Basics of counting?

My first answer was 11121111111 (with 11 one's in the box) which I think works out OK too - if you agree, it might be worth stating that each of the blanks can only contain one digit?

My laziness got the best of me with this one. When I got the solution, I felt extremely stupid.

When I started, I noticed that for small numbers at position 1, it will arrive at a paradoxical answer sooner (i.e., place an answer will cause the answer itself to change).

For example:

0123456789

122... <-- I now have to change it to 3.

These situations are not ideal for this puzzle, so I figured that I should start at 9 for position 1:

0123456789

1932111112 <-- This won't work as there are not enough 1s.

Then I've tried for 8 and it didn't work. At this point I got too lazy and gave up, so I wrote a program to solve it for me: link

That's how I got 1732111211 and the only thing I thought was: why didn't I just go 1 step further with my logic? Anyway, at least computer scientists will approve my answer.

Edit:

I just had another look, I didn't realise all you have to do is "3 trials". Start with:

0123456789

1221111111 <-- Now there are 3 2s, update accordingly.

1232111111 <-- Count the number of 1s, and you get 7, thus:

1732111211 <-- Solution.

My program is far more inefficient than solving this with my own brain.

Workings by "Trial & Error)

Start by assume all 1's (for the numbers that already appear) 11111111

This means you have 1ten 1's (but will loss 2 of them in a second - so enter 8 ones & 2 eights) 181111121

Now you have two 2;s (meaning one less one & the extra 8 becomes a 7)

1721111211 You know have three 2, (& a extra 3)

Answer : 173211211

Sorry if this is hard to follow - I couldn't think of a better way to write it & hope it at least gives the general idea of mine & Sravanth's "method"

We can see that there CANNOT be a zero in ANY of the blanks as all the 10 digits appear at least once in the box. Hence the number of 0's in this box MUST be one as that zero MUST be the only zero in the box.

Most of the numbers in the blanks should also be one where that one occurence is the digits before the blanks. We let the number of times one occurs in the blanks be x so the number in the blank is x where x is at most 9 but definitely greater than 1. The number of times x occurs would be 2 - once in the phrase 'the number of x's' and another time in the blank pertaining to the number of ones. We fill in 2 in the blank pertaining to the number of x.

Noe we see that there have to be at least 3 twos as there are already 2 twos - one in the phrase 'the number of 2's' and another in the blank pertaining to the number of x. If we claim that there are 2 twos in this box, then there is a contradiction as this 2 that we fill in is already a third two.

Suppose there are 3 twos. Then it follows that there are 2 threes - one in the phrase 'the number of 3's' and the 3 in the blank pertaining to the number of twos. We see that we are able to make up exactly 3 twos - one in the phrase 'the number of 2's' , another in the blank pertaining to the number of threes and the third pertaining to the number of x's.

Now determine what is x. We see that we have filled in a value other than one for the number of ones, twos, threes and x's, so 10-4 = 6 blanks will have one. Adding this to the 1 that we see in the phrase 'the number of 1's', x=7.

So we have 1732111211 as our magic number!!!

We need to fill 10 blanks. First we notice that there is already at least one of each digit, so no blank can be filled with a 0, and the number of 0s is 1.

For the remaining blanks we notice this: if the blank for digit d is filled with a number n, that means d appears n times, the one already there and in (n-1) other blanks. So for each blank filled with n, n blanks need to be reserved. This means that there can't be more than 5 blanks filled with numbers equal or larger than 2. But if only 5 blanks can be filled with numbers larger than 1, and no blank is filled with 0, then at least 5 blanks must me filled with 1, and the number of 1s at least 6.

By the same argument, there can't be more than 1 blank filled with numbers equal or larger than 6, so of the blanks for number of 6s, 7s, 8s and 9s, three are filled with 1, and one is filled with 2 (the one corresponding to the number of 1s).

This also tells us that all of the blanks, except the one for the number of 1s, are filled with numbers between 1 and 5. At least 5 of those are 1, leaving at most 4 blanks to be filled with numbers between 2 and 5. This means there isn't enough room for any blank to be filled with a 5, so the number of 5s is 1.

It also means that there isn't enough room for more than 1 of the remaining blanks to be filled with anything larger than 2, so at least one of the blanks for number of 3s and 4s has to be filled with a 1.

Now, we already know that the numbers of 2s is at least 2, and it's easy to see it actually has to be larger, because if the blank for the number of 2s was filled with a 2, it would add to the total number of 2s, making it contradict itself. So the number of 2s is either 3 or 4, and the blanks for number of 3s and 4s have to be filled with a 1 and a 2.

Now we mostly know the contents of all the blanks:

• Number of 0s: 1
• Number of 1s: between 6 and 9
• Number of 2s: 3 or 4
• Number of 3s, 4s: a 1 and a 2
• Number of 5s: 1
• Number of 6s, 7s, 8s, 9s: three 1s and a 2

By counting these we see that we have exactly 6 blanks filled with 1s and exactly 2 blanks filled with 2, so the number of 1s is 7, and the number of 2s is 3. This gives us immediately that the blanks filled with must be for the number of 3s and the number of 7s, and the rest are 1. The only possible answer is then:

1732111211

$\huge1732111211$

I found it by trial and error method, do you think there is any other way? If yes do post it!

For those who are confused: You have; 1×0|1×1|1×2|1×3|1×4|1×5|1×6|1×7|1×8|1×9 leading you to have; 1×0|10=>9×1|1×2|1×3|1×4|1×5|1×6|1×7|1×8|1×9 changing to 1×0|8×1|1×2|1×3|1×4|1×5|1×6|1×7|2×8|1×9 1×0|7×1|2×2|1×3|1×4|1×5|1×6|1×7|2×8|1×9 1×0|7×1|3×2|2×3|1×4|1×5|1×6|2×7|1×8|1×9 first numbers 1732111211

I feel like I started trying solve level 4 problems way too soon but here's what went through my mind:

Since every number is mentioned at least once, 0 will not be filled in. This makes the first space being 1. I was trying to convince a friend of a tactic that I had to go on but as I was talking, I noticed that 9 cannot have a number that its not 1. If it did, the 9 would appear in at least another row leaving 10-3=7 spaces left. This row has a number $n\geq 1$ in it and thus will have to appear 9 times. But there are only 7 spaces plus the time its been mentioned. This would add up to a max of 8.

I wondered if the strategy was then to work upwards testing if the next biggest number could have another number, other than 1, in its blank space. So far we have 8 spaces left to fill in. Putting a number in the 9th row that is not one means that we would have to write 8 somewhere else. This would take up 2 more spaces and so we would have 6 left. But then, if 8 was written in the (n+1)th row, n would have to appear 8 times. This takes all 6 spaces plus the one being mentioned adding to a max of 7. This cannot happen either.

I tried to do what I was doing for 7. That's 6 spaces left. It fails immediately everywhere else except in the 2nd row. I did then and then boxed the remaining spaces. My next question was "do we have to have a different number, other than 2 in the 2nd row?" But trying 3 worked. So I kinda got lucky.

I started by putting '1' in every blank as I knew that each digit had already appeared once in each of the sentences themselves. Then some experimenting is required, where you keep compensating for the numbers you have just added in. This should lead you to 1732111211. Hope this is clear enough.

Let there be n 1's in the box, that means that there will be 2 n's in the box.

That will mean that there will be atleast 2 2's in the box. However If the answer of number of 2's is 2 then the number of 2's will become 3.

So the number of 2's will be 3. Thus the number of 3's itself will be 2 validating the number of 2's.So we have number of 2's and 3's.

For finding n, find the total number of places 1 will be and hence number of 1's.(As there are 7 places (answer to number of 0's, 3's,4's...9's) which can potentially be filled with 1 and one of those is n, the answer to 6 of them is 1.This along with the 1 in the second line give the n=7)

So answer to number of 1's is 7 and number of 7's is 2.Answer to all the remaining fields is 1.

So our answer is 1732111211

Hoping it's clear enough, I've tried to follow a structured progression.

First I noticed that we're looking for $10$ one-digit numbers

• whose sum equals to $20$ (there are $20$ digits in the box in the end),

• and none of them is $0$ (each digit appears in the box).

E.g. $1+1+1+1+1+2+2+3+3+5=20$

I then assumed that there's a majority of $1s$ . In this case the highest number should be the number of $1s$ in the sum plus $1$ (the $1$ in the left column).

Thus I started testing if the number of $1s$ could be $9$ , then $8$ , etc.:

• if there are $7\ 1s$ in the sum, there will be $8\ 1s$ in the box, then we're searching : $1+1+1+1+1+1+1+??+??+8=20$ Then there are $8\ 1s$ , and $2\ 8s$ . One of the missing numbers must be $2$ , but it doesn't match (the last one would be $3$ , and no digit appears exactly $3$ times).

• if there are $6\ 1s$ in the sum, there will be $7\ 1s$ in the box : $1+1+1+1+1+1+??+??+??+7=20$

Since $1+1+1+1+1+1+2 (number\ of\ 7s)+7(number\ of\ 1s)=15$ , we're left with two digits to choose, whose sum equals to $5$ : as we've fixed the number of $1s$ , it's necessarily a $2$ and a $3$ .

Thus, there will be $3\ 2s$ and $2\ 3s$ , and the result is $1732111211$ .

One good clarification might be to say that the numbers included in the sentence "The number of 0's in this box is" also count toward the total (although I could also see that as part of the problem solving element).

We start out knowing that the first digit has to be 1 because there can only be one 0 because since each number appears in the sentence, no blank could possibly be zero so the only zero is the one in the first sentence. We also know that there have a lot of ones because of the way each sentence contains each number. So I left it out as 1X.

Next you can see that the answer for 2 cannot be 2 because then you will have those two 2's and the other 2 that shows up in the row for the X digit. Therefore, there must be three or more 2's. It also becomes clear that it cannot be more than three because it will throw off the rest of our numbers.

So after we establish that it starts off as 1X3, we see that there are two 3s and thus our number starts out 1X32. Then recall that X will correspond to some other number so there will be one more digit with a value of 2. That means in our answer we will have four out of the ten digits equalling a number other than 1. This means there must be six digits in our answer that are 1 plus the 1 in the sentence, and you get seven ones and thus X=7. From there you can plug in the 1's as necessary to find that the answer is 1732111211.

1st don't expect bigger no.s to repeat more than once.2nd the no. of any number should not be the no. itself(example no. of 2s cannot be 2)=(The axioms used). We notice that there is already at least one of each digit, so no blank can be filled with a 0, and the number of 0s is 1.From axiom 1 we can understand that most of the no.s in the blank are 1.If the no. of 1's is x then x should be there twice(this is not favourable to our 2nd axiom) ,using axiom 1 adjusting the no. of 3's is the best way to cope with this problem. Hence the answer.