Please help solve this one!!!!!!!!!!!!!!!!?

Question

The number 1920212223....939495 consists of 154 digits, by placing the numbers 19 to 95 one beside the other. If I were to remove 95 digits in such a way that the resulting number is the greatest possible one, what are the first 19 digits of the balance 59? Please help solve this one.

you are allowed to remove the digits from anywher in the number

Answer 1

I assume you can't reorder the digits, just remove them. So starting at the front, you want to keep as many 9s as possible.

remove 1 (1 digit)
--> keep 9
remove 2021222324252627282 (19 digits)
--> keep 9
remove 3031323334353637383 (19 digits)
--> keep 9
remove 4041424344454647484 (19 digits)
--> keep 9
remove 5051525354555657585 (19 digits)
--> keep 9

Okay, you've removed 77 digits (18 remaining).

So now you need to find the biggest in the next 19 digits, since you can't make it all the way to 9... the biggest is the 8.

So remove 60616263646566676 (17 digits)
--> keep 8

Then you have 1 more digit, 6 or 9.
Remove the 6
--> keep 9
--> and keep the rest (707172...939495)

The biggest 59-digit number will therefore be:
..99,999,897,071,
727,374,757,677,
787,980,818,283,
848,586,878,889,
909,192,939,495

And to answer the question, the first 19 would be:
9999989707172737475

Answer 2

Full Sequence:

1920212223242526272829303132333435363738
3940414243444546474849505152535455565758
5960616263646566676869707172737475767778
7980818283848586878889909192939495

Removing all 95 digits from the left that are not the digit 9 or the larger digit if no 9 is available towards the end of the removal.

The resulting first 19 digits are:

9999989707172737475

(remaining digits are 76777879808182838485868
78889909192939495)

Answer 3

Remove the first digit that is followed by a larger digit.
Repeat this step 95 times.

You will end up deleting
* the first 1
* the nineteen digits 202122...282 (not the 9)
* the nineteen digits 303132...383
* the nineteen digits 404142...484
* the nineteen digits 505152...585
* the seventeen digits 606162...6
* the one 6 in the sequence 697071...

Remaining 59 digits:
99999897071727374757677787980818283848586878889909192939495

Answer 4

9999999999999988888

I got this by removing all the 0's, then the 1's, then the 2's then the 3's, etc. until i deleted had only 59 remain, because as long as you have a certain number of digits, you want the largest digits possible. Then, you want to put them in order from largest to smallest, hence you start with all the ramining 9's, then 8's, until the 6's.

Answer 5

x² = x+6 x² - x - 6 = 0 (x - 3)(x + 2) = 0 So x = 3 or - 2 x² + 2x = 0 x(x + 2) = 0 So x = 0 or - 2 60 = x² - 4x x²- 4x - 60 = 0 (x - 10)(x + 6) = 0 So x = 10 or -6 4x = x² - 45 x² - 4x - 45 = 0 (x - 9)(x + 5) = 0 So x = 9 or -5 -6x = x² + 9 x² + 6x + 9 = 0 (x + 3)(x + 3) = 0 (x + 3)² = 0 So x = -3 -15=x² + 8x x² + 8x +15 = 0 (x + 5)(x + 3) = 0 So x = -5 or -3 -9x = x² + 14 x² + 9x + 14 = 0 (x + 7)(x + 2) = 0 So x = -7 or - 2 Hope that helps. =D

Answer 6

At first glance it would seem that the answer would be pretty simple.
The digits at the first part of the number would be the most important for making the value the highest.
So simply take off the first 95 digits that arent nines.

Answer 7

I'd start with putting the 95 in the front.

Your also going to want to remove the 95 numbers with lower number.

Answer 8

It depends on whether you have to keep them in order when removing the digits. If you have to keep them in order then I would say that the first 19 digits are: 6,8,6,9,7,0,7,1,7,2,7,3,7,4,7,5,7,6,7
If you don't have to keep the numbers in order then I would say that the first 19 digits are: 4,7,4,3,8,8,3,2,9,9,2,6,6,5,7,7,5,4,8
Don't know if this is helpful or not....but I hope it is!

Answer 9

ok, well, first, can you remove the 95 digits from any where or do you have to take all 95 out with one chunk?

Answer 10

I think I am going 2 blew my head