Photo:
THE MIRROR
At first glance, this week's puzzle seems to be easy to solve.
But don't underestimate the task!
In the case of a natural number, the last digit is shifted from the far right to the far left to the new position 1.
This results in a number with the same number of digits, but twice as large as the original number.
We are looking for the smallest natural number with which this digit shift works.
The smallest solution has 18 digits and is:
105,263,157,894,736,842
We represent the number we are looking for as the sum 10a + b.
b is a one-digit natural number greater than zero.
a is an n-digit natural number.
The number 10a + b you are looking for then has n+1 digits.
In order to fulfill the conditions of the task, the following must be true:
(10a + b)*2 = 10n * b + a
20a + 2b = 10n * b + a
a = b*(10n - 2)/19
b is a one-digit natural number greater than zero and therefore not divisible by 19.
It follows that 10n - 2 must be divisible by 19, only then is a an integer.
So when we are looking for the smallest number that satisfies the conditions of the problem, we should first find the smallest number n such that 10n - 2 is divisible by 19.
Or in other words:
10n must leave remainder 2 when divided by 19
.
The search is not as complicated as it first appears.
We don't have to divide gigantic powers of 10 by 19.
If we know the remainder of 10n when divided by 19, we can easily find the remainder of 10n+1 by multiplying the remainder of 10n by 10 and finding the remainder of that number when divided by 19.
Let's start with n=1.
101 when divided by 19 has the remainder 10.
Continue with n=2.
When 102 is divided by 19, it has ten times the remainder of 101 – i.e. 10*10 = 100. We subtract the multiples of 19 from this and get the remainder 5 (5*19=95, 100-95=5).
Now to n=3.
When 103 is divided by 19, it has ten times the remainder of 102 – i.e. 10*5 = 50. From this we subtract the multiples of 19 and get the remainder 12 (2*19=38, 50-38=12).
We'll repeat this process over and over again until hopefully eventually hitting a remainder of 2.
And that actually happens at n=17, as the following table shows (I created it with Excel, but you can also do it by hand):
Back to the original question: We are looking for the smallest number 10a + b that is half the size of 10n * b + a.
In order for this condition to be met, a = b*(10n - 2)/19 must apply.
The number a is the product of b, a one-digit number, and (10n - 2)/19.
n = 17 is the smallest possible number for n for which a is an integer.
The two numbers you are looking for must therefore have at least 18 digits.
If we also find the smallest possible number b, we have the solution.
Let's start with b=1.
For a we get the number
5,263,157,894,736,842
This has 16 digits.
If we add the number b to the right, we get the 17-digit starting number:
52.631.578.947.368.421
Double this 17 digit number is the 18 digit number
105.263.157.894.736.842
At least here we see that this is not a solution.
a would have to be 17 digits so that 10a+b is 18 digits and after moving the last digit all the way to the front there is again an 18 digit number.
If we move the 1 of 52,631,578,947,368,421 all the way to the front, we have to add a 0 before the 5 to make the math work.
With the next larger number b=2, however, we get a solution where everything fits:
a = 10,526,315,789,473,684 (17 digits)
10a + b = 105,263,157,894,736,842 (18 digits)
(10a + b) * 2 = 210,526,315,789,473,684
This challenging task comes from the MindYourDecisions YouTube channel and was suggested by reader Dieter Fritsche.
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