Oxford University Admissions Interview Question with Solution

[@TomRocksMaths]

Check out a longer Oxford Admissions Interview Question here: kzbin.info/www/bejne/nqWlkIF9orV-jKs

[@morrispearl9981]

The way I look at this problem is that from 000 to 999, there are 1,000 numbers, so there are a total of 3,000 digits. Every possible combination of 3 digits appears exactly once. Therefore there are equal quantities of each of the 10 possible digits, therefore each digit appears 1/10 of the time, or exactly 300 times. (300 being 1/10 of 3,000)

[@sohamsud3794]

For this scenario it would probably be a more optimal method, however if you were to alter the question to a different upper limit, it would be much longer to calculate this way.

[@dudewaldo4]

^^^^

[@mildtoadstool]

And alternatively, when a, b, c is a digit of hundred, ten, one respectively then 1×9×9×1×3(if 1 had appeared only once at which digit; a, b, c) + 2×9×1×1×3(if 1 had appeared twice; ab, bc, ca) + 3×1×1×1×1(if all at once; abc)

[@5Stars49]

Very nice 😊

[@dionisispapadimitriou8269]

When counting the hundreds you should start from 100 not from 101. Great question btw really enjoyed solving. Im studying mathematics and ive been really enjoying your videos and content!

[@TomGillardMusic]

I had the same thought! Good spot

[@TomGillardMusic]

I had the same thought! Good spot

[@colinsmith6480]

First thing I noticed does 100 not have a 1 in the hundreds column ?

[@shruggzdastr8-facedclown]

Well, since there was no year-0,.....

[@RC32Smiths01]

The questions are just so interesting man. Of course it'd be for a prestigious institute! Cheers!

[@ThePokeGod]

1:29 amount of numbers is right but should be [100, 199]

[@TomRocksMaths]

yes my bad

[@hoodedR]

A very cool solution I found using symmetry. Assume that all the numbers are written in the 3 digit format where leading zeros are added to the 1 and 2 digit numbers and include 000 in the list as well. (Extending our list by that does not change the number of 1s) Now by symmetry, all digits must occur an equal number of times. (Formally, it is invariant under the transformation of any digit into any other digit. I.e. if you take any number in the list and change its digits to any other digit - it will always be in the list) Let the number of occurrences of any digit be x. (x is also the solution to our problem) Total number of digits = 3 * 1000 since there are 1000 3 digit numbers. Also, total number of digits = 10 * x by counting up number of occurrences of each digit. Hence 10x = 3000 directly gives us the solution x = 300

[@vihaannair5165]

Genius: Starts counting all the one’s from 0-999 by going through every number.

[@johnchessant3012]

As for the similar question of how many numbers from 0 to 999 _contain_ a 1, just count the opposite: There are 9^3 numbers that don't contain a 1, so the answer is 10^3 - 9^3 = 271. Or, a more thematic solution using the principle of inclusion-exclusion: The answer is ({1 ? ?} + {? 1 ?} + {? ? 1}) - ({1 1 ?} + {1 ? 1} + {? 1 1}) + {1 1 1} = (100 + 100 + 100) - (10 + 10 + 10) + 1 = 271

[@benextinction__144]

That first method is really neat.

[@deanej1]

I thought that was the question. That was the answer I got (with a slightly less elegant method). Are we saying the question actually meant 110 counts for two appearances of the digit 1? That was not at all clear.

[@tewseries1234]

I used two different methods and got the answer 271. Professor is wrong.

[@neuralwarp]

Or: There are 1000 numbers from 000 to 999. 000 is excluded but contains no 1 so we can include it for convenience. A 10th of the numbers have a 1 in each of the 3 positions. 100+100+100=300

[@xiaoyang]

This video's answer is 3*(1*10C9*10C9) or (10C1*10C1*10C1)*3/10 [ 10C1 and 10C9 both equal to 10 ] Numberphile's "3 is everywhere" answer is 10^3-9^3, because it exclude number (1*1*10C1)*3 + (1*1*1), but the question in this video is asking how many 1s appear, not how many whole number contain 1.

[@LetoTheGodEmperor]

From 0 to 999, 1/10th of the first digits will be ones, 1/10th of the second digits will be ones and 1/10th of the third digits will be ones. So 1/10th + 1/10th + 1/10th will be 300. I think this is correct and if it were to be generalized I think it would be quite easy. For example, for numbers between 0 and 999'999, 6 digits with 1/10th of the numbers in term of occurrences mean 600'000 ones. We can that for 10 digits, so for numbers between 0 and 10^10, the number of occurrences of "1" is equal to 10^10, meaning the average number of ones in a number is 1, which is quite logic. For 10^n with n>10, the average of ones in a number is higher than 1, and the total is higher than 10^n

[@anish3839]

I thought about it another way. Basically, if you consider all the numbers between 000 and 999 (starting from 000 instead of 001 makes no difference as there are no 1s in 000), you can see there are 1000 numbers. Each of these have three digits so in total, there are 3000 digits. All the numbers from 0 to 9 will occur the same number of times in these digits so the number of 1s is 3000/10 = 300. Just felt like this would be slightly quicker.

[@victoriagrace9059]

I like this idea but I’m not sure if this method works for finding the number of ones for 1 to some number xyz. Take xyz to be 199 instead of 999. Using Toms method, the number 1 appears 100 times in the hundreds column; 20 times in the tens and 20 times in the units. Therefore 1 appears 140 times if you write out the numbers from 1 to 199. Using your method, there are 200 number in this range, 600 digits so one appears 60 times. Contradiction! I believe the problem arises from the statement that “0 to 9 appears an equal number of times in these digits” since this is only true of numbers that are powers of ten (?). Anyway this was a stupidly long reply to a random comment on a maths video so who the fuck cares lmao. Sorry just had to get this out of my head

[@allypezz]

@@victoriagrace9059 It's funny how something insanely logical as the poster's comment is deemed incorrect. I have no doubt the answer is 140 as you say, but it's intriguing how this method (talking about the video now) claims you are not double counting when you are accounting for numbers 111-119 being in both the 'hundreds' and in the 'tens'. You'd imagine that 111 is triple counted when factoring in the units! Maths is a funny beast and of course there'll be an easy answer to this i'll see in hindsight. As is the way.

[@BroArmyCommander]

@@allypezz No, you are not double counting! Because, even though you may count the same number more than once, you are only adding the 1 in the column you are looking at. For example: If I am counting in the column of the tens, 112 is only 1 one, you're not adding the 1 that appears in the hundreds column.

[@BroArmyCommander]

@@victoriagrace9059 It's no really that the method is wrong, it is just not good for generality, even though it is for this particular case. If you were to use the OP's method for your example, you'd have to figure out and adjust the proportionalities for each number in the digits, which would take quite a bit of more work

[@victoriagrace9059]

@@allypezz I think the original post was correct in the statement it made, I was just saying the method wouldn't necessarily work in other cases.

[@callumfaulkner6256]

In the units section there is 10 options for the hundreds choice of what goes in the hundreds column but surely anything that is a 1 in the hundreds digit or 1 in the tens digit is counted twice ? By definition of how youve isolated the problem for example if you choose the hundreds digit to be 1 and the tens digit to be zero whilst fixing the units digit as 1 This gives you the number 101 which fits your condition for all "types" you've described in your solution

[@sadmanislam5111]

Great video. Thank you for helping students. Can I also say loving the hairstyle.

[@TomRocksMaths]

Thank you :)

[@ThePhysicsMathsWizard]

This was indeed fun! I like the approach you used, so cool

[@hjs6102]

faster: including leading zeros 1.000 numbers have 3.000 digits. as they follow each other, every digit has the same amount, so 3.000/10=300

[@fesosorpro1106]

I thought of a 10x10x10 box with 0 to 9 labeled to each column, then each small grid will represent an unique 3 digit number ranged from 000 to 999. We know that the 1s would only appears in 3 layers, and each layer has 10x10=100 grids. Hence the require answer is 10x10x3=300 ones. Quite lengthy as a solution tbh😅

[@yvesxxx5875]

from 001 to 999 u have 1000 numbers so u have 3000 single digets, every number is the same amount represented so it is 3000/10 equals 300 of each number

[@yesdcotchin]

*000 to 999

[@sohammakim9178]

I counted 20 numbers from 1-100, then from 100-200 would be 120 because of the original 20 ones, and then 1 in every hundreds place. Then you just add 20 for every more hundred you go, but be careful not to count 1,000 because then the total would be 301

[@archivist17]

Interesting and fun. I broke it down a tad further, but spotted the pattern. Incidentally, you should start the count in the hundreds column at 100, not 101.

[@jamesrawlings8493]

I thought that also.

[@ParksandPosh]

@@jamesrawlings8493 and this guy is an Oxford professor… the mind boggles

[@BroArmyCommander]

@@ParksandPosh His method includes 100, he just didn't write it as the first term. Content and not appearance

[@tajpa100]

Great solution.

[@rockman1876]

isn't that double counted? when you are counting at the hundreds it included 111 as well as ten and unit?

[@10forever46]

Exactly! I can't figure it out. Does anyone know if it is in fact double counted?

[@rockman1876]

@@10forever46 I think we misunderstood the question, the question was how many one in the series not how many numbers have one , so it is not double counted

[@TomRocksMaths]

yes exactly!

[@khizarsaleem383]

Easily solved by using FPC ( Fundamental principle of counting)

[@roshan1273]

If you would like to take the same question to the next level. Find a formula which allows you to calculate the no. of 1s to n digits. (Answer 👇) No. of 1s = (n/10) * 10^n

[@OnlyTheQuack]

I feel like this should of bin how many numbers have a '1' in them from 0 to 999. (this means that things like 110 are double counted making it more difficult)

[@enigma7791]

It's true...Maths can be fun! Well at least when Tom teaches it anyway.

[@ranakmadan3010]

I'm getting 271 as the answer, as I believe you have overcounted the number of 1's if you take all of those cases together. Considering the numbers 100 - 199, there are 1*10*10 = 100 ones. Considering the numbers m1n, there are 9*1*10 = 90 ones as m is not 1. Considering the numbers pq1, there are 9*9*1 = 81 ones as both p and q are not 1 in this case. This totals up to 271. Another solution is if you consider 000 to 999, there are 10^3 = 1000 numbers in total. Now if you want to find the number of 1's, you can find the number of numbers with at least one 1. This can be done by using complementary counting, and subtracting off the numbers with no 1's in them. This is 9^3 = 729. So 10^3 - 9^3 = 1000 - 729 = 271, giving the same answer again.

[@tewseries1234]

This is correct. The solution can be generalised as 10^n -9^n for n digits. ( works for 1 to 9 )

[@Blobfish3561]

I think you misunderstood the question. 271 is the amount of numbers between 0 and 999 which contain the number 1 but the question which he is asking is how many times does the number 1 appear between 0 and 999.

[@reccossu]

Why not recognize that if you use leading zeros there are 3000 total digits and each is used equally. 3000/10 = 300 uses of 1.

[@sagarmajumder7806]

Sir, what about 100 in your first case? I think you count it but not mentioning it. Like 101to199 there are 99 numbers in hundred place.100 to 199 there are 💯 numbers. Your 2nd case is like (10*10)matrix, in which the number of elements is (no.of rows*no. of columns) which is 10*10=100.

[@lsjkl01]

Can I make sure there are no duplicate numbers？

[@MrGreyprof]

I included 000 and 3 times 10 %=0.3x1000=300

[@arthurdoktor]

Just make a loop in python. total = 0, for i in range(1000): s = string(i), for c in s: if c == '1': total+=1, print(total). XD

[@TomLeg]

So glad you used the same logic I did! :-)

[@shruggzdastr8-facedclown]

So, it's: n(x^n/x) where n is the number of digits and x is 10?

[@secyll5565]

Dear Tom, since I couldn't find this info anywhere else I gotta ask, which books would you personally recommend in order to prepare for University? I started to get interested in mathematics thanks to you and worked through all of my old school books. Sadly I've lost the most important ones from the last 2 years and before I buy any rubbish I'd love to hear your opinion :)

[@TomRocksMaths]

For university I really like 'A concise introduction to pure mathematics' by Martin Liebeck

[@Incognitooo0815]

P.S.: Do you, Tom, at the time being hold lectures by online means such as for example BigBlueButton (BBB) ? Then please provide all of us in time with the respective _links_ to your BBB/zoom/etcetera please !

[@fudgenuggets405]

Cool, I got one right.

[@flsal27]

@1:35 why start @101 and not 100 (although you did count it)?

[@TomRocksMaths]

yes I meant to include that!

[@Dz-iu9nk]

What about 100? if you add that one would you get 301 different numbers?

[@hugohugo37]

How have you not counted 111 three times?

[@brendanchamberlain9388]

Because he isn't counting the number of positive integers less than of equal to 999 which contain 1, if he were then yes you are correct he would overcount certain numbers. He is counting the number of times the digit 1 APPEARS among the first 999 positive integers, and so the case where 1 appears in the hundreds digit is disjoint from the cases where 1 appears in the tens or ones digit.

[@hugohugo37]

@@brendanchamberlain9388 Yes I get it. He's counted 111 three times because 1 appears in that number three times.

[@superdan2593]

@@hugohugo37 Yeah, that also confused me because I first understood the question as: "How many numbers from 1 to 999 contain a 1?". Now it totally makes sense once I understood the real question! :D

[@shukailu6731]

Loved it

[@ashwalker1]

Is 100 counts as hundreds?

[@BT-kf4kx]

When can Oxford takes application for BA of Applied Quantum Mechanic ?

[@Emily-fm7pt]

10^3 - 9^3?

[@Incognitooo0815]

But __look__ , Tom, you could explain the same thing somehow distinctly __easier__ : 000-999 are 1000 numbers, each one with 3 digits, so in total 3000 digits, where each single one of 0,1,2,3,4,5,6,7,8,9 obviously occurs equally often --- namely of course 3000:10=300 times often !

[@Incognitooo0815]

_Tom_ , I would _love_ to support you _further_ in the role of _the_ most largest, most youngest maths professor ever since worldwide even at Oxford ! You find me easily enogh on facebook both under my clearname _and_ alternatively under my chess pseudonyme Siegbert Nimzomüller (with ü-Umlaut) !

[@kimshik.s]

I'm star solve this quest, he is intersting but my answer is a 290 and i didit now were is last 10 ,but then I realized.

[@tchowdhury2087]

I want that t-shirt!!! Drop the link please!!!!!

[@TomRocksMaths]

www.beautifulequation.com/collections/tom-rocks-maths

[@5Stars49]

300

[@deanej1]

I did a quick calculation in my head, but got 271, so could have easily missed something. However, It seems to me that you actually have double counted. For example, when you wrote down your second set of numbers for the tens column, I’d have said that entire row has already been counted when you calculated the numbers that started with 1 in the hundreds column. So, I took a similar approach, and reasoned that you had 100 numbers when considering all the possibilities for the first digit, consequently 90 new cases for the second digit, and 90 less 9 of the remaining numbers (those ending in 1, but not starting with 1) that we had already counted when considering the tens digit. Make sense?

[@zacnelson8732]

That was a premise of the question, not how many numbers have 1 in them but how many 1s appeared. 300 is correct. However, you would be correct if it was how many numbers with 1 appeared.

[@dennisosadchuk]

i thought 144

[@m4gn3tic82]

Why am I still confused lol

[@AsilKhalifa]

yeh toh ncert wala hei

[@alfiecollins5617]

You're definitely a strange fella, but very likable despite the oddness!

[@headred76]

Why 101 and not 100?

[@sayeedahmad8594]

But aren't 110-119 covered in 100-199?

[@allypezz]

How are you not double counting when there's 111 in the 'hundreds' and the 111 in the 'tens'? Same applies to anything between 110 and 119. Having difficulty seeing the logic there...

[@TomRocksMaths]

because we only count the appearance in a specific column. trust me, if you write out all 999 numbers you will count 300 occurrences of the number 1.

[@eternal_riftz8801]

Lol i didn't know they will ask questions like this in uni

[@TomRocksMaths]

we have fun over here

[@marcg2555]

This is great. I'm early

[@BASKETBALL_UNI]

Repent and receive life from Christ and be saved in Jesus mighty name amen.

[@yograjbir9993]

No you have double counted as 101 is there in hundreds counting and 101 again in unit's counting. Please see more carefully. Sir, If i am wrong pleaae correct me...

[@Sepioloidea]

He didn't double count. The question was how many times does the digit 1 appear, not how many unique numbers are there with at least one 1 in them. So 111 should be counted as three occurrences of the digit 1 for example.

[@yograjbir9993]

Okay thanks

[@yoshi-jh1el]

the easiest way to do maf is to have a tattoos more tattoos the more maf you can do

[@paulinebuenavista7689]

Hello Professor, may I ask you a question? What is the difference between Oxford Special Diploma and degree diploma given by Oxford University?

[@charlesarthurs]

He is not a professor.

[@iteerrex8166]

Hummm i somehow got 137 lol

[@kimjong-un8413]

New challenge, how many times does ‘1’ appesr between 1 and 999; as in 101 is two ones and 10 is one one

[@TomRocksMaths]

this is exactly the question

[@WeAsBee]

Numbers with 1only at hundredth place = 1×9×9 = 81 Numbers with 1 only at tenth place = 9×1×9 Numbers with 1 only at unit place = 9×9×1 And total number will be = 81×3 = 243

[@FaridTaba]

10. Not 9.

[@WeAsBee]

@@FaridTaba It has to be 9 so that we do not double count. You see in the video 1 at hundredth place will have 100 numbers but that will include numbers with 1 at tenth place and 1 at unit place. So we remove 1 from our list and can put only 9 numbers at ten and unit place. Hence it should be 9.

[@FaridTaba]

@@WeAsBee There’s no double counting :) I initially made the same mistake, that’s why I searched the comments section and came across yours. :) The question is NOT “How many numbers are there between 1 and 999 that INCLUDE the digit 1?” The question is “How many TIMES does the digit 1 APPEAR?” Give it a little more thought and you’ll get it like I did. I’m sure. Cheers!

[@WeAsBee]

@@FaridTaba Got it. Thank you! 😊

[@jacks5kids]

I simply can't watch this person because of all the vandalism he has done on his body. It makes me puke.

[@nigerrria3854]

Sorry Teacher..the only thing I focus in the whole session is your face..shooo cuttee..how is the solution for this Teacher if I may know?