Can You Solve The Impossible Age And House Number Puzzle?
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- Опубліковано 18 вер 2024
- Thanks to Filippo for the suggestion! This problem is from the 2016 International Championship of Mathematical Games.
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Solution slide 17
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I worked it much more simply. To get the .16 in the average, you had to have 25, 50 or 100 in the denominator of the average. So we were looking at 26, 51, or 101 houses to start. A little trial and error starting with 26 houses quickly got me to the answer. Sum of 1 to 26 is 351. You want something divisible by 25 with a remainder of 4. Subtracting 22 from 351 gives you 329 which fits the bill. 329/4 = 13.16, so that makes Benoit 7.
It can't be 50 or above because anything above 40 would make the average too high. That means that we can figure out right from the start that there are exactly 26 houses.
But otherwise I did it exactly your way. Took a few minutes, but was pretty easy once I figured out what was actually going on.
You must have had a good math teacher. Mine would have said that that's not allowed :)
@@owbu Yes, and I became a math teacher (and now a math teacher coach) to encourage creative thinking in mathematics :)
How do you get that the denominator of the average must be 25, 50, or 100 to get 0.16?
@@maxhagenauer24 0.16 is 4/25, which is an irreducible fraction.
The smallest possible average house number value is 1.5 (1+2, where Benoit lives in house 3). That means Benoit's age has to be less than 18.66, which means his maximum age is 18.
The smallest possible age for Benoit is 1, which means the overall average cannot be created with more than 38 houses, since the average of 39 houses would be 19.5, and the upper limit is 19.16 (20.16 - 1).
The 0.16 fractional leftover has to be evenly created using the number of houses being averaged. 16/100 is not possible because 100 is more than 38 (which we already determined is the maximum number of houses). 8/50 is not possible for the same reason. 4/25 is possible, and it also can't be reduced any further. Thus we have 26 houses on the street, and 25 of those are being averaged together.
The range of possible averages can be found by removing either the 1st or the 26th house, and finding the resulting average. The sum of 1..26 is 351. If you remove the first house, the average is (351-1)/25 = 14.0. If you remove the last house, the average is (351-26)/25 = 13.
Benoit's age thus has to be between (20.16-14) = 6.16 and (20.16-13) = 7.16. The only whole number in that range is 7. Benoit's age is 7.
The house average is thus the remaining 13.16. Multiply by the number of houses in the average (25) and we get 329. Taking the difference between that and the sum of all the houses (351), we get 22, which is Benoit's house number, which wasn't included in the average.
honestly, a much better solution
Yes, I love this approach. ❤
@tiagogarcia4900 Yeah, I love the approach of doing easy math to find upper or lower bounds first, to possibly avoid harder math.
I don't know why, but these solutions where one whittles down upper and lower bounds to get an answer are so satisfying to me.
Commenting before watching the solution:
Today being Benoit's birthday means that his age is not fractional.
The average of all the house numbers except for Benoit's is some fractional number that ends in .16. The question helpfully points out that this is equal to 16/100 which simplifies to 4/25.
So it can be reasonably concluded that there are 26 houses in Benoit's street.
The 25th Triangular number is 325, averaging out to 13. This would be if Benoit lived in house 26. If he lives in 25, the average is 1/25 higher, so if he lives in house 22, the average of every other house's number is 13.16.
This means that Benoit is 7 - a reasonable age for someone to be doing this if you're in a maths challenge. I don't know the quality of the French education system so I don't know if your average French 7-year-old could do this given enough time.
In order to conclude there are 26 houses on the street, you must eliminate the possibility of 25*k+1 houses on the street. This is somewhat trivial because with 51 houses the average will be a minimum of 25.5, which is greater than the actual average PLUS Benoit's age.
I did something like that.
Average is 20.16, so we can't have more than 40 houses (which you kinda assumed)
16/100 simplifies to 4/25. It means that in order to turn 0.16 into a whole number you have to multiply it by 25 (or any multiple of it to be precise, but we have less than 50 houses).
0.16*25=4 -> the last digit of sum of the houses minus the dude's house ends in 4.
25th triangular number is 325, I need to subtract a number so that the last digit is 4, possible numbers are 1, 11 and 21. Only 21 is correct also that is his house.
After that you figure his age
Idea from a South African:
Perhaps he is a Canadian living in Quebec? 😊😊😊😂😂😂
My French boss won't appreciate me saying this though
😂😂😂😂😂😂😂
@lorenzosotroppofigo1641 If you take the 25th triangular number and remove 21, you're only averaging 24 numbers. And last digit being 4 is irrelevant: to get the 4/25 remainder, you need 4 more than a multiple of 25.
The triangular number for 25, 325, is of course a multiple of 25, so we need the sum to be 4 more. We can do so by adding back house #26 and instead removing #22.
@@NicD Oh yeah damn, I did it right the first time, then when I had to do it the second time to write the commentI redid the process counting an house less.
Sorry about that, I swear the first time I checked with the video and everything was fine ahah
After a bit of thinking, I followed what I now see is the same path as several of the other commenters and realized that 0.16 = 4/25 meant that the number of houses had to be a multiple of 25, and since the average of 50 houses is 25.5 which is higher than the answer of 20.16, the number of houses must be 25, and the sum of the house numbers must also give a remainder of 4 when divided by 25. 1+...+26 = 351, which means that you have to subtract house number 22 to get 329 which is 13 + 4/25. This finally gives you Benoits age of 7.
Had a very similar reasoning.
Some of your recent problems have been too simple to be deserving of your channel. This one definitely brings you back on course!! Well done!!
His channel is for all ages and math proficencies, though... (well, maybe not all, but you get the point) but yes, it's definitely fun to get a bit of a challenge now and then.
Omg, Presh's talking about me ! This riddle sounds very pleasant to my ear, thank you.
Now I'll pause the video, and give this problem a try.
Easier way would have been to identify that the decimal part of the average is unaffected by addition of his age to the average because his age is an integer. Now, 0.16 can only become an integer when multiplied by 25 or some multiple of 25 (like 50). This is important because the number of houses cannot be in decimal and will always be an integer. So, the houses are one more than 25 (like 26) or one more than 50 (like 51) and so on. This is because Benoit's own house is skipped here.
Also, Benoit's age and the integer part of the average add up to 20 (X+Y = 20) as the number comes to 20.16.
Let, X be Benoit's age and Y be the integer (whole number) part of the average.
If we were to multiply (Y +0.16) with 25 (say) as explained earlier, it comes to 25Y + 4 which is equal to 26 X 27 / 2 - N
Here, using the formula n(n+1)/2 for addition of 26 houses in total and the subtracting Benoit's house number N from it
Thus, 25Y + 4 = 351 - N
25Y + N = 347
N now needs to be such that Y is an integer, so it will have to be 22 cause then we have 25Y = 325 and Y comes as 13. We cannot choose N to be 47 as we assumed number of houses to be only 26 here so Benoit's house cannot be numbered 47.
If we were to try with 51 houses next, Y comes to 26, which is not possible as Benoit's age and Y are equal to 20 only.
So, Number of houses is 26 and Benoit's house is numbered 22 and because Y is 13, and X+Y = 20, Benoit is 7 year's old.
Sorry, but my birthday is not today…
😆😆😆
But there will come a day, maybe soon , where this comment will be untrue.
Rewatch this video on your birthday
But is it today?
Congrats on turning 7
My solve was so much easier than all that algebra. I just poked around with calculating the case for three or four houses and quickly noticed:
-Removing 1 number from the average can at most change the result by 1
-The amount the average result changes is done in increments of 1/(n-1)
For example for four houses, depending on which house is being removed, the resulting average is 3, 2.67, 2.34, 2. Or in other words, 9/3, 8/3, 7/3, 6/3. Four houses means moving in increments of 1/3.
So, because the end result ends in a .16, or a 4/25, there must be 26 houses to move in increments of 1/25. I did the average of if Benoit was in house 26, which was 13, which is where I probably could've just said that Benoit is 7; but to be sure I calculated the 4th result down from 26 to add the 4/25, and it was 13.16. 7 off from the 20.16 goal.
I did a little short and less complicated. Since the average of the house numbers is the only way to get a fraction in the results, and 16/100=4/25 we can conclude that the number of houses in the street must have a remainder of 1 when dividing by 25. With 51 the minimum average would be reached with Benoits house number being 51, the average of the remaining houses would be (51*50/2)/50=25.5 which is too large already as Benoit can not have a negative age. For all greater numbers the average would increase even more, therefor we know already that the number of houses must be 26. If Benoits house number is x, the average is ((26*25)/2-x)/25=(26*25-2x)/(2*25)=(702-2x)/50.
702 has the remainder 2 when dividing by 50. To get the decimal numbers 16 behind the dot we need 702-2x to have the remainder of 8 when dividing by 50. Therefor 2x needs to have the remainder of -6 which equals the remainder of 44. Since x needs to be less or equal to 26 2x needs to be less or equal to 52, the only possibility for 2x is therefor 44, which means x=22. From her we can easily calculate the average house numbers which equals (702-44)/50=13.16. We can now conclude that Benoits age must be 7 years with 20.16-13.16=7.
So the info that Benoit's birthday is today was given to us to indicate that he's not a newborn baby calculating averages, but is at least 1 year old. Riiight....
I guess it's more to rule out the possibility that his age is not a whole number ...
@@matthiasschulz3569 mmmm, age>=1 actually follows mathematically though. Unless I'm missing something, his age being a whole number does not follow simply from the statement "His birthday is today". All one needs to demonstrate that is to say that 1. It is both possible and commonplace to have "fractional ages" in common parlance (e.g "I'm 3 and a half years old"), and 2. Nothing in the puzzle claims that we should "round" his age. Therefore it should be completely possible to have, say, an age like "3 years 45 minutes". As a question for the people much smarter than me at math, is this problem still readily solvable if one does NOT assume that his age is a whole number? If not, it feels important to state that specifically, rather than making prospective solvers assume that the his age should be assumed to be a whole number.
The day one is born is totally a birthday... I think they said that to hint that the age should be a Natural number, though that still forces us to assume we're not considering the time of birth.
I was referring to the 8:15 mark of the video
@@acarbonbasedlifeform70 so were we
I wrote a Python program to calculate his age:
I'm new to Python, so there are probably better methods.
# y is the house number
# x is the number of houses
for y in range (1,40):
for x in range(1, 40):
sum = 0
count = 0
for i in range(1, x+1):
sum = sum + i
count = count + 1
sum = sum - y
count = count - 1
if count > 0:
average = sum / count
fractional_part = average % 1.0
if fractional_part > 0.159 and fractional_part < 0.161:
age = 20.16 - average
print("Benoit's age is", age)
The average of houses is 20.16-x, so the number of houses (n-1) must be a multiple of 25. n=25k+1
if there are n houses, the average of n-1 houses is at least n/2, at most n/2+1 (if the benoit's house number is 1 or n, factorise and calculate)
when k=2, n=51, n/2=25.5. So k=2 is impossible, k=1
k=1, the average must be 13.16, his age is 7.
At the age of 7 Benoit did that calculation.... wow
Originally I was thinking wow, Benoît has a house? At age 7?
@@parodoxisHis family’s house?
@@cloverisfan818 presumably that's what is meant yeah.
He calculated the average and added his age. He did't solve this puzzle.
At the age of 7 years Benoit does not know yet that it is illegal to add a time to a unit-less number.
The fraction of the sum is .16=4/25. So the number of houses (except his should be multiple of 25). But if it is 50 or more then the average will be 25.5 or greater. That would lead to the age added will be negative. Therefore it is 25. So the total number of houses is 26. The sum of 1 to 26 is 351. But the remainder of the sum divided by 25 should be 5. So his house should be such number that subtracted from 351 would give difference that has 4 as a remainder when divided by 25. Therefore his house is 22 (351-22=329/25=13 4/25 = 13.16). And his age is 20.16-13.16 = 7.
It can be concluded without long equations and paperwork. Actually, fractional part can be only from the average of house numbers and 16/100 is 4/25. So, denominators can be 25, 50, 75... That means the number of houses are 1 above denominator (as Benoit's house is subtracted). 50 or more houses are not possible (actiually, no more than 40) as average must be less than 20. That implies the number of houses are 26, sum of all is 351 (27*26/2). The smaller number that will give the remainder of 4 when divided by 25 is 329 (Benoit's house is 22) and then the average number is 13 4/25. Finally, his age is 7.
5 (FIVE) lines of code in Python, altho *it could technically be done in 3 (THREE), and **_even LESS_* , albeit with reduced readability and increased clutter:
CODE
for houses in range(4,40):
for withoutH in range (1,houses):
total = sum(range(1,houses+1))
spec_avg = float(total - withoutH) / (houses - 1)
if str(spec_avg).split('.')[1] == "16": print(houses, withoutH, spec_avg)
CODE
Running this produces *the output: "26 22 13.16"*
Therefore, *the number of houses is 26* . _Benoit lives in house #22_ . *His age is 7* (20.16 - 13.16).
.
I did it the fun way of charting on excel the total possible number of houses from 3 to 40, and the averages of each number of houses excluding each number up to the total number. And ctrl+F ".16", and there you go, 13.16, in 26th row, 5th cell from the left, so the total number of houses is 26, and it's the 5th house from the end, so 22nd house. And 20.16 -13.16 = 7 years old.
I did the same but I happened see the 13.16 when I quickly glanced at the table to see if it looked about right.
Surely that's the anti-fun way?
I actually figured this out with much less math and some guessing: since the age is a whole number, the .16 in the sum must come from calculating the average. Since 0.16 = 4/25 i guessed that there were 25+1=26 houses. Then the average can be calculated to be 13 + (26-b)/25. If b = 22 then it checks out - the average is 13.16 and in this case the age must be 7.
This method does not check if this is the only answer but it's still nice :)
Just used a simple Excel Visual Basic macro and it solved it in about 1 second 😀 I let n = number of homes and x = Benoit's house number where x
I can succesfully say I understood nothing but knew I should stop as soon as you said it was from the international championship of mathematics. I saved my self a headache. Thanks for the tip
I would've been better off if I just wished Benoit a Happy Birthday and walked away!
(Just watched the video. My solution is basically the same but more logic, less math.)
This is how I worked it out.
The equation for the sum of numbers 1 to n = n(n+1)/2.
If we were taking the average of all the houses then we would divide by n.
Average = (n+1)/2
At the minimum age of 0, this means taken average is around 20.16 and therefore n=~41 at the very most. The taken average does not include his house thus you would divide by (n-1) instead, but since "41/40=1.025" is a insignificant difference you can still say n = ~41 at most.
Next, a big hint is the taken average leaves a decimal of 0.16 which can be used to identify the denominator. With a small range of 1 to 42, you can divide each number by 0.16 to find an integer (or reduce 16/100). Doing so yields that multiples of 25 can yield 0.16, however due to the restricted range only 25 works.
Thus n-1=25, therefore n=26.
The taken average can be written as:
[total - H]/(n-1)
= total/(n-1) - H/(n-1)
where the total would be n(n+1)/2.
Substituting n=26 we get:
14.04 - H/25
At this point H doesn't matter; only the fact that H is between 1 and 26, and therefore H/25 is between 0.04 and 1.04. The only value with a 0.16 decimal between 13 and 14 is 13.16, and that is the average of the houses.
Simple math solves that the added age is 7yrs old.
At this point the question is a bit confusing with "Today is Benoit's birthday."
- Was 20.16 calculated before today? If so he is 8yrs old.
- Was 20.16 calculated today? If so he is 7yrs old.
Thus Benoit is 8yrs old, or if 20.16 was calculated today then he is 7yrs old.
How I solved it:
First, I recognized that the number 20.16 places a limit on the number of houses. To find the limit, I assumed Benoit is 0 years old and lives in the last house, then found the largest number of houses which would produce a number smaller than 20.16. That number was 40.
Second, I realized the decimal portion was a function of dividing a sum of house numbers by the number of houses excluding Benoit's. The only number less than 40 which can do this is 25, which means there are 26 houses including Benoit's.
At this point it was obvious Benoit's age was 7: If Benoit lives in house #1, the average of all the houses is exactly 14 (which is the maximum). And if he lives in house #26, the average of the houses is exactly 13. However, the average has a decimal, so we know Benoit must not live in either house #1 or house #26. In other words, the whole part of the average must be 13, and the only way to reach a whole part of 20 is if Benoit is 7.
We can also find which house Benoit lives in: We know the average of the house numbers is 13.16, because it is 7 less than 20.16. If Benoit lives in house #26, the average is exactly 13.00. We need to somehow add 4 to the numerator so we can get a remainder of 4/25 = 0.16. 26 - 22 = 4, so we know Benoit lives in house #22.
Solved this by other means.
1. (sum(N) - n) / (N - 1) + m = 20.16, where N = number of houses, n - the number of Benoit's house, m - Benoit's age.
2. (sum(N) - n) / (N - 1) must be a number that's result ends with a 0.16 fraction.
3. 0.16 = 4/25, to return an integer, N - 1 must be equal to 26 or 101 or more, but, if N - 1 > 26, there is no solution where n
If you miss the decimal in the thumbnail, he always lives at house number two and his age is variable. (2014 minus the number of houses on the street)
I couldn't come up with an efficient solution but I think I got there....
Benoit is 7! I believe his house number is 22 and he lives on a street with 26 houses, including his own.... I think.
Wow, so this whole time (Chris) Benoit is a 7 year old kid who lives in house no.22 in a subdivision consisting a total of 26 houses.
Another beautiful problem written by mathematicians high enough to leave low Earth orbit.
Your solution seems to be trial and error, checking 7 and then 8. This isn't necessary. Once we've figured out n=26, we can deduce that the average house number must be between the range of 13 and 14, and since we have that .16 to work with, it must be 13.16. From there we can very easily see Benoit is 7, and we can also fairly easily calculate he's in house 22.
Adding an age to a(n average) house number is as illegal as adding a distance to a weight, or a temperature to a power, or a voltage to a number of trees.
Let a = average of house numbers, b = Benoit's age, and h = Benoit's house number.
We have a + b = 20.16
We can find a by using the formula for the sum of consecutive numbers, subtracting h, and averaging by the remaining n - 1 houses:
a = (n(n+1)-2h)/2(n-1)
Since it's Benoit's birthday, b is an integer, meaning the remainder of .16 will come from a. Write a = x + .16, for some integer x. We can determine that 25 divides 2(n-1) by noticing that .16=4/25.
Thus, n is a multiple of 26. Take n = 26. Then:
a = x + .16 = (26(27)-2h)/2(25) = (702 - 2h)/ 50
so 50x + 8 = 702 - 2h
50x = 694 - 2h
25x = 347 - h
We know h is at most n, which forces x = 13. Therefore, b = 20.16 - a = 20.16 - (x + .16) = 20.16 - 13.16 = 7, so Benoit is 7 years old.
In your fifth line, 2(n - 1) must be inside grouping symbols, such as this [2(n - 1)], because it
occurs as a denominator in this horizontal form. For the same reason, in your ninth line,
2(25) needs to be inside grouping symbols, such as this [2(25)], because it occurs in the
denominator.
Always good to have education content on UA-cam
I solved it, but differently. After some messing around with lower values of n I realized that the average of the houses minus the house number Benoit lived in was 0.5n+((n-b)/(n-1)) and because he was adding his age to this 0.5n+((n-b)/(n-1)) had to have a 16 to the right of the decimal point. Also, since n is an integer, 0.5n was going to be either an integer or exactly half way between integers I realized that (n-b)/(n-1) had to have either a 16 or a 66 to the right of the decimal point. After a little trial and error I determined that this occurred when n was 1 greater than a multiple of 25 and that 51 (and therefore every solution to 1+25x where x is an integer greater than 1) would be too high. Now, knowing n=26 it was simple to work out that b=22 was the only way to make 0.5n+((n-b)/(n-1)) equal 13.16 and that that was the only b and n for which there was a 16 after the decimal. Then it was simple subtraction
Also, I think the question is slightly ambiguously worded to show that the sum is exactly 20.16 and not just accurate to the 2nd decimal place, because there are multiple solutions if we only have to end up with 20.16 accurate to the 2nd decimal place. But since there was only one answer that got it exactly, I figured we must be looking for the one that got it exactly. However, n=20, b=17, a=10; n=32, b=27, a=4; n=33, b=12, a=3; n=38, b=32, a=1; and n=39, b=14, a=0 all also get get you 20.16 accurate to the 2nd decimal
It's his birthday, so his age is a whole number.
Therefore, the average of the house numbers must be some number ending in .16.
16/100 simplifies to 4/25.
So the divisor in the average calculation is 25.
So there are 26 houses (since his own isn't in the calculation).
I brute forced the average of "sum of all numbers 1-26 except one" to discover the house number.
If 22 is the missing number, then the result is 13.16. No other missing no. results in a number ending in .16.
20-.16-13.16 =7
He's 7.
20.16 = 20 & 4/25, therefore the houses averaged must be a multiple of 25. The average will be approximately (min+max)/2 with some variation for B living in the middle of the range. The highest increase of the average will be yielded by B living at house 1 and lowest increase by B living at the highest number house. When we check 51 houses with B living at 51, we see it has an average of 25.5 (out of range), therefore the only possible answer is that there are 25 houses averaged (26 total).
When we check the min-max for 25 houses averaged, if B lives at house 1 the average is 14, and at house 26 the average is 13.
13.16 is the only number in this range that can satisfy the question, therefore B is 7. There is no need to determine the specific house (the question does not ask for it). But we know it is 4/25 greater than the min and 21/25 less than the max, this means he lives at house 22.
We know with this method that the answer is unique.
solution given is so much more complicated than needed, if n is even then the fractional part of (21-21/25) must be (b-1)/(n-1) which means (b-1)/(n-1) = 21/25 which means b = 22 and n= 26 which implies that benoit's age is 7. you can prove the odd n doesn't work because only n which satisfy this equation are n = 50k+1 which means the average of house numbers excluding benoit's house is greater than or equal to 25 which already exceeds 21 so its not possible to have odd n.
Picture from smögen in sweden
"Ben-wa"? Didn't he invent some famous balls or something?
At school, I would have been caned for saying eg "forty point thirty-two". We had it impressed on us from a young age to say "forty point three two".
He's 7 years old.
I did not do the math necessarily. I did it the long way, and there are 26 houses.
Congratulations for your pronounciation of Benoît 😂
I brute forced using code and got the right answer. I'm surprised the hand solution was so long
§ 1:
The average house number has no unit. The age has a unit of time. It is illegal to add a time to a unit-less number. So there is no solution. This is the only mathematically correct answer!
§ 2:
But since Benoit is only 7 years old, he does not know about this rule, so we can forgive him.
§ 3:
I hope you can see the logical error of my reasoning in paragraph § 2 and agree with § 1 - at least if you are older than 7 years.
bravo. That was a fun trip.
Benoit needs to get out more often.
A little nitpick at 8:49
318 + 50a = b
If a ≥ 1 then we should have b ≥ 368
Суперская связка, такие редко попадаются в открытом доступе, в основном в приватках и могут подкинуть прям друзья!
10:51: The unit "years" comes out of nowhere. A baseless illegal assumption is made here. He could also be 7 decades old by the same reasoning, for example.
No, he had his birthday that day, so he is atleast 1.
Birthdays are made every year, so his age is in years.
Thala for a reason ?
4:21 Aaagh! "Forty point thirty-two" !! 😨
😱😱😨😨😰
I looked for what value of n would get me a 0.16 fraction. So 16/100 or 4/25 at best. Average is n-1, so it would have to be 26 vs 25. Then just jiggle the numbers, add Magic Powder, stir, shake, sift, then out comes n=26.
20.16 : so benoît is 1 too big.
Min avg : 13 and max avg : 14, so benoit between 6 and 7.
6 not possible so benoit is 7.
Here's my attempt, I got that their street has 26 houses, their house number is 22, and they are 7 years old.
By using triangular numbers we can pretty easilly find that the given number is equal to the following expression:
(N(N+1)-2n)/(2N-2) + a
Where N is the total number of houses and n is the ommitted person's house number. a is the person's age.
To make this easier, im gonna instead try to get an expression for the remainder 'r'=m/k. As such we can automatically ignore 'a' since that is an integer.
(N(N+1)-2n)/(2N-2)
=(N(N-1)+2(N-n))/(2N-2)
=(N-n)/(N-1) + N/2
=(N(N+1)-2n)/(2N-2)
=((N-1)(N+1)-2n+N+1)/(2N-2)
=(N+1-2n)/(2N-2) + (N+1)/2
Assume N is even r!=0:
r = (N-n)/(N-1) = m/k
1-r = (n-1)/(N-1) = (k-m)/k
Assume N is odd:
Assume 2nN+1
2r=(3N-2n-1)/(N-1)
3-2r = (2n+4)/(N-1) = (3k-2m)/k
Note for the given problem we have m=4, k=25. N
I just came to see how you pronounced Benoit
I computed 27*13/25, then tried subtracting integers from 27*13 until I got something ending in .16. That was (27*13-22)/25. Then I subtracted 13.16 from 20.16 and got his age, 7.
He's 7 and lives in house 22.
Huge fan bro make more videos like this one
From the thumbnail:
Benoit's house number = 22 , Benoit's age = 7
_Calculation:_
Let
N = number of houses
K = Benoit's house number, K ≤ N
B = Benoit's age
and N, K, B are non-negative integers.
Then
sum of house numbers (minus Benoit's house number) : S = N(N+1)/2 - K
average of house numbers (minus Benoit's house number) : A = S/(N-1)
total of averaged house numbers and Benoit's age : T = A + B = 20.16
A = 20.16 - B
S/(N-1) = 504/25 - B
Since S and B are integers, (N-1) must be a positive multiple of 25 .
A = S/(N-1) =
= [N(N+1)/2 - K]/(N-1)
= ½N + 1 + 1/(N-1) - K/(N-1) [eq. 1]
... 0 < K ≤ N , so K/(N-1) ≤ N/(N-1) < 2 ...
> ½N + 1 + 1/(N-1) - 2
= ½N - 1 + 1/(N-1)
With A ≤ 20.16 , we must have
½N - 1 + 1/(N-1) ≤ 20.16
N - 2 + 2/(N-1) ≤ 40.32
N + 2/(N-1) ≤ 42.32
N ≤ 42
And since (N-1) is a positive multiple of 25 , we must have N = 26 .
So with N = 26 , K ≤ 26 , and (see equation 1) A = ½N + 1 + 1/(N-1) - K/(N-1) :
A + B = 20.16
½N + 1 + 1/(N-1) - K/(N-1) + B = 20.16
½*26 + 1 + 1/(26-1) - K/(26-1) + B = 20.16
13 + 1 + 1/(25) - K/(25) + B = 20.16
14 + 0.04 - K/25 + B = 20.16
14 + B = 20.12 + K/25
... LHS is an integer, hence K/25 = 0.88 in order to make RHS an integer ==> K = 0.88 * 25 = 22 ...
14 + B = 21
B = 7
Benoit must be a very smart 7 year old.
He didn't solve this problem though... he just counted houses and averaged them. He knew all the variables
Saying forty point thirty two is really sloppy. It’s also incorrect
At 7:39, was the LHS of the equation written incorrectly? If the whole side was supposed to be divided by two then you're need another set of parentheses encapsulating everything to the left of the / wouldn't you? As written, only the k(...) element would be divided by 2.
Edit; to clarify, what I mean is, should it read (1 + k(625k + 50a - 933))/2 = b?
the line above is
2 + k(625k + 50a - 933) = 2b
the first addend is 2, not 1
so when dividing both sides by 2, the first addend of 2/2 is already simplified to 1, so it's not part of the /2 parentheses
@@VeryGoodDeals Right, that makes sense. I overlooked the first component changing from 2 to 1.
JUST SEE THE TITLE OF VIDEO!
HE ASKED THAT IF ANYONE COULD SOLVE IMPOSSIBLE PUZZLE😂😂😅
10:53 So Benoit is just a little boy?! UNBELIEVABLE!
Why not?
@e1woqf With a house? In thus economy?
No, we don't know that he is a "little boy." We find out he is a *young* boy of 7-years-old.
At 7-years-old, he may not be little, so do not use the adjective "little."
@@T.Gundham He doesn't have to be the owner. What sounds a little bit strange though is the fact, that he should have done this calculation.
But *why* did Benoît bother calculating this?? What does this info possibly do for him
I'm guessing that he was bored
@@Becky_Cooling So sad he has nothing to do on his birthday
9:30 is this found by trial and error or how do you show that for k>2 we end up in the same situation as in k=2?
The 625K^2 term is already outpacing the -933K term when K=2 and higher, and the growth rate is much faster than the increase on the upper bound N=25K+1, so all values greater than K=2 will have a house number further and further "out of bounds" than was found for K=2.
@@MichaelOninesThanks! Yes now I see, k^2 scales faster than k 😊
Correct where I am wrong , to get 0.16 fraction we have to divide by 6. i.e the no of houses must be 7 because we have to eliminate benoit house, then by formula the sum is 28 now we have to divide by 6 to a no which gives remainder of 1, so assuming Benoit house no has 3 we get total sum of 25 and its avg is 4.16 , therefore benoit age is 16 year
1/6 = 0.1(6) repeating, not 0.16 exactly. To get a whole number from 0.16, you need to multiply with 25 or a multiple of 25.
What about house = 18, no. of houses = 51 and age 26?
How would you end up with a number smaller than his age when you add his age?
wow shoot me now congrats.
Why would a 7 year old own a house
Here is how I worked this out
Grab 20.16 and divide to 4
huh the answer is supprising close
Just throwing this down here before watching solution but there is no chance I can show my ridiculous amount of factoring work. I figured his age was 8 his house number 21 and his neighborhood has 25 houses.
EDIT: rip bozo off by one everywhere somehow lol
Thank you.❤
This was a good one
Sorry but this was harder than most.
MindYourDecisions, nice content keep it up dude
Indeed amazing.
Not given is whether his age is truncated to integer in the calculation. (We can assume the house numbers are integers.)
True, especially considering birth time, but I think they tried to imply the colloquial floored-integer convention by saying it was his birthday.
Every time there's a new video notification, it becomes the best part of my day!❤️ 0 🕋,
I don't wish to be pedantic, but wouldn't the following sequence of house numbers comply with the question as asked, and make this a lot, lot more difficult, or even impossible to solve.
House 1, house 1.5, house 2.0, house 2.5, house 3.0 // The houses are numbered consecutively, from 1 and finish on an integer greater than 2. They are just not all integer house numbers. !!! It doesn't say they have to be. So oops. Unsolvable.
Who the hell calls their kid 'benoit'
They quickly gave him a weird new name after he started doing strange calculations at a very young age. 🥸
fan of chris benoit, on second thought lets scratch that.
Found the racist
😳
As a wwe fan, this just gave me ptsd
This problem sucks!
26.16=[(26×27)/2 -22]+7
Benoit is 7
and live in house 2
amazing video also Trump 2k24 🎉