles. :oops: I had hoped you would let my use of the decimal slide. You led the way by shifting the FIVE commas in your answer. --- :wink: But seriously, I should have introduced it as a normalized significand (mantissa). Can you explain (or give reference) why the inverse of any prime should hold the answers? | |

Posted by javaguru | 03/06/09 |

Sorry Stil if I got ruffled by your comment, it was a linguistic misinterpretation. I agree with your distinction between puzzle and teaser, but everything on this site is called a teaser so I'm just going with the local lingo. My distinction between puzzle and math curiosity is whether the answer can be discovered through reasoning or only through a non-trivial search. A pure math curiosity may be interesting, but is not a puzzle. Oh, and my commas are formatting added after the rotation. Your decimal point is part of the number. :P Seriously though, I should have restricted the answers to natural numbers. I meant to do this, but I didn't. :-? | |

Posted by javaguru | 03/06/09 |

As to the question regarding the answers being exclusive to inverses of prime numbers, they actually may not be, but this doesn't come into play until you get to much larger numbers. I didn't figure on complicating an already complex explanation with this, but since you asked... It can easily be seen that the longest possible repeating fraction for 1/p is p-1. This is so because the repeating fraction is based on remainders and there are only p-1 distinct remainders. Next, the number of repeating digits in 1/p, where p is prime in this case, must be a factor of p-1. This means that if it isn't p-1, then it can at most be (p-1)/2. In this case, where r is the number of repeating digits, then there will be (p-1)/r different sets of r digits where each digit in each set is the first digit in the decimal fraction exactly once. Proving this is beyond what fits in this comment. If p is composite, then the number of repeating digits will be a multiple of the number of repeating digits r in 1/q where q is one of the prime factors of p-1. The number of repeating digits will not be less than the greatest number of repeating digits for any prime factor of p-1. Further, the number of repeating digits will be m*r where m is the product of one of the factors of each of the prime factors of (p-1)/r. (The factors of a prime number are only 1 and the number itself.) OK, so all that may be interesting, but why is the answer to this teaser limited to being from the repeating fraction of 1/p where the number of repeating digits is p-1? Well, it might not. My theory is that the answers are limited to being from 1/x where the number of repeating digits is greater than (x-1)/2. The first composite number for which this is true is 1/49. 1/7 has six repeating digits and 1/49 has 7 * 6 = 42 repeating digitis. It happens that 49 does not provide a solution, but I don't know that other such numbers can't. At any rate, after finding this result in the manner described in the solution, I did verify that there are no other solutions with 13 or fewer digits. It'll take quite a while longer on my computer to finish checking up to the answer, but I'm confident that none will be found. There's a lot of interesting and weird properties of repeating fractions. 8) | |

Posted by EntangledQuark | 03/16/09 |

I would have never gotten it without the hint. When I saw the hint, I realized that a repeating fraction with 16 digits could provide the answer. Dividing by 17 was the most likely to work. Then I was able to find the answer by working through the different multiples of 1/17. Without knowing the answer was 16 digits I don't think I would ever have thought to use repeating fractions, but certainly I wouldn't have thought the number was that big. Fantastic puzzle and really interesting mathematical trivia. :D | |

Posted by javaguru | 03/23/09 |

After thinking about Stil's question regarding why the answer need to be the inverse of a prime, I realized that there is a much easier way to find the answer. In fact, you can generalize the problem such that you can find all solutions for "1/X again as big" (i.e. 1/2 again as big, 1/6 again as big, etc.) First, there can only be one answer starting with any specific digit. This number can be determined by successive approximation similar to Stil's approach above with 150% of the decimal fraction. So to get half again as big starting with a 1, the next digit must be 1 x 1.5 rounded down (1) or 1 x 1.5 + 1 rounded down (2). If the first two digits are 11, then the second and third digits must be 11 x 1.5 or 11 x 1.5 + 1, rounded down (16 or 17). Where D is the first digit, this can be generalized as the fraction: D / (10 - 3/2) = D / (8.5) so for D = 1 this is 1 / 8.5 = 2/17. This can be further generalized for any 1/X again as big as: D / (10 - (X+1)/X) = D / ((10X - (X+1)) / X) = D / ((9X - 1) / X) = D * X / (9X - 1) where 9X - 1 is not divisible by either 2 or 5. The requirement for non-divisibility by 2 or 5 is because those fractions will have a non-repeating portion of the fraction, and thus can't be rotated because the non-repeating portion of the fraction will be rotated. This gives a solution for every even value of X that does not end with a 4. The first several smallest values for rotating one to the left and getting a number that is 1/X larger are: 2 = 2/17 = 1176470588235294 x 3/2 = 1764705882352941 6 = 6/53 = 1132075471698 x 7/6 = 1320754716981 8 = 8/71 = 11267605633802816901408450704225352 x 9/8 = 12676056338028169014084507042253521 8) | |

Posted by javaguru | 03/23/09 |

By the way, one other consequence of the approach above is that the next smallest (6th smallest) answer for half-again-as-big is 11764705882352941176470588235294 x 3/2 = 17647058823529411764705882352941 8) | |

Posted by Paladin | 04/09/09 |

Wow... tricky and challenging. Great explanations though. | |

Posted by c3po | 04/20/09 |

When I first read the answer, I thought this was going to be a case of just having to know the answer, but I see that it was possible to figure this out. Just not possible for me. :) Very nice teaser. The best of the really hard math teasers. | |

Posted by TZCastor | 10/05/09 |

Waaaaaay too confusing. :o | |

Posted by dalfamnest | 10/13/09 |

Very tough, but fascinating. Well written and explained! :) In fairness, we are warned about difficulty levels before we open these teasers, so why criticise a writer for making something too difficult? It's not really designed as a TOTD, of course! I am interested to know, javaguru, how you knew how precious1026 rated your teaser. I thought our ratings would be confidential! :-? | |

Posted by javaguru | 10/28/09 |

To dalfamnest: I knew because I happened to refresh the page at almost the same time precious1026 left the comment. From one minute to the next the difficulty rating went down and only one additional rating was cast. (As a puzzle creator, you get to see how many people have rated the puzzle.) | |

Posted by rodolforiverol | 05/03/10 |

I took a different initial approach. I let our number be A(n)*10^n+A(n-1)*10^(n-1)+...+A(1)*10+A(0), where the A(i) are digits between 0 and 9. Then the 3/2 time constraint can be written as: 3*(A(n)*10^n+...+A(0)) = 2*((A(n-1)*10^(n-1)+...+A(0))*10+A(n)) (RHS) = (20*A(n-1)*10^(n-1)+...+20*A(0) +2*A(n)), so collecting around powers of 10 we get: A(n)*(3*10^n-2)=17*(A(n-1)*10^(n-1)+...+A(0)). Since 17 is prime, and A(n) lies between 1 and 9 we must find the smallest n so that 17 divides 3*10^n-2. This turns out to be n=15 (the easiest way to see this is to use modular arithmetic). Clearly, to minimize the number choose A(15)=1. The smallest number is then 10^15+(3*10^15-2)/17. To write out the explicit digits again use modular arithmetic. It's also easy to check the initial 1.5 times constraint using the 10^15+(3*10^15-2)/17 expression. | |

Posted by javaguru | 05/18/10 |

to rodolforiverol: That's a cool alternate approach. I thought about trying to solve it as an algebraic equation, but I didn't think it would reduce to something solvable. Very nice! :D | |

Posted by wearymachine | 06/11/10 |

not too bad once i saw it...if you take an algebraic approach to it, it's actually kinda easy...just turns into 16 simple iterations... | |

Posted by lunarrainbow | 06/14/10 |

Couldn't it have been 0? | |

Posted by wearymachine | 06/17/10 |

hah no...that's too obvious...and it's kinda implied that it can't be 0 since it involves moving the left digit to the right...i guess you could turn 0 into 00 and make that one work too... another trick to make it easier is to realize that for every digit you add...if you use algebra...you get (n-1) digit number = (2999___8 / 17) x 1 digit...last digit of the dividend is 8, the one before it is 9...so for the next iteration...jack up the remainder from the current iteration by one, bring the 8 down...also realize that 17 x 4 = 68...this is key bc the last digit of the dividend is 8...so when your remainder of an iteration is 5, you're all set...jack up the 5 by 1 to get 6...bring down the 8 to get 68, and then 68 / 17 = 4... also you have to check to see if 2999__8 / 17 can be multiplied by a single digit number to make it rational, before you add another digit...for this problem, turns out that wasn't the case...16th iteration worked perfectly (which you can see once you get the remainder of the 15th iteration)... | |

Posted by SRB_1807 | 08/18/11 |

:o :o :o :o | |

Posted by algor-heuris | 01/15/12 |

Here's how I solved it. I think this is similar to rodolforiverol's solution. Let n be the number you're trying to find. Let x be the leading digit of n let g be the place value of x. Ex: n=5000, x = 5, g = 3 (10^3 is the thousandths place) Mathematically, rotating left is: 10(n - 10^g*x) + x (n - 10^g*x) will remove the leading digit multiplying by 10 will make the second digit the leading digit and shift everything else adding x will append x to the ones place. After rotating, the number will be 3/2 as big as the original, so: 10(n - 10^g*x) + x = (3/2)n Solving for n: (17/2)n = 10^(g+1)*x - x n = x(2/17)[10^(g+1) -1] So, because we're looking for integers, [10^(g+1) -1] needs to be divisible by 17, and it will be a long string of 9's (because you're subtracting 1 from something like 100000....). So, I took out my calculator and kept using the modulus function. mod(999, 17) = 13 mod(9999, 17) = 3 . . . mod(9999999999999999, 17) = 0 So, g = 15 and: n = x(2/17)(9999999999999999) = 1,176,470,588,235,294*x Now, x needs be be a single digit and it multiplying it by 1,176,470,588,235,294 needs to yield a number with the leading digit equal to x. This means x can be any integer between 1 and 5. To get the lowest value of n, let x=1 Check by multiplying by 3/2 and compare it to rotating. Thanks for the teaser javaguru! |

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