Solutions to IMO Test 1 1. Prove that for any integer n, n 30 n14 n18 n 2 is divisible by 46410. Solution: We factorize the expression n 30 n14 n18 n 2 into n 2 (n12 1)( n16 1) . Note that 46410 2 3 5 7 13 17 . It suffices to prove p | n 2 (n12 1)( n16 1) for p 2, 3, 5, 7, 13, 17 . If p | n , the statement is obviously true. Now, we may assume gcd( p, n) 1 . For each possible value of p, p 1 is a factor of either 12 or 16. By Fermat Little Theorem, n p 1 1 (mod p) . So n12 1 (mod p) or n16 1 (mod p) , this completes our proof. 2. (a) Let a, n, k be positive integers. Prove that a 4 n k and a k have identical first digits in their decimal representations. (b) Find the first digit of the decimal representation of the number 21999 71999 91999 . Solution: (a) We are asked to prove 10 | a k (a 4 n 1) . Clearly a k and a 4 n 1 have different parity, one of them must be even. It remains to prove 5 | a k (a 4 n 1) . This statement is obviously true if 5 | a , otherwise gcd( a, 5) 1 and a 4 1 (mod 5) in accordance with Fermat Little Theorem. The result follows. (b) Since 1999 4 499 3 , by result of part (a) we have 21999 71999 91999 2 3 7 3 9 3 0 (mod 10) . So the first digit of the number is 0. Page 1 of 4 3. Let ABC be an acute triangle. Suppose a circle 1 , with center O1 , touches the sides BC produced at E, AC produced at G, and AB at C . Suppose also that another circle 2 , with center O2 , touches the sides AB produced at H, BC produced at F, and AC at B . Let the extensions of EG and FH intersect at P. Prove that PA BC . Solution: H G A M N I E B L I’ C F Let L, M, N be respectively the feet of perpendicular from A, B, C to the segments BC, FH, EG. Note that L, M, N are points lied on the sides of IBC , and LA, MH, NG are lines through L, M, N which perpendicular to the sides of IBC . We want to prove LA, MH, NG are concurrent, by Carnot Theorem, it is equivalent to ( BL2 LC 2 ) (CN 2 NI 2 ) ( IM 2 MB 2 ) 0 --------------- (1) Note that CN 2 NI 2 (CE 2 EN 2 ) ( IE 2 EN 2 ) CE 2 IE 2 . Similarly we have IM 2 MB 2 IF 2 BF 2 . Adding these two equalities and using the relation BF CE to make a simplification, we obtain CN 2 NI 2 IM 2 MB 2 IF 2 IE 2 -------------------- (2) Compare (1) and (2), it remains to prove IE 2 IF 2 BL2 LC 2 ------------------- (3) Denote I be the foot of perpendicular from I to the segment EF and p ( BF CE ) be the semi-perimeter. Then I E CE CI p ( p c) c (recall that c denote the length of AB). Similarly I F b . So IE 2 IF 2 I E 2 I F 2 AB 2 AC 2 BL2 LC 2 , this proved (3). Hence, the lines LA, MH, NG are concurrent at a point. This point is P and follows that PA BC . Page 2 of 4 4. Let a, b, c, d , l , m, n, r be given (fixed) positive numbers. Suppose x, y, z , w are positive numbers satisfying ax by cz dw 2 . Determine the minimum of F l m n r . x y z w Solution: By Cauchy-Schwarz Inequality, 2F ( So F al bm cn dr )( ax by cz dw) ( al bm cn dr ) 2 . ax by cz dw 1 ( al bm cn dr ) 2 . 2 l m n r . With the given condition : : : a b c d ax by cz dw 2 , we can solve for the corresponding values of x, y, z , w . Equality holds when x : y : z : w In fact, these values are x z 2 al bm cn dr 2 al bm cn dr l , a y n , c w Hence, the minimum value of F is 5. 2 al bm cn dr 2 al bm cn dr m , b r . d 1 ( al bm cn dr ) 2 . 2 For a graph G, G = min{k: G has a k-coloring}, G = max{degrees of vertices of G} (a) Prove that for any simple connected graph G, G G 1 . (b) A regular graph is a graph that every vertex has the same degree. Prove that if a simple connected graph G is not regular, then G G . Solution: (a) We are going to prove there is a ( G 1) -coloring for any simple graph G (not necessary connected). Without loss of generality, we use induction on n, the number of vertices of graph G. For n G 1, clearly there is a ( G 1) -coloring. Page 3 of 4 Assume the statement is true for a positive integer k G 1 . For example, any simple graph G with k vertices has a ( G 1) -coloring. Now, consider a graph G with k 1 vertices. We may choose an arbitrary vertex v and then remove it from the graph (of course, we remove the edges incident with v at the same time). This gives a new graph G with k vertices. By assumption, the graph G has a ( G 1) -coloring and follows that it has also a ( G 1) -coloring (since G G ). Then we put back the vertex v to the graph, it adjacent with at most G vertices and there must be a suitable color that we can assign this color to the vertex v. This gives a ( G 1) coloring to the graph G. By principle of mathematical induction, any simple graph G has a ( G 1) coloring. Hence, G G 1 . (b) Similar to part (a), we need to prove there is a G -coloring for any simple non-regular graph G. (In fact the proof is almost the same as part (a)) With the same notation as part (a), we use induction on n. For n G 1, since the graph is not regular, we may choose a vertex v, with deg v G , from the graph and first arbitrarily assign color to the other vertices (of course, adjacent vertices are assigned by different colors). Then there must be a color which is unused, we may assign this color to vertex v in order to give a G -coloring to the graph. Assume the statement is true for a positive integer k G 1 . For example, any simple non-regular graph G with k vertices has a G -coloring. Now, consider a graph G with k 1 vertices. Since the graph G is not regular, we may choose a vertex v from the graph with deg v G . Remove this vertex from the graph, this gives a new graph G with k vertices. By assumption, the graph G has a G -coloring. So it has also a G -coloring. Next, we put back the vertex v to the graph, it adjacent with at most G 1 vertices and there must be a suitable color that we can assign this color to the vertex v. This gives a G -coloring to the graph G. By principle of mathematical induction, any simple non-regular graph G has a G -coloring. Hence, G G . ~ The End ~ Page 4 of 4

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