Divide 30 utensils into $30-10+1 = 21$ groups of 10 consecutive utensils (groups 1-10, 2-11, etc).

Let $f(n)$ be a function which returns the number of forks in the $n^{th}$ group of utensils. The function is obviously continuous since the difference between number of forks of two consecutive groups cannot be greater than 1.

Consider now $f(1)$ , $f(11)$ and $f(21)$ and let's say that none of them equals 5. In that case, since $f(1) + f(11) + f(21) = 15$ , at least one of them must be greater than 5 and one of them less than 5. Then, we can apply logic of Intermediate value theorem and say that there must be some $n$ between these so that $f(n) = 5$ .

We can partition the $30$ utensils into thirds of $10$ utensils each. Assuming none of them have $5$ forks, then there will be at least one adjacent pair where one has more than $5$ forks and the other less than $5$ . If any scoop has $x$ number of forks, then an adjacent scoop one utensil over will change the number of forks by $-1$ , $0$ , or $1$ . Hence, there exists at least one scoop within that adjacent pair that will have exactly $5$ forks.

Try this random sequence

$0,1,1,0,1,0,0,1,0,0,0,0,1,0,1,0,1,0,0,0,1,1,1,0,1,0,1,1,1,1$

where the 1st partition has $4$ forks, the 2nd has $3$ forks, and the 3rd has $8$ forks. A good scoop would be $0,1,0,1,0,0,0,1,1,1$

Suppose a configuration without this pattern was possible. Consider some arbitrary setup of spoons and fork. Now, consider only the first 10 elements, and count the number of spoons, X. It of course can't be 5, so WLOG, let's say X<5. Now, consider the 2nd-11th elements in the setup. So, 1 element is removed, and 1 is added. The new number of spoons must be either X+1, X-1, or X. However, X can't equal 5, and since X<5, X must continue to be less than 5, for every possible sub sequence of spoons and forks.

However, this cannot be possible, as then we would not have seen enough spoons. Throughout our process, we've looked at (25-10+1) = 16 different sequences of 10 elements. At most, we have seen 16*4 = 60 spoons (16 sequences with 4 spoons each). Now, it is not enough to say that, since that is less than half, we have proven it impossible, as the spoons on the end will be counted in the sequences less often than the ones in the middle. To prove this is too low, I must go a bit more in depth. To minimize the number of spoons used, we want to position them in the edges. However, we can only place at most 4 spoons on each edge, because X is always <5. On each edge, in the best case, we count the spoons (1+2+3+4) = 10 times. Then we have 7 spoons that must go in the middle, which we count 10 times each. So in total, we must count at least 90 spoons - but we've said we only count 16! This is a contradiction, so the setup must not exist.

NOTES: 1). It is interesting that 90 is more than half the total spoons that have been counted. If this setup truly existed (it does not need to exist, since I am just setting a lower bound), it would mean there is a setup with only 70 forks counted, making it seem that the 90 was derived from a logical error, since we can get 70. However, since we've assumed X<5, the situation is no longer symmetric, and the logic holds.

2). This type of pattern is not always impossible. Consider 24 utensils (12 each), no 5|5 pattern. SSSSFFFFFFSSSSFFFFFFSSSS

If the answer is "No", that would require a setup where every possible set of 10 utensils contains more of "one" then the "other"

If you started on the left with $N_{spoons} > N_{forks}$ and you move by one position to the right, you cannot transition to the situation where $N_{spoons} < N_{forks}$ without hitting the intermediate state of $N_{spoons} == N_{forks}$

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Lemma1 (Inspired by comment from Tom Verhoeff ) : It is impossible to have a setup of M utensils that contains equal amounts of spoons and forks and where M mod 10 == 0, so that every possible subset of 10 utensils contains more spoons then forks.

Proof: Let's split a setup on a consequent subsets of non intersected tens (1 to 10, 11 to 20, ... ). Since M mod 10 == 0, there is no utensils left after the split. Let's assume that it is possible to forge a setup where every subset contain $N_{spoons} > N_{forks}$ . In such case after summing them together we will have to conclude that $\sum_{1}^{M mod 10} N_{spoons} > \sum_{1}^{M mod 10} N_{forks}$ which is a contradiction.

Let's try to find a solution where Nidhi won't be able to find 10 consecutive items with 5 of them being forks and the other 5 being spoons.

We'll want to get as close as possible to 50% of the items being forks and the other 50% being spoons as possible, as we only gave 15 of each.

Let's start with a fork (K) and a spoon (S). If we repeat the pattern as long as possible, we end up with KSKSKSKSK. Adding another spoon would result in 5 being forks and the other 5 being spoons. Thus, we should add another fork to get KSKSKSKSKK. Making sure there aren't 5 forks/spoons in a row, we can continue the pattern like this: KSKSKSKSKKKSKS... having a group of 3 forks every now and then. However, if we try to continue this pattern, we end up running out of forks before we run out of spoons!

If we try other patterns (e.g. KKKKSSSSSS), we'll still end up with the same result - we run out of one utensil far before we run out of the other. Therefore, it's logical to conclude that there will always be 10 consecutive items, in which 5 are forks and the other 5 are spoons.

mb it's not the right answer, but i said yes, because u didn't state she can't just choose whatever she wants, and she probably wants what she needs.

Let's clear the confusion since I too had faced the same mistake, I can feel the annoyance related to the question.

In such Questions where a possibility is concerned, it is customary to assume the possibility of the existence of a path to the solution rather than the solution itself. This adheres to the uniqueness of the Problem.

You could simplify the problem by dividing each number in the problem by 5. The question would then read:

There's a randomly ordered line of 3 forks and 3 spoons on the counter. Chef Nidhi only needs 1 fork and 1 spoon.

Can she always pick up exactly what she needs by selecting 2 consecutive utensils starting from somewhere in the line-up?

This makes the problem easier to visualise, and you can easily determine the answer is yes , because at least one fork and spoon must sit next to each other in the lineup. The answer will still be yes if we scale the problem up.

Please correct me if my logic is not sound

Suppose there exists configuration where it's impossible to choose 10 consecutive utensils with 5 spoons and 5 forks. Then each 10 consecutive utensils must include 0-4 or 6-10 spoons. And when we move from 10 consecutive utensils 1 step left or right, the number of spoons can only increase/decrease by one or remain the same. If then the first 10 utensils have number of spoons in range 0-4, then all other 10 consecutive utensils have number of spoons also in that range, because otherwise to get to 6 spoons we would have to get there from a row 5 spoons in a row of 10 utensils. Let's then consider 1-10, 11-20, 21-30 - three groups of consecutive utensils. Note, that those 3 groups are mutually exclusive. And each of them have at most 4 spoons, that means that we have in total at most 12 spoons, which is less than 15. Then our assumption about impossibility must have been not true. If first 10 utensils have 6-10 spoons, then there are 4-10 forks and the argument goes the same.

3B1B has a wonderful video about this: https://youtu.be/FhSFkLhDANA

We start by selecting the first 10 utensils. The difference between forks and spoons will be an even number, or zero. If we slide our frame to the right one utensil, the difference will change by either zero or two. Since at some point during the process of moving across the entire group of utensils, we either have zero every time, or we go from above zero to below zero, always moving by even numbers through the even numbers, at some point it is clear that the difference has to be zero.

She can always look at what she's picking up. It's not in the dark and she isn't blindfolded.

I would like to draw your attention to the view that this exercise is analogous to a theorem that states that if a function is continuous in an interval/convex area then the function takes every value between the function's values at any two points of this interval/area. Therefore if at one point the function value is positive and at another point it is negative, then somewhere it is 0, i.e. there is a zero-point. Although here the domain of the function is a finite set (the first of the 10 consecutive utensils can be 1..21), in a wider sense it is analogous. It cannot be that all 21 groups contain more than 5 spoons, because then the 1-10, 11-20, 21-30 groups would all contain more, so their sum, which covers the whole 30, would contain more than 15. As two adjacent groups have 9 overlapping (i.e. same) elements and only differ in their left/right 1 element, that 1 element can only make a 0 or 1 difference in the value of the function (the number of spoons out of the 10 consecutive utensils starting from the xth utensil). If we do not start from 5 spoons in the first or last group, but from more than 5, and some group's spoon number must be less than 5, and can only change by +-1 or 0 at a step, then somewhere it must be 5, just like when a continuous function must go through 0 somewhere if it must go positive from negative (or vice versa). The generalised theorem would be that if the difference between the function values of any two adjacent points is a constant or zero, then every integer number of steps=differences will be taken between any two function values. E.g. if somewhere the number of spoons is 10 out of 10 consecutive utensils, then somewhere it must be 9, somewhere it must be 8, 7, 6, 5, 4: every step must be taken till the other side of the average 5.

We will assume that there are no successful scoops and derive a proof by contradiction.

Label the utensils consecutively, left to right, with the integers 1 to 30. The first 21 of these labels let us refer to the different scoops which we can make. Scoop number one takes utensils 1 to 10, scoop number four takes utensils 4 to 13 and so on. We want to show that there is (at least!) one scoop with an equal number of forks and spoons.

By our assumption scoop 1 fails.

So suppose however that scoop one has more forks (it might have more spoons, then just swap 'spoons' and 'forks' in the argument that follows) and move on to consider scoop number two. Now scoop number two might still have more forks, so we might fail again. But could it fail by having more spoons? No, because scoop number one must have at least two more forks than spoons (6 forks and 4 spoons or an even worse imbalance). In moving from scoop one to scoop two we could loose a fork from the left hand end, and gain a spoon at the right hand end. This might not lead to success, but it certainly can't lead to scoop two failing by having too many spoons.

So if we assume that there is no successful scoop we can repeat this argument and move through the scoops asserting that each one fails by having too many forks.

But this leads to an easy contradiction! If all the scoops have more forks than spoons then in particular scoops 1,11 and 21 have more forks than spoons. But these three scoops contain between them all 30 utensils, implying there are more forks than spoons, contrary to the conditions of the problem.

So, invoking the principle of proof by contradiction, there must be at least one successful scoop.

EXTENSION

With very slight modifications my proof actually solves two harder problems -

With only 10+10=20 utensils it is still true that with one scoop of ten you can get five of each type!

Also with an imbalance of 17 to 13 utensils it is still possible to get five of each in a single scoop. Because if not you can get six more forks (or spoons) in the three disjoint scoops which is impossible.

I think that this should be a 'comment' on its own rather than merely a reply to another comment. The comment that I replied to used the fact that 10 is divisible to 30. And so the 'first 10' plus the 'second lot of 10' plus the '3rd lot of 10' had to be the whole lot of 15 forks and 15 spoons. But what if the question had been such that the total number of forks and spoons were not exactly divisible to the number of forks and spoons chosen? Regards, David Ps I was supposing that the answer would have still been 'yes it is possible to find a place to start where the number of forks equals the number of spoons. But I would like to know how to show this?

Consider the number of forks among the first ten consecutive items. If that number is five, we can stop. Let $F(j)$ represent the number of forks in the sequence that goes from j $(1 \le j \le 21)$ to $j+9$ .

Suppose, without loss of generality, that $F(1)>5$ (that is, there are more forks than spoons). That implies there are at least eleven spoons among the last twenty items. Note that either $F(11)<5$ or $F(21)<5$ (if $F(11)\ge 5$ , that implies there are at least six spoons in the sequence from the 21st to the 30th item, meaning $F(21)<5$ ).

Furthermore, note that, for every $1 \le j \le 20$ , $F(j)-1 \le F(j+1) \le F(j)+1$ , that is, upon shifting the beginning of the sequence one spot foward, the number of forks does one of three things:

• Increases by 1 (if the $j$ 'th item is a spoon and the $(j+10)$ 'th item is a fork),
• remains constant (if the $j$ 'th and the $(j+10)$ 'th item are the same),
• or diminishes by 1 (if the $j$ 'th item is a fork and the $(j+10)$ 'th item is a spoon).

Therefore, if $k>j$ , $F(j)>5$ and $F(k)<5$ , there exists a value $j<l<k$ such that $F(l)=5$ .

Therefore, if $F(21)<5$ or $F(11)<5$ , then there exists a value of $1<j<21$ such that $F(j)=5$ .

Relevant wiki: Parity of Integers

Assume that there exist such a figure where we can't find 5 spoons in one scoop of 10 consecutive utensils. Number the places of the utensils by ${1, 2, 3, ..., 30}$ . Let's define $f(x)$ as #spoons - #forks in the places ${x, x+1, x+2, ..., x+9}$ . Initially, since #forks $\ne$ #spoons let's assume W.L.O.G that #forks > #spoons in the first scoop. $\Rightarrow f(1) = k < 0$ Note that $k$ must be even because $f+s = 10$ and $f-s, f+s$ have the same parity (f = #forks, s = #spoons). Now $f(x+1)$ changes from $f(x)$ by ${-2, 0, 2}$ this is simply from taking cases of the change when shifting the scoop by one utensil.

Starting with $f(1)$ negative even number then $f(x)$ is even $\forall x = {1, 2, ..., 20}$ . If $f(x) < 0$ for all values of $x$ we get a contradiction by noting that the total number of spoons would be smaller than the total number of forks. Which means that at some point we must reach $f(y) > 0$ but the change is even each time, and we started with an even number so there must be point which $f(x) = 0$ which gives us the answer.

Intermediate Value Theoren

forks must be < 5 at one point and > 5 at another, so it must cross 5 at some point since it can only change by 0 or 1 each time

Look at the leftmost 10 utensils: Either they're 5:5, or there are more of one than the other. The difference between the two numbers must be even, as you can't make 10 by adding an odd and an even number.

Now remove the far left utensil, and add in the next one along. The difference either stays the same, increases by 2, or decreases by 2. Continue this pattern until you find a 5:5 set.

If there are more of the same type in every set of 10 utensils (the only way you could avoid encountering a 5:5 - you can't go directly from a 4:6 ratio to a 6:4 ratio by removing and adding a single utensil), there must be more of that utensil than the other in the full set. As it is established that there are 15 of each, that means at some point, there will be more of the other utensil in the set of 10, and so you will pass over a set of 10 with a 5:5 ratio.

there's an excellent solution and an expantion to this problem in the link here [https://www.youtube.com/watch?v=FhSFkLhDANA)

random e.g: 0101110110 0111001101 1001010000 (divide by order)

count 1: 6,6,3 count 0:4,4,7

Let's put the series end by end. We can always move split line deducting 1 and let count -1.

When split line moves to right by one number

count 1:6+0,6+1,6-1 count 0:4+0,4-1,7+1

again moves by one number, we notice that when count 1 doing +1,count 0 doing -1.

count 1:6+0,7-1,5+1 count 0:4+0,3+1,8-1

again

count 1:6+1,6-1,5+0 count 0:4-1,4+1,7+0

and now for middle series we have 5 1s and 5 0s -- 1001101100

No matter what series we have we can always using +1,0,-1 to get the middle one 5 1s and 5 0s.

I used an intuition to reduce the problem to smaller one, don't know if it's correct. Instead of 10 consecutive from a set of 15 forks and 15 spoons I'll take 2 consecutive from a 3 forks 3 spoons set. In the smaller scaled case you can see that you will always have a transition from fork to spoon where you can pick there your consecutive utensils.

We have 30 objects. If we count +1 for each spork and -1 for each spoon, the sum of all packs of 10 objects (1-10, 11-20, 21-30) will be 0. Furthermore one pack will have an even value. If all of these packs have a value unlike zero, then there will be a pack with a negative value and a pack with a positive value. Let us assume, the pack with the negative value is left and the pack with the positive value is right. Let us move the pack one object to the right. The value of the pack will not change if there will be exchanged the same objects or the value will be changed by 2 if the exchanged objects differ. But we know, that we will reach a positive value if reach our pack from our assumption. So the value changed from negative to positive only changing by 2 and only using even value. This is only possible, if we have a pack with value 0. So there must be a pack with value 0.

Let $f_n$ and $s_n$ be the amount of forks and spoons amongst the utensils from position $n$ to $(n+9)$ , where $1\leq n\leq 21$ .

Let $d_n=f_n-s_n$ . Consider $d_1$ , $d_{11}$ and $d_{21}$ . If one of these three values is equal to 0, we’re already done.

Suppose now that none of these 3 values is equal to 0. Notice it is impossible that all 3 values are positive or negative, since this would imply that there are more forks than spoons or the other way round. Contradiction!

Thus, without loss of generality, we may assume that $d_1<0<d_{11}$ . (Analogue argumentation is applicable to the other cases.)

Since $d_n=f_n-s_n=f_n-10+f_n=2f_n-10$ is an even number and $d_n$ and $d_{n+1}$ differ by either -2 or 0 or 2, the sequence $d_k$ where $k={2,3,...,10}$ must take on any even integral value between $d_1$ and $d_{11}$ , in particular the value 0.

Take these 10 consecutive utensils and we are done.