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 Beginning and End (Posted on 2015-07-11)
When the digit is placed at the beginning and end of a positive integer N, the new number is 99*N.

Find the smallest value of N.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 More precision with UBASIC computer program | Comment 2 of 6 |
(In reply to We're going to need more precision (spoiler) by Steve Herman)

If d is the digit in question and p is one more than the length of N:

99*N = d*(10^p + 1) + 10*N
so N=d*(10^p+1)/89

10   for P=2 to 99
20   for D=1 to 9
30     N=D*(10^P+1)//89
40     if N=int(N) then print P,D,N
50   next
60   next

produces values of N that work if the length of its representation is p - 1.

The table below shows the smallest value of N is 112359550561797752809.

`   p   d    N  22   1   112359550561797752809   22   2   224719101123595505618   22   3   337078651685393258427   22   4   449438202247191011236   22   5   561797752808988764045   22   6   674157303370786516854   22   7   786516853932584269663   22   8   898876404494382022472 x 22   9   1011235955056179775281   66   1   11235955056179775280898876404494382022471910112359550561797752809   66   2   22471910112359550561797752808988764044943820224719101123595505618   66   3   33707865168539325842696629213483146067415730337078651685393258427   66   4   44943820224719101123595505617977528089887640449438202247191011236   66   5   56179775280898876404494382022471910112359550561797752808988764045   66   6   67415730337078651685393258426966292134831460674157303370786516854   66   7   78651685393258426966292134831460674157303370786516853932584269663   66   8   89887640449438202247191011235955056179775280898876404494382022472 x 66   9   101123595505617977528089887640449438202247191011235955056179775281 `

Those marked with x do not work as the length of N is too great (i.e., equal to p).

Edited on July 11, 2015, 9:40 am
 Posted by Charlie on 2015-07-11 09:36:36

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