Take the last digit of this number and move it to the front.
The resulting number will be exactly double that of the original.
What is the smallest number with this property?
(Merry Xmas)
AiY
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What is the smallest positive number with this property
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Re: What is the smallest positive number with this property
There are two kinds of people. Those who have to think about it, and those with an instant answer.
Binary, of course. Double 01 and you get 10.
Binary, of course. Double 01 and you get 10.
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Re: What is the smallest positive number with this property
UncleEbenezer wrote:There are two kinds of people. Those who have to think about it, and those with an instant answer.
Binary, of course. Double 01 and you get 10.
Would you mind if I waited for some more answers please? It may be, on this occasion only, that instant answers aren't quite what is being sought.
I can give you a small clue though. It's a large number.
AiY
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Re: What is the smallest positive number with this property
AsleepInYorkshire wrote:It's a large number.
01 (Base 2)
Julian F. G. W.
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Re: What is the smallest positive number with this property
AsleepInYorkshire wrote:Take the last digit of this number and move it to the front.
The resulting number will be exactly double that of the original.
What is the smallest number with this property?
A spoiler that assumes the usual conventions for such puzzles, namely that we're working in decimal and leading zeros are not allowed (other than for the number 0 itself):
Suppose a positive n-digit number x has the property and that its last digit is d. Note that d is not 0 because taking the last digit and moving it to the front would produce a smaller number even if we were to allow that use of a leading zero.
Then x with its last digit removed is (x-d)/10. Putting d at the front of that adds d*10^(n-1) to it, so we have:
2x = d*10^(n-1) + (x-d)/10
Multiplying that through by 10 gives:
20x = d*10^n + x - d
and then subtracting x from both sides gives:
19x = d * (10^n -1)
Since 19 is prime and doesn't divide any of 1-9 (the possible values of d), it must divide (10^n - 1) - i.e. the remainder of 10^n when divided by 19 must be 1. Calculating the remainders of powers of 10 when divided by 19 goes:
Remainder of 10^0 = 1 when divided by 19 is 1
So remainder of 10^1 when divided by 19 is the same as the remainder of 1 * 10 = 10 when divided by 19, which is 10.
So remainder of 10^2 when divided by 19 is the same as the remainder of 10 * 10 = 100 when divided by 19, which is 5.
So remainder of 10^3 when divided by 19 is the same as the remainder of 5 * 10 = 50 when divided by 19, which is 12.
So remainder of 10^4 when divided by 19 is the same as the remainder of 12 * 10 = 120 when divided by 19, which is 6.
So remainder of 10^5 when divided by 19 is the same as the remainder of 6 * 10 = 60 when divided by 19, which is 3.
So remainder of 10^6 when divided by 19 is the same as the remainder of 3 * 10 = 30 when divided by 19, which is 11.
So remainder of 10^7 when divided by 19 is the same as the remainder of 11 * 10 = 110 when divided by 19, which is 15.
So remainder of 10^8 when divided by 19 is the same as the remainder of 15 * 10 = 150 when divided by 19, which is 17.
So remainder of 10^9 when divided by 19 is the same as the remainder of 17 * 10 = 170 when divided by 19, which is 18.
So remainder of 10^10 when divided by 19 is the same as the remainder of 18 * 10 = 180 when divided by 19, which is 9.
So remainder of 10^11 when divided by 19 is the same as the remainder of 9 * 10 = 90 when divided by 19, which is 14.
So remainder of 10^12 when divided by 19 is the same as the remainder of 14 * 10 = 140 when divided by 19, which is 7.
So remainder of 10^13 when divided by 19 is the same as the remainder of 7 * 10 = 70 when divided by 19, which is 13.
So remainder of 10^14 when divided by 19 is the same as the remainder of 13 * 10 = 130 when divided by 19, which is 16.
So remainder of 10^15 when divided by 19 is the same as the remainder of 16 * 10 = 160 when divided by 19, which is 8.
So remainder of 10^16 when divided by 19 is the same as the remainder of 8 * 10 = 80 when divided by 19, which is 4.
So remainder of 10^17 when divided by 19 is the same as the remainder of 4 * 10 = 40 when divided by 19, which is 2.
So remainder of 10^18 when divided by 19 is the same as the remainder of 2 * 10 = 20 when divided by 19, which is 1.
...
Since the number of digits of a positive number must be at least 1, that means that n is at least 18. Can we have n = 18? If so, we have:
19x = d * (10^18 - 1)
from which it follows that:
x = d * (10^18 - 1) / 19 = d * 52631578947368421
So the smallest x might be is 1 * 52631578947368421 = 52631578947368421, but that doesn't work because it's a 17-digit number. The next smallest it might be is 2 * 52631578947368421 = 105263157894736842, which is an 18-digit number, and it has the property because 2 * 105263157894736842 = 210526315789473684. So we've established that n is at least 18, that we can have n = 18 and that 105263157894736842 is the smallest 18-digit number with the property. And since any n-digit number for larger values of n must be larger than 105263157894736842, that's enough to establish that 105263157894736842 is the smallest positive number with the property.
Gengulphus
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Re: What is the smallest positive number with this property
Gengulphus wrote:AsleepInYorkshire wrote:Take the last digit of this number and move it to the front.
The resulting number will be exactly double that of the original.
What is the smallest number with this property?
A spoiler that assumes the usual conventions for such puzzles, namely that we're working in decimal and leading zeros are not allowed (other than for the number 0 itself):
Suppose a positive n-digit number x has the property and that its last digit is d. Note that d is not 0 because taking the last digit and moving it to the front would produce a smaller number even if we were to allow that use of a leading zero.
Then x with its last digit removed is (x-d)/10. Putting d at the front of that adds d*10^(n-1) to it, so we have:
2x = d*10^(n-1) + (x-d)/10
Multiplying that through by 10 gives:
20x = d*10^n + x - d
and then subtracting x from both sides gives:
19x = d * (10^n -1)
Since 19 is prime and doesn't divide any of 1-9 (the possible values of d), it must divide (10^n - 1) - i.e. the remainder of 10^n when divided by 19 must be 1. Calculating the remainders of powers of 10 when divided by 19 goes:
Remainder of 10^0 = 1 when divided by 19 is 1
So remainder of 10^1 when divided by 19 is the same as the remainder of 1 * 10 = 10 when divided by 19, which is 10.
So remainder of 10^2 when divided by 19 is the same as the remainder of 10 * 10 = 100 when divided by 19, which is 5.
So remainder of 10^3 when divided by 19 is the same as the remainder of 5 * 10 = 50 when divided by 19, which is 12.
So remainder of 10^4 when divided by 19 is the same as the remainder of 12 * 10 = 120 when divided by 19, which is 6.
So remainder of 10^5 when divided by 19 is the same as the remainder of 6 * 10 = 60 when divided by 19, which is 3.
So remainder of 10^6 when divided by 19 is the same as the remainder of 3 * 10 = 30 when divided by 19, which is 11.
So remainder of 10^7 when divided by 19 is the same as the remainder of 11 * 10 = 110 when divided by 19, which is 15.
So remainder of 10^8 when divided by 19 is the same as the remainder of 15 * 10 = 150 when divided by 19, which is 17.
So remainder of 10^9 when divided by 19 is the same as the remainder of 17 * 10 = 170 when divided by 19, which is 18.
So remainder of 10^10 when divided by 19 is the same as the remainder of 18 * 10 = 180 when divided by 19, which is 9.
So remainder of 10^11 when divided by 19 is the same as the remainder of 9 * 10 = 90 when divided by 19, which is 14.
So remainder of 10^12 when divided by 19 is the same as the remainder of 14 * 10 = 140 when divided by 19, which is 7.
So remainder of 10^13 when divided by 19 is the same as the remainder of 7 * 10 = 70 when divided by 19, which is 13.
So remainder of 10^14 when divided by 19 is the same as the remainder of 13 * 10 = 130 when divided by 19, which is 16.
So remainder of 10^15 when divided by 19 is the same as the remainder of 16 * 10 = 160 when divided by 19, which is 8.
So remainder of 10^16 when divided by 19 is the same as the remainder of 8 * 10 = 80 when divided by 19, which is 4.
So remainder of 10^17 when divided by 19 is the same as the remainder of 4 * 10 = 40 when divided by 19, which is 2.
So remainder of 10^18 when divided by 19 is the same as the remainder of 2 * 10 = 20 when divided by 19, which is 1.
...
Since the number of digits of a positive number must be at least 1, that means that n is at least 18. Can we have n = 18? If so, we have:
19x = d * (10^18 - 1)
from which it follows that:
x = d * (10^18 - 1) / 19 = d * 52631578947368421
So the smallest x might be is 1 * 52631578947368421 = 52631578947368421, but that doesn't work because it's a 17-digit number. The next smallest it might be is 2 * 52631578947368421 = 105263157894736842, which is an 18-digit number, and it has the property because 2 * 105263157894736842 = 210526315789473684. So we've established that n is at least 18, that we can have n = 18 and that 105263157894736842 is the smallest 18-digit number with the property. And since any n-digit number for larger values of n must be larger than 105263157894736842, that's enough to establish that 105263157894736842 is the smallest positive number with the property.
Gengulphus
Err ... wow
AiY
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