Problem solving/elementary solutions for the ukmt imc 2025

Below are some fairly random comments and elementary solutions for the ukmt imc 2025. Full paper and solutions are at
https://ukmt.org.uk/competition-papers/jsf/jet-engine:free-past-papers/tax/challenge-type:70/
1) As per solutions, there is little structure in the question so simply evaluate, so A.
2) As per solutions, note each digit pair is divisible by 13 apart from trailing digit 4, so D. Note doing long division wouldnt eat much time and you may spot this property and while there are divisibility rules for 13, theyre not that useful here.
3) As per solutions, the square and equilateral triangle define equal lengths so two congruent isosceles triangles. <TSR=150 so <TRP=30. So E.
4) As per solutions, 3 extra legs so 4 cats so 8 ears, so D.
5) As per solutions, 5/5=1 and 5/(1/5) = 25. So E.
6) Choose n=1, so C.
7) As per solutions, so D.
8) By inspection A as 34=2*17 and 1/7 = 2*1/14
9) Another way of asking about triangular numbers so rows t_2=2+1=3 and columnns t_5=5+4+3+2+1=15. So A. Simply colour the rows/columns in to spot the triangular number pattern.
10) Could simply evaluate for an easier number so 4^2-4-3=9=3^2, so 2024 so C. A wordy way of doing the actual question is to note you have
11) As per solutions m+n+1 is the difference 40 so m+n=39 so A.
12) Could note the number of dogs must be divisible by 3 and 5 so E.
13) It must be 3*48 (as per solutions) so E. Note the main thing is to get the 3=360/120, then its roughly 3*50.
14) Could note that if the 0.4... was 0.3... then the sum would be 1, so A. Or the (partial) sum of the 2nd, 3rd and 4th terms is 1 so A.
15) Could do using dots so (3^3-3^2)(3^3+3^2) / 2(2^4-2)(2^4+2) = 18*36/(2*14*18) = 9/7 so D.
16) As per solutions, he must have bought 44 books 100-56 (pence) and a bit of simple reasoning on the pounds gives B as 57-44=13 (1.99 books).
17) Similar to solution, assume there are 100 bits of paint, then 36+16 must be dark blue so 52:48 so D
18) Note the diagram is not to scale, the circles should overlap more and if the length was 39 (3/2*26) then it would be Euclids famous proprosition
https://www.maths.tcd.ie/~dwilkins/Euclid/Elements_Lardner1855/BookI/EuclidElementsBk01Prop01_Lardner1855.php
which is similar to a hard gcse edexcel from a couple of years ago. Using this, the overlap would be > 13sqrt(3) which leaves only 24. Alternatively as soon as you note the radius is 13 (26/2), the pythagorean triple 5:12:13 should come to mind and by drawing the radii, chords, joining centres you get two such right triangles so 24. So E. A different non elementary (but simple) way would be to use intersecting chords, so in one circle the horizontal chord is 18 & 8, so the vertical chord length is 2*sqrt(8*18 ) = 24
19) As per solution D.
20) As theyre in decreasing order, you could evaulate them and see which one gives first. You must have 2y=180+x. 180 is not divisble by 71 but it is by 45 so B.
21) As per solutions, look at the last digit and spot the repeating pattern for 8^1, 8^2, 8^3, .... and its 6/5 so B.
22) The diagram as per solutions but its a 3:4:5 pythagorean triple problem (scaled by 4) so 12:16:20 so D. Touching circles (the 3d part/spheres is irrelevant) with a common tangent line is a famous problem/way of asking questions involving pythagoras, and the (R+r)^2 and (R-r)^2 are fundamental in deriving pythagorean triples/am-gm.
23) Pretty much as solution, though note the famous relationship w^2 + x^2 = y^2 + z^2, so D.
https://en.wikipedia.org/wiki/British_flag_theorem
24) Pretty much as solution, its a bit of a grind on 3 simple simultaneous equations. Simply adding them together gives 5:4:4 so its nearly what you want but you have to get one less r ... You could note that each item must be about 2 quid and once you have 5;4:4=26.50, simply subtract 2 and divide by 4 and its a shade over 6, so approximate.
25) An similar but alternative way is to try and work out the area of the rectangle, then subtract 17. So youd have from origin in bottom left and rectangle a, b (length height) and x, y along a, b
1/2 xy = 6 so xy = 12
1/2 (a-x)b=5 so ab - xb = 10 or x = (ab-10)/b
1/2 a(b-y)=6 so ab - ay = 12 or y = (ab-12)/a
Combine so
(ab-10)(ab-12) = 12ab
(ab)^2 - 34ab + 120 = 0
ab = 30, 4
so 13 so B. Note if youd guessed that the ans was B as that would give a total area of 30 and note that the 6 & 5 white triangle areas occur along the sides and 30=6*5, then that might suggest a=6,b=5. So x=4 and y=3, gives a valid answer for the 3 white triangles and total area. So a bit of guessing of the integer solution works. Another way to guestimate the ans is to draw a horizontal line from the point where the shaded triangle touches the left border and a vertical line from the point where the shaded triangle touoches the bottom border up ot the horizontal line. This produces 3 rectangles where the shaded area is 6 in the bottom left and 6 in the top one if we rob a shaded congruent triangle from the bottom right rectangle. There is only a small shaded area left in the bottom right rectangle so we might guess its about 1, so 13 in total. Obv this assumes the diagran is accurate (which it appears to be) by comparing the different areas given.