Delete the Digits!

Logic Level 3

$\large 1\ 2\ 3\ 4\ 5\ 6\ \ldots \ 37\ 38\ 39\ 40$

Delete 60 digits from the number above in such a way as to make the resulting number as small as possible. Submit the resulting number as your answer.

Note: The resulting number must not start with a 0.

Also try this problem: Locate the Digits!

The answer is 10001233330.

4 solutions

Head Jump
Oct 3, 2015

The whole number would have 71 digits, so taking out 60 would leave us with 11.

We must start with 1, as we cannot begin with 0.

There are only four 0's available for us to keep. The first three go immediately after the 1, however since the fourth 0 is the last digit in the number, it must be last in our answer. So we need 6 digits immediately before that last 0.

There are no 0's in between the third and fourth 0, so we go with the only available 1, then the only available 2.

Now, because we are in the thirties range, we have enough 3's to fill up the rest of the number.

10001233330

Jesse Nieminen
Sep 23, 2015

Start by writing the 71 digit number.

Starting from left iterate trough the number.

If current digit is greater than the digit on the right side remove it and go back to first digit. (If you are on first digit don't remove it if the second digit is 0) Otherwise go to next digit.

Repeat 60 times.

1234567890111213141516171829202122232425262728293031323334353637383940 123456780111213141516171829202122232425262728293031323334353637383940 12345670111213141516171829202122232425262728293031323334353637383940

...

10111213141516171829202122232425262728293031323334353637383940 1011113141516171829202122232425262728293031323334353637383940

...

10001233330

The original 71 digit number is 12345678910111213141516171819202122232425262728293031323334353637383940.... The final number cannot start with a zero.... Thus, to make it as small as possible it must necessarily start with a 1.... And the digits to its right should be zeroes for as long as possible in order to make the place value as small as possible..... Applying this logic,

Iteration 1: 101112131411516171819202122232425262728293031323334353637383940

Iteration 2: 10021222322425262728293031323334353637383940

Iteration 3: 100031323334353637383940

Now, it is no longer possible to obtain any more zeroes as the final zero is the right most...... So next step would be to obtain a 1.

Iteration 4: 10001323334353637383940

Again, no more 1's so go for 2.

Iteration 5: 1000123334353637383940

Again, more 2's left. so go for 3.

Iteration 6: 100012333333330

By now, we are left with a 15 digit number [i.e. we have eliminated 56 digits so far]. We can delete 4 more. The only option left is delete 4 among the 8 3's.

Iteration 7: 10001233330

Which is our smallest 11 digit number.

K.R. Manish - 5 years, 8 months ago

Actually,(Considering there were no specific guidelines to how to go about the process of elimination in the original question) If you were to go through and remove all of the largest numbers first, there is no possible way to delete 60 numbers and still be left with a single "3" or "2". The only numbers to remain would be seven "1's" and four "0's". Thus making the SMALLEST possible number without starting with zero......... 10000111111.

Matt Williams - 5 years, 8 months ago

Yes, but you need to start from number 1234567890111213141516171829202122232425262728293031323334353637383940 and delete 60 digits. You cannot alter the positions of the digits by any other means. Therefore the answer is 10001233330.

Jesse Nieminen - 5 years, 8 months ago

Where in the original question does it say, "You cannot alter the positions of the digits by any other means."?

Matt Williams - 5 years, 7 months ago

@Matt Williams – You start with the number: 1234567890111213141516171829202122232425262728293031323334353637383940. You are told to DELETE 60 digits, and therefore you cannot rearrange them.

Jesse Nieminen - 5 years, 7 months ago

10000111111

prasad mahadik - 5 years, 8 months ago

You are not allowed to rearrange the digits.

Jesse Nieminen - 5 years, 8 months ago

A Former Brilliant Member
Feb 5, 2021

The number above is constitued of 31 numbers of 2 digits (from 10 to 40) & 9 numbers of 1 digit (from 1 to 10) mean total of 71 digits. After deletion, 11 digits remain which are 10001233330 :

1 of 1

0 of 10

0 of 20

0 of 30

1 of 31

2 of 32

33 of 33

3 of 34

3 of 35

0 of 40

Sidd Namachivayam
Oct 11, 2015

Well clearly the original number has 71 digits. Since we can't start with a 0 we want to start with a one, followed by as many 0's as possible. This means we have the number 1000_ _ _ _ _ _ _ where there are 7 blanks. Well clearly we need to start choosing from number in the 30's, so we can take the 1 from 31, the 2 from 32, the 3's from 33 through 36 and finish with the 0 from 40, to attain our final answer which is 10001233330.

Delete the Digits!

Also try this problem: Locate the Digits!

The answer is 10001233330.

4 solutions

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