Problem solving/elementary solutions for the ukmt smc 2024
21
Like last year, below is an attempt to do this years smc in an elementary / problem solving way as possible. Aim is to show that in about half the questions you can quickly bodge a solution with fairly elementary knowledge.
1. The ans is ~0.1 as its 0.4 of .252525... and it must be recurring so a) or e). 5/2 == 2.5 and multiplying e) by that gives the original number so e). Of course multiplyig 4 by 0.0252525 gives 0.10101... by doing 4*25 for each two digit part.
2. The exact value is irrelevant as all ans lead with 27. So 0.000018 == 1.8*10^-5 and 4800 == 48*10^2, so the division is ~27*10^7 so a)
3. Pretty much as the solutions so sum of two dice = 6*7 (triangular/rectangular numbers) and 42-33 == 9. So number is 9 == n + 7 -> n ==2. So b)
4. a) and b) are out as too small for the largest angle. As the ans are integers and the ans are squares, only c) or e). Test c) and its 10+70+100 == 180 so c)
5. As solutions b)
6. As solutions d)
7. Dots so (3^4+1)(3^4-1) == 82*80. Both are even but 82 == 2*41 so a)
8. As solutions d)
9. As solutions. e)
10. Pretty much as the solutions, as the number of factors of a number expressed as prime factors so 2^3 * 11 and 2^2 * 23 -> (3+1)*1 + (2+1)*1 == 7 d)
11. The highest power must be even so b) or d) and its not b) so d). Though subbing n==1, 2 as per the solutions is a good strategy as well.. Similar bmo question a few years ago.
12. As solutions a)
13. Pretty much as the solutions, but you could note the 2nd bottom row could be written as 1+z,z,10 rather than having x and y, as edge values move up and central values double. So you get to 11+3z == 2024. So c)
14. As solutions b)
15. As solutions c)
16. Count: 2 reds in corners on same line + 2 reds in corners on different lines + 3 reds in corners: 2*2 + 4*4 + 4 == 24 b)
17. Just try numbers 4 reds means p==1, 3 reds means p==3/4*2/3==1/2 so 1 white ball, so p==0 of 2 balls being white. e)
18. Square proportional relationship between r(adius) and a(rea). r==1, a==4 (k==4) so r==sqrt(14)/2, a==14. So x == sqrt(14)/2-1 so c)
19. As solutions c)
20. Similar to the famous 1/2 == 1/3 +1/6 (and others) then x==21 gives y==420. d)
21. As solutions b)
22. By comparing/ratio of areas IH is 1//4 of FH and K is the midpoint (1/2) of FG so 32/8 == 4 so a)
23. Pretty much as solution as kites are 60-90-120 (angles) and made from two (right) 1-sqrt(3)-2 (sides) triangles (area sqrt(3)), so just count the sides e)
24. As solution e)
25. P (bottom shaded) is a 30 sector so area 4. Q (left shaded) is also a 30 sector area 4 as the vertical chord means the minor segment can be moved to the base. R (top right shaded) is also a 30 sector area 4 as the horizontal chord means the minor segment can be split into two and arranged on the left. So a). Similar to the famous euclid 2 intersecting circles - equilateral triangle / hexagon inscribed in circle.
Paper and fuller solutions are at
https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:76/#free-past-paper-anchor
https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:76/#free-past-paper-anchor
1. The ans is ~0.1 as its 0.4 of .252525... and it must be recurring so a) or e). 5/2 == 2.5 and multiplying e) by that gives the original number so e). Of course multiplyig 4 by 0.0252525 gives 0.10101... by doing 4*25 for each two digit part.
2. The exact value is irrelevant as all ans lead with 27. So 0.000018 == 1.8*10^-5 and 4800 == 48*10^2, so the division is ~27*10^7 so a)
3. Pretty much as the solutions so sum of two dice = 6*7 (triangular/rectangular numbers) and 42-33 == 9. So number is 9 == n + 7 -> n ==2. So b)
4. a) and b) are out as too small for the largest angle. As the ans are integers and the ans are squares, only c) or e). Test c) and its 10+70+100 == 180 so c)
5. As solutions b)
6. As solutions d)
7. Dots so (3^4+1)(3^4-1) == 82*80. Both are even but 82 == 2*41 so a)
8. As solutions d)
9. As solutions. e)
10. Pretty much as the solutions, as the number of factors of a number expressed as prime factors so 2^3 * 11 and 2^2 * 23 -> (3+1)*1 + (2+1)*1 == 7 d)
11. The highest power must be even so b) or d) and its not b) so d). Though subbing n==1, 2 as per the solutions is a good strategy as well.. Similar bmo question a few years ago.
12. As solutions a)
13. Pretty much as the solutions, but you could note the 2nd bottom row could be written as 1+z,z,10 rather than having x and y, as edge values move up and central values double. So you get to 11+3z == 2024. So c)
14. As solutions b)
15. As solutions c)
16. Count: 2 reds in corners on same line + 2 reds in corners on different lines + 3 reds in corners: 2*2 + 4*4 + 4 == 24 b)
17. Just try numbers 4 reds means p==1, 3 reds means p==3/4*2/3==1/2 so 1 white ball, so p==0 of 2 balls being white. e)
18. Square proportional relationship between r(adius) and a(rea). r==1, a==4 (k==4) so r==sqrt(14)/2, a==14. So x == sqrt(14)/2-1 so c)
19. As solutions c)
20. Similar to the famous 1/2 == 1/3 +1/6 (and others) then x==21 gives y==420. d)
21. As solutions b)
22. By comparing/ratio of areas IH is 1//4 of FH and K is the midpoint (1/2) of FG so 32/8 == 4 so a)
23. Pretty much as solution as kites are 60-90-120 (angles) and made from two (right) 1-sqrt(3)-2 (sides) triangles (area sqrt(3)), so just count the sides e)
24. As solution e)
25. P (bottom shaded) is a 30 sector so area 4. Q (left shaded) is also a 30 sector area 4 as the vertical chord means the minor segment can be moved to the base. R (top right shaded) is also a 30 sector area 4 as the horizontal chord means the minor segment can be split into two and arranged on the left. So a). Similar to the famous euclid 2 intersecting circles - equilateral triangle / hexagon inscribed in circle.
(edited 9 months ago)
Reply 1
19
Original post by mqb2766
Like last year, below is an attempt to do this years smc in an elementary / problem solving way as possible. Aim is to show that in about half the questions you can quickly bodge a solution with fairly elementary knowledge.
1. The ans is ~0.1 as its 0.4 of .252525... and it must be recurring so a) or e). 5/2 == 2.5 and multiplying e) by that gives the original number so e)
2. The exact value is irrelevant as all ans lead with 27. So 0.000018 == 1.8*10^-5 and 4800 == 48*10^2, so the division is ~27*10^7 so a)
3. Pretty much as the solutions so sum of two dice = 6*7 (triangular/rectangular numbers) and 42-33 == 9. So number is 9 == n + 7 -> n ==2. So b)
4. a) and b) are out as too small for the largest angle. As the ans are integers and the ans are squares, only c) or e). Test c) and its 10+70+100 == 180 so c)
5. As solutions b)
6. As solutions d)
7. Dots so (3^4+1)(3^4-1) == 82*80. Both are even but 82 == 2*41 so a)
8. As solutions d)
9. As solutions. e)
10. Pretty much as the solutions, as the number of factors of a number expressed as prime factors so 2^3 * 11 and 2^2 * 23 -> (3+1)*1 + (2+1)*1 == 7 d)
11. The highest power must be even so b) or d) and its not b) so d). Though subbing n==1, 2 as per the solutions is a good strategy as well.. Similar bmo question a few years ago.
12. As solutions a)
13. Pretty much as the solutions, but you could note the 2nd bottom row could be written as 1+z,z,10 rather than having x and y, as edge values move up. So you get to 11+3z == 2024. So c)
14. As solutions b)
15. As solutions c)
16. Count: 2 reds in corners on same line + 2 reds in corners on different lines + 3 reds in corners: 2*2 + 4*4 + 4 == 24 b)
17. Just try numbers 4 reds means p==1, 3 reds means p==3/4*2/3==1/2 so 1 white ball, so p==0 of 2 balls being white. e)
18. Square proportional relationship between r(adius) and a(rea). r==1, a==4 (k==4) so r==sqrt(14)/2, a==14. So x == sqrt(14)/2-1 so c)
19. As solutions c)
20. Similar to the famous 1/2 == 1/3 +1/6 (and others) then x==21 gives y==420. d)
21. As solutions b)
22. By comparing/ratio of areas IH is 1//4 of FH and K is the midpoint (1/2) of FG so 32/8 == 4 so a)
23. Pretty much as solution as kites are 60-90-120 (angles) and made from two 1-sqrt(3)-2 (sides) triangles (area sqrt(3)), so just count the sides e)
24. As solution e)
25. P (bottom shaded) is a 30 sector so area 4. Q (left shaded) is also a 30 sector area 4 as the vertical chord means the minor segment can be moved to the base. R (top right shaded) is also a 30 sector area 4 as the horizontal chord means the minor segment can be split into two and arranged on the left. So a). Similar to the famous euclid 2 intersecting circles - equilateral triangle / hexagon inscribed in circle.
Paper and fuller solutions are at
https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:76/#free-past-paper-anchor
https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:76/#free-past-paper-anchor
1. The ans is ~0.1 as its 0.4 of .252525... and it must be recurring so a) or e). 5/2 == 2.5 and multiplying e) by that gives the original number so e)
2. The exact value is irrelevant as all ans lead with 27. So 0.000018 == 1.8*10^-5 and 4800 == 48*10^2, so the division is ~27*10^7 so a)
3. Pretty much as the solutions so sum of two dice = 6*7 (triangular/rectangular numbers) and 42-33 == 9. So number is 9 == n + 7 -> n ==2. So b)
4. a) and b) are out as too small for the largest angle. As the ans are integers and the ans are squares, only c) or e). Test c) and its 10+70+100 == 180 so c)
5. As solutions b)
6. As solutions d)
7. Dots so (3^4+1)(3^4-1) == 82*80. Both are even but 82 == 2*41 so a)
8. As solutions d)
9. As solutions. e)
10. Pretty much as the solutions, as the number of factors of a number expressed as prime factors so 2^3 * 11 and 2^2 * 23 -> (3+1)*1 + (2+1)*1 == 7 d)
11. The highest power must be even so b) or d) and its not b) so d). Though subbing n==1, 2 as per the solutions is a good strategy as well.. Similar bmo question a few years ago.
12. As solutions a)
13. Pretty much as the solutions, but you could note the 2nd bottom row could be written as 1+z,z,10 rather than having x and y, as edge values move up. So you get to 11+3z == 2024. So c)
14. As solutions b)
15. As solutions c)
16. Count: 2 reds in corners on same line + 2 reds in corners on different lines + 3 reds in corners: 2*2 + 4*4 + 4 == 24 b)
17. Just try numbers 4 reds means p==1, 3 reds means p==3/4*2/3==1/2 so 1 white ball, so p==0 of 2 balls being white. e)
18. Square proportional relationship between r(adius) and a(rea). r==1, a==4 (k==4) so r==sqrt(14)/2, a==14. So x == sqrt(14)/2-1 so c)
19. As solutions c)
20. Similar to the famous 1/2 == 1/3 +1/6 (and others) then x==21 gives y==420. d)
21. As solutions b)
22. By comparing/ratio of areas IH is 1//4 of FH and K is the midpoint (1/2) of FG so 32/8 == 4 so a)
23. Pretty much as solution as kites are 60-90-120 (angles) and made from two 1-sqrt(3)-2 (sides) triangles (area sqrt(3)), so just count the sides e)
24. As solution e)
25. P (bottom shaded) is a 30 sector so area 4. Q (left shaded) is also a 30 sector area 4 as the vertical chord means the minor segment can be moved to the base. R (top right shaded) is also a 30 sector area 4 as the horizontal chord means the minor segment can be split into two and arranged on the left. So a). Similar to the famous euclid 2 intersecting circles - equilateral triangle / hexagon inscribed in circle.
You make me feel so stupid...
Reply 2
mqb2766
OP
21
Original post by sdfj
You make me feel so stupid...
Not really. The extended solutions are good, but perpetuate the myth that you need to do full solutions to solve the questions, whereas spotting patterns, thinking about basic properties and a good dose of problem solving mean that the solutions can be quite brief/elementary.
(edited 9 months ago)
Reply 3
the only reason I solved q24 so quick was because it was an oxford interview q I had seen before
as soon as I saw that q16 was a counting q, I skipped it immediately.
Some guy in my year apparently drew all 24 cases and it only took them a few mins, so they got the right ans by that.
as soon as I saw that q16 was a counting q, I skipped it immediately.
Some guy in my year apparently drew all 24 cases and it only took them a few mins, so they got the right ans by that.
Reply 4
mqb2766
OP
21
Original post by mosaurlodon
the only reason I solved q24 so quick was because it was an oxford interview q I had seen before
as soon as I saw that q16 was a counting q, I skipped it immediately.
Some guy in my year apparently drew all 24 cases and it only took them a few mins, so they got the right ans by that.
as soon as I saw that q16 was a counting q, I skipped it immediately.
Some guy in my year apparently drew all 24 cases and it only took them a few mins, so they got the right ans by that.
Yeah theres an oxford interview video with tom rocks which is very similar to q24, so plug numbers in and hope youre going to get back to f(3) so you have a small number of simultaneous equations to solve.
Its similar to the functional equations slides in drfrost as well.
ukmt / oxbridge interviews seem to like counting questions as its easy to describe but the constraints (corners and on both lines here) make it non standard. So its really a case of thinking how to split it up to incorporate the constraints, which isnt too hard in this case if you consider what breaks them
(edited 9 months ago)
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