Problem solving the smc 2023
From @DFranklin post in another thread, Ive gone through this years smc
Paper: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Paper.pdf
Model solutions: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Solutions.pdf
and tried to do it in a simple problem solving / elementary way. I doubt Ive done an "optimal" approach in some cases, but the usual points that
* There is little A level content. Up to (and including) 22 was reasonably straightforward.
* Know your pythagorean triples, and 45-45-90 and 30-60-90 side ratios and spot them in questions/answers
* Round stuff if the answers are sufficiently different
* Subbing answers is a reasonable solution in some questions
* hard questionns 20-22 were fairly write down answers with a bit of simple reasoning.
* completing the square and dots are popular
* a decent sketch with tangents .... added is very useful.
* exam time shouldnt be too much of an issue, though doing full algebraic working for each question will eat up time.
----------------------------------------
1) Either note its a ukmt “year - prime factors” question so 2023 = 7*17^2, or note the answers are reasonably different (square of the values) so 2023/7 ~ 2100/7 ~ 300 and 17^2 = 289 ~ 300. So C)
2)The difference is a recurring decimal so C) or E) and the latter is too small. So C)
3) The multiplier is 1.2*0.85 > 1. So D) or E). Simple enough to evaluate so 0.85+0.17. So D)
4) Answers are reasonably different so 5km in ⅓ hr so 15km/h. So C)
5) The possible areas are integers, so pythagorean triples are likely involved. Using the given 5-5-6 isosceles triangle, so bisect 6 to form 2 congruent right triangles which are 3-4-5, so area 12, so B).
6) After trying a few examples, you should note theyre all c-d + c + c+d = 3c. So 8+8+8, so B). Really you just note that there are 2 examples per row/col and a few on the diagonal as the answers are sufficiently different and it must be > 16 and < 30.
7) Just plugging in a few more values, it goes down 2 every 3 (as the model solution notes) so 2023-16=2007, so D).
8) The difference is 0.050505. Multiply by 100 gives ~ 5. So 5, so A)
9) Starting with 1 down, the only possibility is 2*243 = 486. 1 across must be 21^2=441 as 20^2 would mean 2 down would start with a 0 digit and 22^2 would need an 8 option in the answers so D)
10) Simple factorisation. See ukmt model solution.
11) See ukmt model solution
12) Simple factorial counting. See ukmt model solution
13) Simple sketch/angle chasing. See ukmt model solution. You could note that when you mark the right angle, the smallest angle of the triangle is 18 = 108-90, so the ratio (sum) must be a multiple of 10 as 180/18 = 10, but chasing the other two angles is probably as simple.
14) Pythagoras / ukmt model solution isnt hard but you could reason d cant be negative as 2d-6 would be too large (magnitude) so not A) or B) and not C) or D) as d<=4 is too small as the distances of the two points from the origin would be too different. So E).
15) Sub answers in, start small. Overlap 0. Then neither : football : basketball = 1:2:4 but 30 isnt divisible by 7. Overlap 5, then 30+5 is divisible by 7 so 5:10:20. So D). Note that overlap 19 A) is also divisible by 7, but the numbers dont work (7:14:28).
16) Fairly simply the base is 4 (square length to diagonal is 1:sqrt(2)) and top is half that so the average is 3. So A) is looking likely. Putting another cube on the top, the perp height is half of sqrt (2*2sqrt(2))^2 + 2^2) = 3, so A)
17) See ukmt model solution
18) Negate and sub z=2sin(x) so z^3+z^2-z-1=0, so a cubic in z and has half the number of “x” solutions as z=+/-2 are not roots and for |z|>2, the cubic term dominates, so all solutions |z|<2. z=1 is one solution as the coeffs sum to zero so (z-1)(z^2+2z +1) = (z-1)(z+1)^2, so 2 (distinct) “z” roots and 4 “x” solutions, so C.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
20) Shrink r to zero, then the only option for which q=0 is C)
21) See the model solution, but the key thing is to know the 30-60-90 side ratios and “always” connect tangent points with circle centres (OX). Doing that, the question is straightforward
22) Sub x=0 gives y=1 and its the max value so C).
23) See the model solution. Note it reduces to finding the area of the triangular overlap as base * perpendicular height / 2. A good sketch is key, the actual algrebra is relatively simple.
24) Similar to the ukmt model solution, the point where the shapes touch corresponds to the intersection of the lines y=sqrt(3)x and y=-x+1, where the origin is the left most point of the hexagon. Elimminate x and get y = sqrt(3)/(sqrt(3)+1). Rationalise and double, though its fairly obviously going to give B). A longer alternative would be trig heavy using sin(75) as a half angle of sin(30) and sin rule on the 60-45-75 triangle.
25) Similar to a question a few years ago, once youve factored out xy^2 it should be clear you want to complete the square in x and y as they are very similar (constant terms cancel) so you can root it as youll get an area bounded by lines. The model solution does dots which is similar. Writing down the first factorisation is really the hard part and then completing the square or dots should be clear.
Paper: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Paper.pdf
Model solutions: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Solutions.pdf
and tried to do it in a simple problem solving / elementary way. I doubt Ive done an "optimal" approach in some cases, but the usual points that
* There is little A level content. Up to (and including) 22 was reasonably straightforward.
* Know your pythagorean triples, and 45-45-90 and 30-60-90 side ratios and spot them in questions/answers
* Round stuff if the answers are sufficiently different
* Subbing answers is a reasonable solution in some questions
* hard questionns 20-22 were fairly write down answers with a bit of simple reasoning.
* completing the square and dots are popular
* a decent sketch with tangents .... added is very useful.
* exam time shouldnt be too much of an issue, though doing full algebraic working for each question will eat up time.
----------------------------------------
1) Either note its a ukmt “year - prime factors” question so 2023 = 7*17^2, or note the answers are reasonably different (square of the values) so 2023/7 ~ 2100/7 ~ 300 and 17^2 = 289 ~ 300. So C)
2)The difference is a recurring decimal so C) or E) and the latter is too small. So C)
3) The multiplier is 1.2*0.85 > 1. So D) or E). Simple enough to evaluate so 0.85+0.17. So D)
4) Answers are reasonably different so 5km in ⅓ hr so 15km/h. So C)
5) The possible areas are integers, so pythagorean triples are likely involved. Using the given 5-5-6 isosceles triangle, so bisect 6 to form 2 congruent right triangles which are 3-4-5, so area 12, so B).
6) After trying a few examples, you should note theyre all c-d + c + c+d = 3c. So 8+8+8, so B). Really you just note that there are 2 examples per row/col and a few on the diagonal as the answers are sufficiently different and it must be > 16 and < 30.
7) Just plugging in a few more values, it goes down 2 every 3 (as the model solution notes) so 2023-16=2007, so D).
8) The difference is 0.050505. Multiply by 100 gives ~ 5. So 5, so A)
9) Starting with 1 down, the only possibility is 2*243 = 486. 1 across must be 21^2=441 as 20^2 would mean 2 down would start with a 0 digit and 22^2 would need an 8 option in the answers so D)
10) Simple factorisation. See ukmt model solution.
11) See ukmt model solution
12) Simple factorial counting. See ukmt model solution
13) Simple sketch/angle chasing. See ukmt model solution. You could note that when you mark the right angle, the smallest angle of the triangle is 18 = 108-90, so the ratio (sum) must be a multiple of 10 as 180/18 = 10, but chasing the other two angles is probably as simple.
14) Pythagoras / ukmt model solution isnt hard but you could reason d cant be negative as 2d-6 would be too large (magnitude) so not A) or B) and not C) or D) as d<=4 is too small as the distances of the two points from the origin would be too different. So E).
15) Sub answers in, start small. Overlap 0. Then neither : football : basketball = 1:2:4 but 30 isnt divisible by 7. Overlap 5, then 30+5 is divisible by 7 so 5:10:20. So D). Note that overlap 19 A) is also divisible by 7, but the numbers dont work (7:14:28).
16) Fairly simply the base is 4 (square length to diagonal is 1:sqrt(2)) and top is half that so the average is 3. So A) is looking likely. Putting another cube on the top, the perp height is half of sqrt (2*2sqrt(2))^2 + 2^2) = 3, so A)
17) See ukmt model solution
18) Negate and sub z=2sin(x) so z^3+z^2-z-1=0, so a cubic in z and has half the number of “x” solutions as z=+/-2 are not roots and for |z|>2, the cubic term dominates, so all solutions |z|<2. z=1 is one solution as the coeffs sum to zero so (z-1)(z^2+2z +1) = (z-1)(z+1)^2, so 2 (distinct) “z” roots and 4 “x” solutions, so C.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
20) Shrink r to zero, then the only option for which q=0 is C)
21) See the model solution, but the key thing is to know the 30-60-90 side ratios and “always” connect tangent points with circle centres (OX). Doing that, the question is straightforward
22) Sub x=0 gives y=1 and its the max value so C).
23) See the model solution. Note it reduces to finding the area of the triangular overlap as base * perpendicular height / 2. A good sketch is key, the actual algrebra is relatively simple.
24) Similar to the ukmt model solution, the point where the shapes touch corresponds to the intersection of the lines y=sqrt(3)x and y=-x+1, where the origin is the left most point of the hexagon. Elimminate x and get y = sqrt(3)/(sqrt(3)+1). Rationalise and double, though its fairly obviously going to give B). A longer alternative would be trig heavy using sin(75) as a half angle of sin(30) and sin rule on the 60-45-75 triangle.
25) Similar to a question a few years ago, once youve factored out xy^2 it should be clear you want to complete the square in x and y as they are very similar (constant terms cancel) so you can root it as youll get an area bounded by lines. The model solution does dots which is similar. Writing down the first factorisation is really the hard part and then completing the square or dots should be clear.
(edited 6 months ago)
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Original post by mqb2766
From @DFranklin post in another thread, Ive gone through this years smc
Paper: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Paper.pdf
Model solutions: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Solutions.pdf
and tried to do it in a simple problem solving / elementary way. I doubt Ive done an "optimal" approach in some cases, but the usual points that
* There is little A level content. Up to (and including) 22 was reasonably straightforward.
* Know your pythagorean triples, and 45-45-90 and 30-60-90 side ratios and spot them in questions/answers
* Round stuff if the answers are sufficiently different
* Subbing answers is a reasonable solution in some questions
* hard questionns 20-22 were fairly write down answers with a bit of simple reasoning.
* completing the square and dots are popular
* a decent sketch with tangents .... added is very useful.
* exam time shouldnt be too much of an issue, though doing full algebraic working for each question will eat up time.
----------------------------------------
1) Either note its a ukmt “year - prime factors” question so 2023 = 7*17^2, or note the answers are reasonably different (square of the values) so 2023/7 ~ 2100/7 ~ 300 and 17^2 = 289 ~ 300. So C)
2)The difference is a recurring decimal so C) or E) and the latter is too small. So C)
3) The multiplier is 1.2*0.85 > 1. So D) or E). Simple enough to evaluate so 0.85+0.17. So D)
4) Answers are reasonably different so 5km in ⅓ hr so 15km/h. So C)
5) The possible areas are integers, so pythagorean triples are likely involved. Using the given 5-5-6 isosceles triangle, so bisect 6 to form 2 congruent right triangles which are 3-4-5, so area 12, so B).
6) After trying a few examples, you should note theyre all c-d + c + c+d = 3c. So 8+8+8, so B). Really you just note that there are 2 examples per row/col and a few on the diagonal as the answers are sufficiently different and it must be > 16 and < 30.
7) Just plugging in a few more values, it goes down 2 every 3 (as the model solution notes) so 2023-16=2007, so D).
8) The difference is 0.050505. Multiply by 100 gives ~ 5. So 5, so A)
9) Starting with 1 down, the only possibility is 2*243 = 486. 1 across must be 21^2=441 as 20^2 would mean 2 down would start with a 0 digit and 22^2 would need an 8 option in the answers so D)
10) Simple factorisation. See ukmt model solution.
11) See ukmt model solution
12) Simple factorial counting. See ukmt model solution
13) Simple sketch/angle chasing. See ukmt model solution.
14) Pythagoras / ukmt model solution isnt hard but you could reason d cant be negative as 2d-6 would be too large (magnitude) so not A) or B) and not C) or D) as d<=4 is too small as the distances of the two points from the origin would be too different. So E).
15) Sub answers in, start small. Overlap 0. Then neither : football : basketball = 1:2:4 but 30 isnt divisible by 7. Overlap 5, then 30+5 is divisible by 7 so 5:10:20. So D). Note that overlap 19 A) is also divisible by 7, but the numbers dont work (7:14:28).
16) Fairly simply the base is 4 (square length to diagonal is 1:sqrt(2)) and top is half that so the average is 3. So A) is looking likely. Putting another cube on the top, the perp height is half of sqrt (2*2sqrt(2))^2 + 2^2) = 3, so A)
17) See ukmt model solution
18) Negate and sub z=2sin(x) so z^3+z^2-z-1=0, so a cubic in z and has half the number of “x” solutions as z=+/-2 are not roots and for |z|>2, the cubic term dominates, so all solutions |z|<2. z=1 is one solution as the coeffs sum to zero so (z-1)(z^2+2z +1) = (z-1)(z+1)^2, so 2 (distinct) “z” roots and 4 “x” solutions, so C.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
20) Shrink r to zero, then the only option for which q=0 is C)
21) See the model solution, but the key thing is to know the 30-60-90 side ratios and “always” connect tangent points with circle centres (OX). Doing that, the question is straightforward
22) Sub x=0 gives y=1 and its the max value so C).
23) See the model solution. Note it reduces to finding the area of the triangular overlap as base * perpendicular height / 2. A good sketch is key, the actual algrebra is relatively simple.
24) Similar to the ukmt model solution, the point where the shapes touch corresponds to the intersection of the lines y=sqrt(3)x and y=-x+1, where the origin is the left most point of the hexagon. Elimminate x and get y = sqrt(3)/(sqrt(3)+1). Rationalise and double. A longer alternative would be trig heavy using sin(75) as a half angle of sin(30) and sin rule on the 60-45-75 triangle.
25) Similar to a question a few years ago, once youve factored out xy^2 it should be clear you want to complete the square in x and y as they are very similar (constant terms cancel) so you can root it as youll get an area bounded by lines. The model solution does dots which is similar. Writing down the first factorisation is really the hard part and then completing the square or dots should be clear.
Paper: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Paper.pdf
Model solutions: https://ukmt.org.uk/wp-content/uploads/2023/10/SMC-2023-Solutions.pdf
and tried to do it in a simple problem solving / elementary way. I doubt Ive done an "optimal" approach in some cases, but the usual points that
* There is little A level content. Up to (and including) 22 was reasonably straightforward.
* Know your pythagorean triples, and 45-45-90 and 30-60-90 side ratios and spot them in questions/answers
* Round stuff if the answers are sufficiently different
* Subbing answers is a reasonable solution in some questions
* hard questionns 20-22 were fairly write down answers with a bit of simple reasoning.
* completing the square and dots are popular
* a decent sketch with tangents .... added is very useful.
* exam time shouldnt be too much of an issue, though doing full algebraic working for each question will eat up time.
----------------------------------------
1) Either note its a ukmt “year - prime factors” question so 2023 = 7*17^2, or note the answers are reasonably different (square of the values) so 2023/7 ~ 2100/7 ~ 300 and 17^2 = 289 ~ 300. So C)
2)The difference is a recurring decimal so C) or E) and the latter is too small. So C)
3) The multiplier is 1.2*0.85 > 1. So D) or E). Simple enough to evaluate so 0.85+0.17. So D)
4) Answers are reasonably different so 5km in ⅓ hr so 15km/h. So C)
5) The possible areas are integers, so pythagorean triples are likely involved. Using the given 5-5-6 isosceles triangle, so bisect 6 to form 2 congruent right triangles which are 3-4-5, so area 12, so B).
6) After trying a few examples, you should note theyre all c-d + c + c+d = 3c. So 8+8+8, so B). Really you just note that there are 2 examples per row/col and a few on the diagonal as the answers are sufficiently different and it must be > 16 and < 30.
7) Just plugging in a few more values, it goes down 2 every 3 (as the model solution notes) so 2023-16=2007, so D).
8) The difference is 0.050505. Multiply by 100 gives ~ 5. So 5, so A)
9) Starting with 1 down, the only possibility is 2*243 = 486. 1 across must be 21^2=441 as 20^2 would mean 2 down would start with a 0 digit and 22^2 would need an 8 option in the answers so D)
10) Simple factorisation. See ukmt model solution.
11) See ukmt model solution
12) Simple factorial counting. See ukmt model solution
13) Simple sketch/angle chasing. See ukmt model solution.
14) Pythagoras / ukmt model solution isnt hard but you could reason d cant be negative as 2d-6 would be too large (magnitude) so not A) or B) and not C) or D) as d<=4 is too small as the distances of the two points from the origin would be too different. So E).
15) Sub answers in, start small. Overlap 0. Then neither : football : basketball = 1:2:4 but 30 isnt divisible by 7. Overlap 5, then 30+5 is divisible by 7 so 5:10:20. So D). Note that overlap 19 A) is also divisible by 7, but the numbers dont work (7:14:28).
16) Fairly simply the base is 4 (square length to diagonal is 1:sqrt(2)) and top is half that so the average is 3. So A) is looking likely. Putting another cube on the top, the perp height is half of sqrt (2*2sqrt(2))^2 + 2^2) = 3, so A)
17) See ukmt model solution
18) Negate and sub z=2sin(x) so z^3+z^2-z-1=0, so a cubic in z and has half the number of “x” solutions as z=+/-2 are not roots and for |z|>2, the cubic term dominates, so all solutions |z|<2. z=1 is one solution as the coeffs sum to zero so (z-1)(z^2+2z +1) = (z-1)(z+1)^2, so 2 (distinct) “z” roots and 4 “x” solutions, so C.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
20) Shrink r to zero, then the only option for which q=0 is C)
21) See the model solution, but the key thing is to know the 30-60-90 side ratios and “always” connect tangent points with circle centres (OX). Doing that, the question is straightforward
22) Sub x=0 gives y=1 and its the max value so C).
23) See the model solution. Note it reduces to finding the area of the triangular overlap as base * perpendicular height / 2. A good sketch is key, the actual algrebra is relatively simple.
24) Similar to the ukmt model solution, the point where the shapes touch corresponds to the intersection of the lines y=sqrt(3)x and y=-x+1, where the origin is the left most point of the hexagon. Elimminate x and get y = sqrt(3)/(sqrt(3)+1). Rationalise and double. A longer alternative would be trig heavy using sin(75) as a half angle of sin(30) and sin rule on the 60-45-75 triangle.
25) Similar to a question a few years ago, once youve factored out xy^2 it should be clear you want to complete the square in x and y as they are very similar (constant terms cancel) so you can root it as youll get an area bounded by lines. The model solution does dots which is similar. Writing down the first factorisation is really the hard part and then completing the square or dots should be clear.
Interesting, I think for question 23 I manually folded a piece of A4 paper (one we were given for workings).
Original post by My name's Cy
Interesting, I think for question 23 I manually folded a piece of A4 paper (one we were given for workings).
Agree. Questions in the past about letter reflections etc have been "done" by looking through the question paper at a light.
Original post by mqb2766
Agree. Questions in the past about letter reflections etc have been "done" by looking through the question paper at a light.
Might also be worth editing your original post to make it more obvious that it's worth learning the prime factors of the year - I feel like there's always a question related to it.
Original post by My name's Cy
Might also be worth editing your original post to make it more obvious that it's worth learning the prime factors of the year - I feel like there's always a question related to it.
Its hardly a secret that that sort of question pops up every 2 or 3 years. But often its like this year so almost as easy to do using conventional methods (round to 2100 so 300 and you should know your squares so 17^2=289).
Original post by mqb2766
Its hardly a secret that that sort of question pops up every 2 or 3 years. But often its like this year so almost as easy to do using conventional methods (round to 2100 so 300 and you should know your squares so 17^2=289).
I suppose - on an unrelated note... any thoughts on this?
https://github.com/ystael/chicago-ug-math-bib/blob/master/README.md
Obviously it's more of a general maths progression guide and other textbooks are just as useful, but is it good?
Original post by My name's Cy
I suppose - on an unrelated note... any thoughts on this?
https://github.com/ystael/chicago-ug-math-bib/blob/master/README.md
Obviously it's more of a general maths progression guide and other textbooks are just as useful, but is it good?
https://github.com/ystael/chicago-ug-math-bib/blob/master/README.md
Obviously it's more of a general maths progression guide and other textbooks are just as useful, but is it good?
Ill have a look later, but what level / background / ... are you asking about?
Original post by mqb2766
Ill have a look later, but what level / background / ... are you asking about?
Personally, I'm in Year 11, but if you look at the link it's more a collation of books (from elementary to advanced).
Original post by My name's Cy
Personally, I'm in Year 11, but if you look at the link it's more a collation of books (from elementary to advanced).
Sure, I did have a very quick scan, and there are a lot of decent/good maths books. Are you asking for you, and if so whats your background / what are you looking to do. Im not sure Ill want to go through them all and make other recommendations etc.
Original post by mqb2766
Sure, I did have a very quick scan, and there are a lot of decent/good maths books. Are you asking for you, and if so whats your background / what are you looking to do. Im not sure Ill want to go through them all and make other recommendations etc.
I want to take maths and further maths at a level and compete in olympiads, although I'm leaning towards an economics degree. This "course" would be mostly out of my own interest (for fun).
Original post by My name's Cy
I want to take maths and further maths at a level and compete in olympiads, although I'm leaning towards an economics degree. This "course" would be mostly out of my own interest (for fun).
Cant really comment on the economics part, but there are obviously a fair number of resources on aops (all levels) and ukmt (topic books are largely ~bmo2, though some of the competition books have chapters about the topics/tips etc). Apart from those I like
* Posamentier as an introduction to problem solving https://www.worldscientific.com/worldscibooks/10.1142/9478#t=aboutBook,
* Gardner Essence of maths as a way to explore the fundamentals a bit beyond gcse and has an smc-type flavour https://www.openbookpublishers.com/books/10.11647/obp.0168
* Zawaira Primer for maths competitions and a bit more advanced Zeitz art and craft of problem solving
* Decent history book so somehting like Burton History of Mathematics
* Variety of pop maths books like Maor Beautiful Geometry ..
* Martin Gardners (many) recreational maths books
They concentrate on the olympiad stuff which is largely not A level and there are 101++ others that are good and the reality is that its more important to get stuck in than worry about which is "best". The nrich problems are decent, as are the frost mat/step/smc/bmo reimann zeta slides.
The reality is that for smc/bmo1 you need little reading and getting stuck in (practice) is more important.
(edited 6 months ago)
Original post by mqb2766
Cant really comment on the economics part, but there are obviously a fair number of resources on aops (all levels) and ukmt (topic books are largely ~bmo2, though some of the competition books have chapters about the topics/tips etc). Apart from those I like
* Posamentier as an introduction to problem solving https://www.worldscientific.com/worldscibooks/10.1142/9478#t=aboutBook,
* Gardner Essence of maths as a way to explore the fundamentals a bit beyond gcse and has an smc-type flavour https://www.openbookpublishers.com/books/10.11647/obp.0168
* Zawaira Primer for maths competitions and a bit more advanced Zeitz art and craft of problem solving
* Decent history book so somehting like Burton History of Mathematics
* Variety of pop maths books like Maor Beautiful Geometry ..
* Martin Gardners (many) recreational maths books
They concentrate on the olympiad stuff which is largely not A level and there are 101++ others that are good and the reality is that its more important to get stuck. The nrich problems are decent, as are the frost mat/step/smc/bmo reimann zeta slides.
The reality is that for smc/bmo1 you need little reading and getting stuck (practice) is more important.
* Posamentier as an introduction to problem solving https://www.worldscientific.com/worldscibooks/10.1142/9478#t=aboutBook,
* Gardner Essence of maths as a way to explore the fundamentals a bit beyond gcse and has an smc-type flavour https://www.openbookpublishers.com/books/10.11647/obp.0168
* Zawaira Primer for maths competitions and a bit more advanced Zeitz art and craft of problem solving
* Decent history book so somehting like Burton History of Mathematics
* Variety of pop maths books like Maor Beautiful Geometry ..
* Martin Gardners (many) recreational maths books
They concentrate on the olympiad stuff which is largely not A level and there are 101++ others that are good and the reality is that its more important to get stuck. The nrich problems are decent, as are the frost mat/step/smc/bmo reimann zeta slides.
The reality is that for smc/bmo1 you need little reading and getting stuck (practice) is more important.
So would it perhaps be worth reading through the elementary books overtime and focus on doing problems?
Original post by My name's Cy
So would it perhaps be worth reading through the elementary books overtime and focus on doing problems?
Certainly do a reasonable number of problems and think about different ways of doing them. Jobbings ukmt imc problem solvers handbook has a fair bit of basic stuff advice as well as a fair number of questions. For the smc he has a couple of senior problems book but I dont know how much advice is in there in addition ot the questions. But get an interesting book or two to read (history/pop/....) as well as just hitting problems.
Original post by mqb2766
Certainly do a reasonable number of problems and think about different ways of doing them. Jobbings ukmt imc problem solvers handbook has a fair bit of basic stuff advice as well as a fair number of questions. For the smc he has a couple of senior problems book but I dont know how much advice is in there in addition ot the questions. But get an interesting book or two to read (history/pop/....) as well as just hitting problems.
Thank you!
Original post by mqb2766
Certainly do a reasonable number of problems and think about different ways of doing them. Jobbings ukmt imc problem solvers handbook has a fair bit of basic stuff advice as well as a fair number of questions. For the smc he has a couple of senior problems book but I dont know how much advice is in there in addition ot the questions. But get an interesting book or two to read (history/pop/....) as well as just hitting problems.
A further question - are there any books you would recommend to solidify extremely basic concepts?
Original post by My name's Cy
A further question - are there any books you would recommend to solidify extremely basic concepts?
Basic gcse concepts or ...?
Original post by mqb2766
Basic gcse concepts or ...?
Basic mathematical concepts, more so a book that consolidates the basics of arithmetic, algebra and geometry - the foundations.
Original post by My name's Cy
Basic mathematical concepts, more so a book that consolidates the basics of arithmetic, algebra and geometry - the foundations.
That tends to be a less covered area in the book market. Its probably not that popular (sales) at your level as opposed to writing a pop maths or competition maths or ... university textbook which have a clear market, even though what you mention is fundamental. If the stuff below doesnt 1/2 cover what youre after, what do you think youre missing?
The aops introductions etc
https://artofproblemsolving.com/store/index.php
maybe 1/2 what youre after? They have a fair coverage of the basics and lead to the more advanced (competition) stuff. Similarly the previously mentioned Gardner Essence of Maths (free download) has some good stuff in. These would cover stuff from a solving problems viewpoint.
Similarly many maths history books would cover this sort of stuff with the stories behind it. So why do we use base 10 place value and what others have used in the past, how algebra has developed, .... Again the previously mentioned Burton is a good one.
Original post by mqb2766
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
19) Alternative to model solution, so m = (7n+12)/(2n+3) so (rearrange and multiply by 2
4nm + 6m - 14n = 24
Factorise and subtract 21 from both sides
(2n+3)(2m-7) = 3
The potential values of (2m-7) are +/-factors of 3 and as they’re symmetric about 0 the sum of the 4 values of m is (7/2)*4 = 14, so E). m,n are integers for all these values as 2# = odd - odd.
Slight variation for Q19
By algebraic division, (7n + 12)/(2n + 3) = 7/2 + (3/2)*(1/(2n + 3)
This can only have integer values if 2n + 3 = +/-1 or +/-3
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