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UKMT Senior Kangaroo 2016

I haven't found a thread for this so I thought I'd make one :biggrin:

Just done the Kangaroo, and wondered how everyone else found it?

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Reply 1
Original post by LaurenLovesMaths
I haven't found a thread for this so I thought I'd make one :biggrin:
Just done the Kangaroo, and wondered how everyone else found it?

I thought it was ok! Needed more time tho! Got up to question 15. Can you remember any of your answers? 10 was a merit last year so really hoping to get one! How did you find it?xx
The time for that test is insane. 3 mins per question is just too little.
I can remember the first answer I put down; 121 :colondollar:
Original post by jadeemma
I thought it was ok! Needed more time tho! Got up to question 15. Can you remember any of your answers? 10 was a merit last year so really hoping to get one! How did you find it?xx


Ahh well done! If I have a look at the questions then I'll probably be able to remember my answers, I can never remember which is which though! I'd be happy if 10 questions were needed for a merit, though I found this paper easier than last year's.

I think I did about 15 questions properly, and on others I semi-guessed bits. Also, pretty annoyed because I'm sure question 18 asked for the square of a number and I found that number but forgot to square it :frown:
Original post by suejiaiuo
The time for that test is insane. 3 mins per question is just too little.
I can remember the first answer I put down; 121 :colondollar:


Agreed. I think it should be an hour and a half long. I'm pretty certain I got 121 for the first question too :biggrin: Though I found that one, and the second one, really weird for a Kangaroo paper as they just required a bit of addition...
Reply 5
I got up to question 15. Did anyone get 103 for that prime factor one (I think it was question 10)
Original post by Trapz99
I got up to question 15. Did anyone get 103 for that prime factor one (I think it was question 10)


Yes, I did :biggrin:
Original post by .Reality Check
Lauren I want to know when you will stop loving maths.

I say first year undergrad.


Hahaha let's hope that doesn't happen, but we'll see :biggrin:
I thought the time was actually pretty reasonable, but maybe because I'd practised with a couple of past papers.

The teacher invigilating us said we were allowed to keep our question papers and scrap paper, and unlike BMO the Kangaroo papers doesn't say you aren't allowed to talk about the answers online, so:

1. [sum of 1 + 3 + 5 + ... + 21] 121 - they were hoping you'd work out it's 11^2 based on the pattern they gave you, but it's also a simple AP
2. [two rows have the same sum] 950
3. [containers of base area 4dm^2 and 36dm^2] 027
4. [how many 4 digit numbers with solely odd digits are divisible by 5] 125
5. [area] I got 018 but I'm not too sure
6. [diagram with circle and tangents] 170 degrees
7. [how many edges does a prism have - between 310 and 320, and odd] 315, I think
8. [number] 007
9. [two concentric circles; AB is 32cm] unsure
10. [largest] 103
11. [number of days with same parity in 2 years] 183?
12. [right-angled triangles by joining 3 vertices of regular 18-sided polygon] unsure
13. [years] 143
14. [perimeter] 035
15. [shaded triangle inside 36 small equilateral triangles] didn't get an answer in time but now I think it might be 120
16. [3f(x)] 087
17. [1 to 10 is written out 10 times; students rub 2 numbers out and replace with their sum + 1] 649
18. [sum] 21 is the smallest number so answer is 21^2 = 441
19. [mean of 3-digit primes that are 21 less than a square] (379 + 463)/2 = 421
20. [barcode with width 12, black strips at either end and black and white strips of length 1/2] unsure

But given that I spent about 3 minutes on each question, there's likely to be some mistakes in my answers. If anyone wants the exact wording of any of the questions, I can type them up.
Original post by ShatnersBassoon
I thought the time was actually pretty reasonable, but maybe because I'd practised with a couple of past papers.

The teacher invigilating us said we were allowed to keep our question papers and scrap paper, and unlike BMO the Kangaroo papers doesn't say you aren't allowed to talk about the answers online, so:

1. [sum of 1 + 3 + 5 + ... + 21] 121 - they were hoping you'd work out it's 11^2 based on the pattern they gave you, but it's also a simple AP
2. [two rows have the same sum] 950
3. [containers of base area 4dm^2 and 36dm^2] 027
4. [how many 4 digit numbers with solely odd digits are divisible by 5] 125
5. [area] I got 018 but I'm not too sure
6. [diagram with circle and tangents] 170 degrees
7. [how many edges does a prism have - between 310 and 320, and odd] 315, I think
8. [number] 007
9. [two concentric circles; AB is 32cm] unsure
10. [largest] 103
11. [number of days with same parity in 2 years] 183?
12. [right-angled triangles by joining 3 vertices of regular 18-sided polygon] unsure
13. [years] 143
14. [perimeter] 035
15. [shaded triangle inside 36 small equilateral triangles] didn't get an answer in time but now I think it might be 120
16. [3f(x)] 087
17. [1 to 10 is written out 10 times; students rub 2 numbers out and replace with their sum + 1] 649
18. [sum] 21 is the smallest number so answer is 21^2 = 441
19. [mean of 3-digit primes that are 21 less than a square] (379 + 463)/2 = 421
20. [barcode with width 12, black strips at either end and black and white strips of length 1/2] unsure

But given that I spent about 3 minutes on each question, there's likely to be some mistakes in my answers. If anyone wants the exact wording of any of the questions, I can type them up.


Thanks for sharing, I got the same answers for you for a lot of those! Still annoyed about Q18 where I found it to be 21 but forgot to square it! Realised as soon as time was up :frown:

Also, how did you do Q16, out of interest? :smile:
Original post by LaurenLovesMaths
Also, how did you do Q16, out of interest? :smile:
Functions is one of my favourite topics so I spotted this trick straight away:

If you sub in x = n and then x = 2016/n, you can get a pair of simultaneous equations to work out what f(n) and f(2016/n) are.

So since they wanted f(8) [note]:

[Let] 3f(8) + 7f(252) = 16
[Let] 3f(252) + 7f(8) = 504

Then it's just simultaneous equations, albeit tricky ones without a calculator:
49f(8) + 21f(252) = 3528
9f(8) + 21f(252) = 48
So 40f(8) = 3480 and f(8) = 3480/40 = 87
Original post by ShatnersBassoon
Functions is one of my favourite topics so I spotted this trick straight away:

If you sub in x = n and then x = 2016/n, you can get a pair of simultaneous equations to work out what f(n) and f(2016/n) are.

So since they wanted f(8) [note]:

[Let] 3f(8) + 7f(252) = 16
[Let] 3f(252) + 7f(8) = 504

Then it's just simultaneous equations, albeit tricky ones without a calculator:
49f(8) + 21f(252) = 3528
9f(8) + 21f(252) = 48
So 40f(8) = 3480 and f(8) = 3480/40 = 87


Ahh that's clever, well done :smile:
Original post by ShatnersBassoon
I thought the time was actually pretty reasonable, but maybe because I'd practised with a couple of past papers.

The teacher invigilating us said we were allowed to keep our question papers and scrap paper, and unlike BMO the Kangaroo papers doesn't say you aren't allowed to talk about the answers online, so:

1. [sum of 1 + 3 + 5 + ... + 21] 121 - they were hoping you'd work out it's 11^2 based on the pattern they gave you, but it's also a simple AP
2. [two rows have the same sum] 950
3. [containers of base area 4dm^2 and 36dm^2] 027
4. [how many 4 digit numbers with solely odd digits are divisible by 5] 125
5. [area] I got 018 but I'm not too sure
6. [diagram with circle and tangents] 170 degrees
7. [how many edges does a prism have - between 310 and 320, and odd] 315, I think
8. [number] 007
9. [two concentric circles; AB is 32cm] unsure
10. [largest] 103
11. [number of days with same parity in 2 years] 183?
12. [right-angled triangles by joining 3 vertices of regular 18-sided polygon] unsure
13. [years] 143
14. [perimeter] 035
15. [shaded triangle inside 36 small equilateral triangles] didn't get an answer in time but now I think it might be 120
16. [3f(x)] 087
17. [1 to 10 is written out 10 times; students rub 2 numbers out and replace with their sum + 1] 649
18. [sum] 21 is the smallest number so answer is 21^2 = 441
19. [mean of 3-digit primes that are 21 less than a square] (379 + 463)/2 = 421
20. [barcode with width 12, black strips at either end and black and white strips of length 1/2] unsure

But given that I spent about 3 minutes on each question, there's likely to be some mistakes in my answers. If anyone wants the exact wording of any of the questions, I can type them up.


Did the Kangaroo after the Olympiad this morning, my brain was fried haha :s-smilie:

I agree with your answers mostly:

5. 009 I think
9. I got 256, you don't need to work out the radii of the two circles.
10. You're right ahhhh, I completely ignored BIDMAS and added brackets i.e. (1+2)x3x4x5x6=1080 therefore 005.
11. I'm glad you got 183 because that's what I put and I wasn't checking my counting.
12. Gonna look at this later, couldn't think of a neat way of doing it earlier.
15. Yeah I got 120 using trig
17. Do you mean 064?
18. Don't be too upset Lauren, I squared 21 and got 442. Like how is that even a possibility hahahaha oops.
20. Ran out of time, will probably look at that if I get time later.
Original post by danielley
Did the Kangaroo after the Olympiad this morning, my brain was fried haha :s-smilie:

I agree with your answers mostly:

5. 009 I think
9. I got 256, you don't need to work out the radii of the two circles.
10. You're right ahhhh, I completely ignored BIDMAS and added brackets i.e. (1+2)x3x4x5x6=1080 therefore 005.
11. I'm glad you got 183 because that's what I put and I wasn't checking my counting.
12. Gonna look at this later, couldn't think of a neat way of doing it earlier.
15. Yeah I got 120 using trig
17. Do you mean 064?
18. Don't be too upset Lauren, I squared 21 and got 442. Like how is that even a possibility hahahaha oops.
20. Ran out of time, will probably look at that if I get time later.


Haha, silly mistakes are just so annoying!! Looks like you did pretty well so well done :biggrin: (also I agree that Q5 is 9)
Pretty sure the answer to 20 is 116

...and I also forgot to square 21
(edited 7 years ago)
Would any of you be able to post photos of the paper up?
I did the BMO, but I'm interested to see what the Kangaroo problems were.
I didn't sit the kangaroo but my brother did and he gave me the paper so I had a go at all the questions I could do

These are my answers (may not be right):

1. 121 (11 * 11)
2. 950 (the first set apart from 1050 equals 55 so the bottom set before * equals 155 therefore * = 1050 - 100 = 950)
3. 027 (216 / 8 = 27)
4. 125 (number must end in 0 or 5 to be divisible by 5 - but 0 is not odd so it must end in 5; then there are 5^3 = 125 possibilities for the other numbers)
5. 012 (by a sketch and subtracting areas)
6. 160 (angles inside quadrilateral = 70 degrees, add right angle = 160)
7. 315 (has 3n edges - only odd multiple of three between 310 and 320 is 315)
8. 007 (values can be written as a set of sets {{k, -k}...}: 2^3 = 8 possibilities but we cannot distinguish between 0 and -0 so the answer is 7)
9. 256 (call R the radius of the big circle and r the radius of the smaller circle; then the shaded area = pi(R^2 - r^2) so k = R^2 - r^2. Drawing in R up to points A and B forms two right angled triangles, each of which have base length 16, side length r and hypotenuse R - by Pythagoras' theorem, R^2 - r^2 = k = 16^2 so k = 256)
10. 103 (replace with multiply sign apart from between 1 and 2; then 1 + 2 * 3 * 4 * 5 * 6 = 721 = 7 * 103: 103 is prime so is largest prime factor)
11. 183 (assume first year is odd and second year is even (does not matter which order); then over a thirty day period, there are 15 days of either parity - and for 31 days, there are 16 odd (for 28 there are fourteen of each); so in the first year, we take the odd months, which have respectively 31, 31, 31, 31, 30, 30 days so there are (4 * 16 + 2 * 15) = 94 days she can go swimming in this year; then repeat the process for the second year to find 89 days so the total is 183 days she can go swimming)
12. could not do this question
13. 143 (rearrange for q = 2/5p so p + q = 7/5p and the year will be a multiple of 7; also rearrange for p and similarly find that the year must be a multiple of 7; between 2000 and 3000, the first year that is a multiple of 7 is 2002, and the last is 2996 - so the total number of years is (2996 - 2002)/7 + 1 = 143, adding 1 because we are counting in the list)
14. 035 (105 = 1 * 105 = 3 * 35 = 5 * 21 = 7 * 15... the only numbers that allow the sum of two sides to be greater than the remaining side are 7 and 15 so the total perimeter is 35)
15. could not do this question
16. 087 (As 2016/8 = 252, 3f(8) + 7f(252) = 16 and 3f(252) + 7f(8) = 504; solve simultaneously for f(8) to find that 40f(8) = 3480 and so f(8) = 87)
17. 649 (the final number is the sum of all the initial numbers (550) added to the number of times we delete the two integers and write on the sum; as deleting two numbers and writing on a number is equivalent to removing one number, we find that this happens to get from 100 to 1 ninety nine times, so the total is 550 + 99 = 649)
18. 441 (n2+(n+1)2+(n+2)2(n+3)2=(n+4)2+(n+5)2+(n+6)2n^2 + \left(n+1\right)^2 + \left(n+2\right)^2 \left(n+3\right)^2 = \left(n+4\right)^2 + \left(n+5\right)^2 + \left(n+6\right)^2; So 4n2+12n+14=3n2+30n+77n218n63=0(n+3)(n21)=04n^2 + 12n + 14 = 3n^2 + 30n + 77 \Longrightarrow n^2 - 18n - 63 = 0 \Longrightarrow \left(n+3\right)\left(n-21\right) = 0. As nn is positive, n2=212=441n^2 = 21^2 = 441)
19. 421 (prime numbers are odd, so n^2 - 21 must be odd and so n must be even; 12^2 - 21 is the first number in the three-digit range, and 30^2 - 21 is the last; but we can also narrow down the possibilities because 12^2 - 21 = 123 which is a multiple of 3, so adding 6 to the number i.e. (18^2 - 21 etc) will also be a multiple of three; then test out all the remaining numbers and see if they are prime, to find that 20^2 - 21 and 22^2 - 21 are the only prime numbers that satisfy the conditions, the mean of which is 421)
20. 116 (after one strip of one colour, we then get a strip of the other colour; we may call the number of possibilities for a black-starting sequence BnB_n and for a white-starting sequence, WnW_n; then Bn=Wn1+Wn2B_n = W_{n-1} + W_{n-2} and Wn=Bn1+Bn2W_n = B_{n-1} + B_{n-2}. So Bn=Bn2+2Bn3+Bn4B_n = B_{n-2} + 2B_{n-3} + B_{n-4}: this will be defined for n5n \geqslant 5; so we define by inspection B1=1B_1 = 1, B2=1B_2 = 1, B3=1B_3 = 1, B4=3B_4 = 3; then by using the formula for BnB_n by recursion we find that B12=116B_{12} = 116 - there probably is a quicker way of doing this though)

Hope this helps :smile:
Original post by Superdjman1
Would any of you be able to post photos of the paper up?
I did the BMO, but I'm interested to see what the Kangaroo problems were.


I haven't got any pics of the paper but I'm pretty sure that UKMT will post them online, with solutions, tomorrow morning (unless I'm wrong aha)
Original post by danielley
Did the Kangaroo after the Olympiad this morning, my brain was fried haha :s-smilie:

I agree with your answers mostly:

5. 009 I think
Yes, Wolfram Alpha agrees with you. Oops.

11. I'm glad you got 183 because that's what I put and I wasn't checking my counting.
That's good news; I was counting things in such a disorganised way that I reckoned it was quite likely I had made a mistake.

17. Do you mean 064?
No I do not, although perhaps that means I completely misunderstood the question. So the sum of the numbers on the board is 10*(1+2+3+...+10) = 10*55 = 550.

Every time a student comes up to the board, they remove two numbers, but write another number such that the sum increases by one. So the overall effect is that there is one number less but one more in the total sum of numbers on the board.

After 99 repetitions of this, there is only 1 number left, and the sum of numbers on the board (and therefore the only number left) is 550+99 = 649.
Original post by ShatnersBassoon
Yes, Wolfram Alpha agrees with you. Oops.

That's good news; I was counting things in such a disorganised way that I reckoned it was quite likely I had made a mistake.

No I do not, although perhaps that means I completely misunderstood the question. So the sum of the numbers on the board is 10*(1+2+3+...+10) = 10*55 = 550.

Every time a student comes up to the board, they remove two numbers, but write another number such that the sum increases by one. So the overall effect is that there is one number less but one more in the total sum of numbers on the board.

After 99 repetitions of this, there is only 1 number left, and the sum of numbers on the board (and therefore the only number left) is 550+99 = 649.


I thought it seemed a bit too easy! I missed the "each written ten times" part. That does make it 649

Original post by Superdjman1
Would any of you be able to post photos of the paper up?
I did the BMO, but I'm interested to see what the Kangaroo problems were.


Don't want to pollute this thread, but how did you find the BMO?

Original post by LaurenLovesMaths
Haha, silly mistakes are just so annoying!! Looks like you did pretty well so well done :biggrin: (also I agree that Q5 is 9)


Tell me about it argh. I could barely think, I made some bad mistakes, well done to you! Noice, so does Desmos:smile:
(edited 7 years ago)

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