Oxford University Admissions Question - Can You Solve It?

University of Oxford Mathematician Dr Tom Crawford presents a question used in the 2021 Maths Admissions Interviews.

For more maths content check out Tom's website

You can also follow Tom on Facebook, Twitter and Instagram @tomrocksmaths.

Get your Tom Rocks Maths merchandise here:

[sevret]

It appears 10% of the time in each position. So 10% of 1000 is 100, take that three times to get 300.

[kieranh]

I did it like so... ANS at bottom. Also example means 1 digit.

1- 9 = 1 example
10 - 19 = 11 examples
20 - 29 = 1 example
It then repeats 1 example for
30 - 39, 40 - 49, ... 90 - 99

For 1 - 99 there are 20 examples.

This also applies for 200 - 999

So 1 - 99 examples + 200 - 999 examples = 180.

The 100 - 199 is a unique part as more examples appear

100 - 109 = 11 examples
111 - 119 = 21 examples
120 - 129 = 11 examples

For 120 - 199, 21 + (11 x 9) = 120

So total the two sets of examples which is 180 + 120 = 300.

300 is answer =)

[ryankelly]

If “a” and “b” are integer digits 0 to 9, there are 100 combinations of “ab1”, 100 combinations of “a1b” and 100 combinations of “1ab”. Therefore, a total of 300 appearances of 1

[michaelpatison]

Not really being a math person:
There’s 10 sets of ones per 100 numbers, plus 10 ones in the tens place per 100 numbers. That’s 100+10*10=200. Then just add 100 for the hundreds: 100->199. 200+100=300

[maxryder]

From 000 to 999, the total number of digits appearing is 3x1000=3000 (as 1000 numbers appear, each of 3 digits). Then by symmetry, this is divided evenly between each of 0, 1, …, 9, so 1 appears 300 times

[jamiehaldane]

I first found there was 20 1s between 1 and 99. Multiply this by 10 we get 200. Then add 1 for every digit between 100 and 199. Therefore 200+100=300

[geezer]

000 to 999 Is 1000 digits

10% of all possible first digit is one
10% of all possible second digit is one
10% of all possible third digit is one

Therefore 30% of numbers have a chance of being one.

30% of 1000 digits is 300.

Answer is 300.

[GaborRevesz_kittenhuffer]

Answer: 300. If we start with 000 and fill in the all the leading zeros, then {000, 001, ..., 998, 999} is just the set of all 3 digit long base-10 codes. That's 10³ codes, 3 digits long each, which makes 3·10³ digits in total. By symmetry, an equal number of them are 0s, 1s, ..., 8s, 9s. So one-tenth of them, i.e. exactly 3·10³/10 = 300 of them are 1s (... and also 2s, and 3s, ..., and 9s).

What about 0? How many 0s appear in 1, 2, ..., 998, 999? Well, 300 overcounts by the leading 0s. There are 99 of them in the 100s place and 9 of them in the 10s place. So there are only 300 – 99 – 9 = 192 zeros.

[Abhay-cw]

We can do thought pnc also, 1 digit, 2 digit, 3 digit where 1 appear, for 1 digit =1, 2 digit=19, 3 digit=280, so add all = 300

[willgordge]

Initially I used the combinations method, (1bc, a1c, ab1), but there’s a slightly longer intuitive method. If you pictured the sequence 1-99 on a grid that runs each decimal interval underneath, you will get two crosses of ones, the number 11 being the intersection of the ‘1’ row and ‘1’ column. That’s 10 + 10 1’s = 20 1’s per century. You multiply this by the number of ‘100’ grids in 999 (10) and get 200, finally you add the 100 ones present before each number in the 100-199 portion of the sequence. Answer = 300

[BenWilkinson-ibyx]

I'm thinking about the general case now and where to go with it:

What is the number of occurences of the digit d in the sequence 1, 2, ..., k.

Let f(d, k) equal this.

Show that for k=10^n - 1, then f(k, d)=n 10^(n-1) for any digit d

Maybe explore other interesting things for different combinations of k and d.

[pbp]

1 century in the hundreds column;
10 decades in the tens column;
100 singles in the ones column.

[greatday]

My approach was:

for 1-9, it appears 1x
for 10-99, it appears 10x in tens position, 1x9 in one's position
in total: 20x from 1-99
for 100-999, it appears 100x in hundreds position, 9x 20 as the numbers 1-99 cycle 9x from 100-999
in total: 1 + (10 + 9) + (100 + 9x20) = 300

[mkiiespi]

I got 300, used combinatorics







9C2x3! + 9x3:2 + 9x3!(x2:2) + 3 = 300
I count every number from 1 to 999 as numbers with 3 digits, so 1 is 001, 69 is 069, so on so forth
- for numbers where 1 appears once, I need to pick out the other 2 digits to fit in the number, then arrange the numbers. 9C2 is for when the other 2 digits aren't the same. The +9 is for the 9 cases when they are. 3! is for arrangement; but for the latter, since the 2 other digits are the same, there is duplicate, so we divide by 2
- same logic for where 1 appears twice, although simpler: just 9 cases. Multiply by 2 because we're looking for how many times 1 is written, and divide by 2 for duplicates, which cancel each other out.
- lastly, 111 is +3

[mahfuz]

Well in the units, we get 10 lots of 1s per 100 interval so 100 so far, and there are 3 units so 3 x 100 = 300. I hope i got that right, im signing up for oxford this year hope i see you sir 😊

[DominicFinlayson]

10 numbers that end with 1 between 1 and 99. Add 10 for the numbers that start with 1. Multiply the total by 20, then add 100 since each 1xx counts 1 time as well.

[Ramon]

The last (least significant) digit appears 1000 times. A tenth of them contain a one. The second last digit appears 1000 - 10 times, a tenth of those contain a 1. You know where I am going. I'm not going to calculate it out right now but that's how I would solve it. Simplified, if you also take 0s at the start of the numbers into account, the are a total of 1000 combinations containing 3 numbers each, of which a tenth is a 1.

[MQTate]

Got there eventually! Ace stuff Tom 😁

Numbers with 1 in units column: (1, …, 91), (101, …, 191), …, (901, …, 991) = 100 members
Numbers with 1 in tens column: (11, …, 19), (111, …, 119), …, (911, …, 919) = 100 members
Numbers with 1 in hundreds column: (100, …, 199) = 100 members

100 + 100 + 100 = 300 ∎

[Nikita]

People like me: solving it by old but gold counting...

[brochocho]

My solution might be a bit messy:
I asked myself: a) how often does the digit 1 appear 3 times in a number, b)how often does the digit 1 appear 2 times, c)how often does it appear one time.
For a): There is only 111, so 3 digits,
For b) There is 011(...) to 911, 110 (...) to 119, and 101(...) to 191 where you have to exclude 1 for each row, so 3*9=27. Since there are 2 digits for each number it is 54 digits in total.
For c) Similar to b) but there are now 2 positions that can vary in the columns, for examle 1xx if you put 1 in the first column, so 81 cobinations of that (9*9, since we exclude 1).You do the same for 1 in the second and third column a get 81*3=243
If we add a) and b) and c), we get 3+54+243=300