Number doubles
Can you find a number such that if its last digit (unit's place) is moved to the first digit*, the number doubles ?
* for example , if the original number be 5678 new number will be 8567 so the new number should be equal to two times the original number.
* for example , if the original number be 5678 new number will be 8567 so the new number should be equal to two times the original number.
Labels: mathemagic, trickofmind
posted by Rajesh Lal at
9:07 AM
18 Comments:
Alas, I couldn't find any DECIMAL numbers that have this characteristic.
However, in HEXIDECIMAL, #10842 (67650 in DEC) becomes #21084 (135300 in DEC), and #10842 / 2 = #21084 (135300 / 2 = 67650 in DEC).
I only checked numbers between 1 and 1048576, so there may be a higher number that I haven't found. I also checked Binary and Octal bases too, but didn't find any there.
You thought it was easy ?
No clue so far , there is some trick i am sure
5678* I FOund 1 XD
easy way? the number is 0...
move 0 to 0 and 2x0 is 0
eurica!!!! the # IS 0
u guys are laem...this question is so ez. the number is 216 see
216 x 2
=432
...wait, no, i did it wrong hold on...
2468 x 2
=4936
wait! crap, no. that was close, tho
ummm...let's try 136842
136842 x 2
=27 AAAAAAA!
Okay, I'll get back to you on this one...
This is what I got:
263,157,894,736,842,105 x 2 =
526,315,789,473,684,210
Which - unless I'm on drugs and my math is imaginary - totally works.
The funnest part about this was the process. I started with a little long division with a 2 as divisor, and then some numeral as dividend:
2/5
As I went on, I took the last number in my quotient and placed it to the right of my dividend:
++2
2/52
Repeat until I have a comfortable place to put the first numeral in my quotient at the end of my dividend.
Going into it, though, I thought the answer would be something really small. Zero's a good one. But I still think there's got to be an integer the follows this phenomenon smaller than two hundred sixty-three quadrillion, one hundred fifty-seven trillion, eight hundred ninety-four billion, seven hundred thirty-six million, eight hundred forty-two thousand, one hundred and five. I mean, honestly.
Frank you are the real genius
i could not sleep whole night multiplying every imaginable number by 2 could not reach one
DARN
Unless the number is 0. There are no such decimal number that I know of. Say the number we need to find is abcd...z then a10^x+b10^(x-1)+...+z=(z10^x+a10^(x-1)+...+y)/2
After calculation:
2z(10^(x+1) - 1)=19 zabcd...y
z(10^(x+1) - 1)/19=zabcd..y/2
10^(x+1) - 1 is a divident of 19 or 10^something -1 is a divident of 19
I don't know I give
Well I've been trying to figure it out all day and i came to the conclusion and many of my work mates agree with me that this question is mathematically impossible no matter how high the numbers go. We spent 3 hours today trying it each with a different group of numbers all the way up to a googol using our company's software programs so i hate to say it the question has no answer unless the answer is 00
frank, I wanna marry your brain
ANSWER
---------------------
105263157894736842
Another RIGHT answer by FRANK
263,157,894,736,842,105 x 2 =
526,315,789,473,684,210
The unique thing about the number is its 18 digits and if you add
the number with its double each of them will give 9 .
split 105263157894736842 in half into two numbers (105263157 and 894736842) and note that these two numbers equal 999999999 when summed.
GOT RIGHT
--------------------
FRANK
For all those guys who tried , this was definetly a tough question. but the solution in itself is pretty interesting
Another interesting answer i got recently for the question was
01 in BINARY
Here is the general solution:
The key is to solve the following. If the number is XY, where Y is a single digit, then
19X = Y(10^n - 2), where n is log(X)+1 rounded down.
since Y is a single digit and isnt divisible by 19, 10^n - 2 must be divisible by 19. The smallest n for this is =17
Therefore X = Y(5,263,157,894,736,842)
Since log (X) + 1 rounded down must = 17, X must be a 17 digit number.
This means any value for Y of 2,3,4,5,6,7,8,9 will give an answer for X that will work. (Y=1 will make X a 16 digit number so will not work)
E.g. for Y=2, X = 2*5,263,157,894,736,842 = 10526315789473684
and the number is 105263157894736842, which is Rajesh's answer
E.g. for Y=3, X=3*5,263,157,894,736,842 = 15789473684210526
and the number is
157,894,736,842,105,263
E.g. for Y=4, X=4* 5,263,157,894,736,842 = 21052631578947368
and the number is
210,526,315,789,473,684
and so on for Y = 5,6,7,8,9
Note that frank gave the number for Y=5
not to be annoying or anything but the question states that moving the first and last digits around is double the first number, so Frank has almost got it rite but if you double Franks answer, and look at it carefully, there is as the second digit and the last digit is 4 not 1 but i reakon youve got the closest answer yet
I appologize it was Rajesh Lal's answer is was referring to.
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