Sách đánh thức tài năng toán học 3 gồm hai phần: Hướng dẫn tư duy và luyện tập Nội dung kiến thức có chiều sâu,được sắp xếp xuyên suốt với mức độ luyện tập tăng dần giúp trẻ dễ học tập và rèn luyện. Gồm nhiều dạng toán rèn luyện tư duy logic và khả năng phân tích, tổng hợp 60% kiến thức bộ sách tương đồng chương trình giáo dục của Việt Nam. 30% là các dạng toán tư duy logic, thực tế… rèn luyện trí thông minh, khả năng tư duy sáng tạo, phân tích tổng hợp. 10% còn lại là nội dung kiến thức đặc trưng Singapore.
A Word by the Author Russian mathematicians have suggested that mathematics is the gymnasium ofthe brains Likewise, in China, Mathematics Olympiad is ferociously practised by the large student population, all for a coveted place in one of the elite high schools MATHS OLYMPIAD- UNLEASH THE MATHS OLYMPIAN IN YOU! (INTERMEDIATE) First Edition 2015 Children who exhibit certain traits and penchant for numbers at the age of or ye_ars old, or even earlier, have great potential to be the mathematical olympians among their peers - provided they are groomed via a systematic, rigorous and routinized training Singapore was ranked 3'd in Mathematics in a recent TIMSS survey, after Hong Kong and Taiwan Notably, China was not among the list of countries surveyed © Singapore Asia Publishers Pte Ltd Published and Distributed by: The most prestigious competition locally is RIPMWC (Raffles Institution Primary Mathematics World Competition) Meant for primary students or younger, the top 50 to 60 or so participants are selected from Round to compete in Round Thereafter, top participants emerge to take part in the world competition for primary school mathematics in Hong Kong Another popular competition, also meant for primary students, is APMOPS (Asia Pacific Mathematical Olympiad for Primary Schools), which is organized by Hwa Chong Institution since 1989 The following awards are being given at the end of two rounds of competition: Platinum, Gold, Silver and Bronze Singapot·e Asia Publishers Pte Ltd 219 Henderson Road #10-04 Henderson Industrial Park Singapore 159556 Tel : +65 6276 8280 Fa x : +65 6276 8292 Email: info.sg@sapgrp.com Website: www.sapgrp.com Facebook: Singapore-Asia-Publishers At primary 5level, the yearly NMOS (National Mathematical Olympiad of Singapore) competition has also captured the attention of parents since 2006, who eye NUS High as their most preferred high schooL ALL RIGHTS RESERVED The first series of books Maths Olympiad: Unleash the Maths Olympian in You! published in 2007 and 2008, has served as an ideal companion to students looking to establish a strong foundation in Mathematics- be it for PSLE preparation or in hope that they might one day take part in the various local and international competitions The books are, therefore, also first-choice materials for parents of primary students looking for quality content in gifted programme training All rights reserved No part ofthis publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publishers ISBN-13 ISBN-I 978-98 1-4672-14-6 981-4672-14-9 Printed in Singapore SAP Global Partners' Network: Antigua & Barbuda Australia Bahrain Bangladesh Barbados Bhutan Bostwana Brazil Brunei Cambodia Canada China Egypt Fiji Island Ghana Grenada Guyana Hong Kong India Indonesia Jamaica Japan Laos Lebanon Macau Malaysia Maldives · Mauritius Myanmar Namibia Nepal New Zealand Nigeria Pakistan Papua New Guinea Philippines Portugal Qatar Rwanda Saudi Arabia Seychelles Singapore Solomon Islands Sri Lanka Saint Lucia Saint Vincent & the Grenadines South Africa South Korea Syria Taiwan Tanzania Thailand Trinidad & Tobago United Arab Emirates United Kingdom· United States of America Vietnam Zambia Zimbabwe In this new edition you will find the following additions: Fraction • Ratio Percentage Angles and Triangles Enhancements have also been made to the following: Write Simple Equations Problems from Page Numbers Pigeonhole Principle Square Numbers and Value of Ones Digits • Speed The objective is to cater to increasingly smarter children who have been exposed to a wide variety of topics Some of these topics, which overlap the local mathematics syllabus, have also been adopted by schools here for students to practice on I feel extremely privileged and honoured to be able to continue serving students in this field My latest series Wicked Mathematics! is currently out on shelves For related courses and workshops, please visit www.terrychew.org For international business enquiries, email ibg@sapgrp.com www.sapgrp.com Terry Chew (2015) (C) 42 X 75 + 42 X 25 = 42 X (75 + 25) = 42 X 100 = 4200 PRACTICE Compute the following (a) 456 + 88- 56 (b) 374+56-74 (c) 3035- 998- 997 (d) 999 + + 98 + 998 + + (e) 636-567-99 + 367 (f) (g) 123 + 456 + 544 + 877 (h) 3456 + 4567 + 6544 + 5433 Use a simple method to compute the following (a) 210-:-(5x6) = 210-:-5-:-6 = 42-:- =7 (b) 748-:-(17 x 11) = 748-:- 11-:- 17 = 68-:- 17 =4 (c) 4000-:- 125-:- (d) 56 X 198 -7- 11 -7- = 198 -7- 11 X 56 -7- = 18 X = 144 = 4000 -7- (125 X 8) = 4000 -:- 1000 =4 (d) 33 X 126-33 X 26 =33 X (126-26) = 33 X 100 =3300 Use a simple method to compute the following -~ (a) 333 X222-7-666 = 333 X X 111 -7- 666 = 666 X 111 -7- 666 = 111 (b) 4444 X 2222 -7- 8888 = 4444X X 1111 X 8888 = 8888 X 1111 -7- 8888 = 1111 (c) 454 545 450-:- 9-:- (d) 999 + = 999 = 999 = 999 = 454 545 450 -7- (9 X 5) = 454 545 450 -:- 45 = 10 101 010 999 X 999 X (1 + 999) X 1000 000 5034 - 997 - 998 - 999 Compute the following (a) 2208- (208 + 139) Comp~te the following (a) 37 + 397 + 3997 + 39 997 (b) 733- (33 + 320) (b) 298 + 2998 + 29 998 + 299 998 (c) 1306- (406- 258) (d) 945 + (372- 245)- 172 (c) + 99 + 999 + 9999 + 99 999 (e) 644- (243- 156) + 143 (f) 717- (617- 225) Compute the following (a) (2 + + + · · · + 2006)- (1 + + + · · · + 2005) (g) 937- (137 + 185) + 85 (h) 1732- (732- 257) (b) 88 - 87 + 86 - 85 + + 4- + 2- (i) 788 - (288 + 546) + 346 (c) 100-98+96-94+···+8-6+4-2 Compute the following (g) 64 X 25 X 125 X 16 (h) 16 000 -7- 25 (a) 360-357 + 354-351 + · + 300-297 (i) (b) 2006- - 2- - 4- - 48 - 49- 50 (c) 280-276 + 272-268 + · + 200- 196 125 X 25 X X 64 G) 101 X 1001- 101 Use a simple method to compute the following (a) 5000+8+125 (b) 6000+25+4 (c) 30 000 + + + + 125 (d) 16000+125+4+8 (e) 32 000 + 125 (f) (g) 1400 -7- 25 (h) 72 OOD + 125 Use a simple method to compute the following (a) 56 x 8+88 x (C) 56 (e) X 33 + 44 198 x 56 X (b) 1600+25 33 (d) 73 (f) X X 12 + 27 64 X 25 X X 12 125 X 97 5300 -7- 25 Use a simple method to compute the following (a) 89 X 11 + 11 (C) 58 X + 84 X (e) 63 X + 74 (g) 44 X + 78 Use a simple method to compute the following (a) 35 (b) 29 X + 42 (d) 58 X 30 + 84 X 15 (C) 287 X (f) 74 X + 152 X (e) 69 X (h) 56 X 16 + 72 X 32 (g) 132 X 11 X X X 128-28 X 12- 187 36-38 X X X 36- 196 35 X 12 18 X (b) 46 (d) 897 (f) 12 X 74 X (h) 156 234-134 X 46 30-297 X 30 X 54-48 X X 48- 124 27 X 12 10 Use a simple method to compute the following (a) 3333 x 3333 + 9999 Problems introduced in this chapter are most commonly solved using tables, which can be tedious at times Instead, we attempt to solve problems of this nature by asking, "If all wefe " (b) 2222 X 9999 -7- 3333 A farmer has 36 chickens and rabbits There are 96 legs altogether How many chickens does the farmer have? How many rabbits does the farmer have? -~· Solution: (c) 99 999 x 88 888 + 11 111 M ethod 1: Solve by Assuming If all are chickens, there will be 36 x = 72 legs 96-72 = 24 All the rabbits have been counted as if they are chickens 4-2=2 The difference in the number oflegs between a chicken and a rabbit is 24 + = 12 rabbits 36- 12 = 24 chickens (d) 6666 X + 4444 X 13 Method 2: Solve by Assuming If all are rabbits, there will be 36 x = 144legs 144-96 = 48 There will be an extra 48 legs 4-2=2 The difference in the number oflegs between a rabbit and a chicken is 48 + = 24 chickens 36-24 = 12 rabbits The farmer has 24 chickens and 12 rabbits Asia Publishers Pte Ltd A cashier collected $312 from the sale of 50 tickets An adult ticket cost $8 A child ticket cost $4 How many tickets of each type did the cashier sell? simil~r basketballs and similar volleyballs cost $325 A basketball costs $10 more than a volleyball What is the price of a basketball? Solution: Solution: Method 1: Solve by Assuming If all were adult tickets, 50 x $8 = $400 would be collected Method 1: Solve by Assuming $325 + $400-$312 = $88 5+6=11 $8-$4 = $4 The total number of basketballs will then be 11 385 -7- 11 = $35 The price difference between an adult ticket and a child ticket was $4 $88 + $4 = 22 child tickets 50 - 22 = 28 adult tickets Method 2: Solve by Assuming $325- Method 2: Solve by Assuming If all were child tickets, 50 x $4 = $200 would be collected The total number of volleyballs will then be 11 ·:> $275 -7- 11 = $25 $25 + $10 = $35 $8-$4 = $4 The price difference between an adult ticket and a child ticket was $4 The cashier sold 28 adult tickets and 22 child tickets There are 30 questions in a mathematics competition marks are awarded for each correct answer and marks are deducted for each incorrect answer If Colin scores 122 marks for the mathematics competition, how many incorrect answers does he give? Solution: If Colin gets all the answers correct, he will get 30 x = 150 marks 150 -.122 = 28 There is an extra 28 marks 5+2=7 The difference in marks between a correct answer and an incorrect answer is marks 28 + = incorrect answers 10 Singapore Asia Publishers Pte Ltd $10 = $275 5+6=11 There would be a shortfall of $112 $112 + $4 = 28 adult tickets 50 - 28 = 22 child tickets X If all are volleyballs, the total price will be $275 $312-$200 = $112 He gives incorrect answers $10 = $385 If all are basketballs, the total price will be $385 There would be an extra $88 X The price of a basketball is $35 Mr Brannon has 58 pieces of two-dollar, five-dollar and ten-dollar notes altogether in his wallet The total value of these notes is $322 The number of two-dollar and ten-dollar notes is the same How many notes of different denominations does he have? Solution: We 'create' a new denomination, which is a ($10 + $2) + = $6 note Method 1: Solve by Assuming If all are six-dollar notes, the total value will be 58 x $6 = $348 $348 - $322 = $26 The difference in the total value of all the notes is $26 $6-$5 = $1 The difference in value between a six-dollar note and a five-dollar note is $1 $26 + $1 = 26 pieces of five-dollar notes 58- 26 = 32 pieces of two-dollar and ten-dollar notes 32 + = 16 pieces for each two-dollar and ten-dollar notes Method 2: Solve by Assuming If all are five-dollar notes, the total value will be 58 x $5 = $290 $322- $290 = $32 A farmer has 45 chickens and rabbits There are 140 legs altogether How many chickens does the farmer have? How many rabbits does the farmer have? A spider has legs A dragonfly has legs 28 spiders and dragonflies have 200 legs altogether How many spiders are there? How many dragonflies are there? Andy spent $55 in all to purchase 20 pieces of two-dollar and five-dollar stamps How many two-dollar stamps did he buy? How many five-dollar stamps did he buy? The difference in the total value of all the notes is $32 $6-$5 = $1 $32 + $1 = 32 pieces of two-dollar and ten-dollar notes 32 + = 16 pieces for each two-dollar and ten-dollar notes 58- 32 = 26 pieces of five-dollar notes He has 16 two-dollar notes, 26 five-dollar notes and 16 ten-dollar notes Asia Publishers Pte Ltd There are 40 cars and motorcycles in a park There are 116 wheels altogether How many cars are there? How many motorcycles are there? An adult ticket to a concert cost $35 and a concession ticket cost $18 Mr Walter paid $598 in all for 20 such tickets How many adult tickets did he buy? How many concession tickets did he buy? Jeff has 20 pieces of five-dollar and ten-dollar notes in his wallet The total value of these notes is $125 How many notes of each denomination does he have? A scien~e quiz consists of 15 questions marks are awarded for every correct answer and mark is deducted for every wrong answer Kelly scores 21 marks for the science quiz How many questions does she answer correctly? There are 30 questions in a mathematics competition marks are awarded for each question answered correctly and marks are deducted for each wrong answer Rena scores 126 marks for the mathematics competition How many questions does she get wrong? A mathematics quiz consists of30 questions The first 20 questions are worth marks each The last 10 questions are worth marks each No marks will be deducted for each wrong answer Justin scores 124 marks for the mathematics quiz How many of the first 20 questions and how many of the last 10 questions does he answer wrongly? ::::: ::::::::::::::::::::::::::::::ggggggg;~~~ ~ ~ ~~~~~~~: ••••• •• • • • • • • • •• • • • • • • ••• • • • • • •• •••••••••••••••••••• • • ••• •••••••••••••••• ••••••••• ••• • 12 + 11 X 11 X 0 [ • • • • • • • • • • • • • • • • • ~t;ljf~ ~~~- JJ!_;:.ttJJ ~~ ~ = 12 X 12 = 144- 12 = 132-:-11 =12 ~ O + 4] -:- = * 8· + = 32 _ 7 1+0 32 28 _ 1o CD~D~D2D~D~ 50 =60-4 :::::::::::::::::::::::::::::::::::::::::::::::::: :; ::::::: :: CD~D~D~D-=!1 75 ABC Telco carried 400 mobile phones at first =56 11 CD~D~ [ ~ D + 3] -:- = ~ The actual difference would be 168 if he had read the numbers correctly + = 32 _ 9 1+D 32 12 Working backwards, 3-1=2 2x2=4 x 2=6 4-1=3 X = 10 6-1=5 X = 18 10- = 18-1=17 17 X = 34 There were 34 marbles in the bag at first 27 _ -9-=9-9-9 + =5 =5-1 =4 {[( - 6) X 6] + 6}-:- = [( - 6) X 6] + = X ( - 6) X = 36-6 -6 = 30-:-6 =5+6 = 11 The value of this number is 11 {[( + 6) X 13 32 Beatrice 32 Chloe 32 14 X = [( + 6) X 3] -:- = + ( + 6) X = X 8 X ~= 12 +6=72-:-3 =24-6 = 18 The number is 18 Maths Olympiad- Unleash The Maths Olympian In You! (Intennediate) Alison 16 16 64 56 32 12 X = 16 16 X = 20 20 X%= 24 24 X ~= 28 28 commuters boarded the bus at the bu s interchange 59 Chapter 52 28 16 Alison had 52 books, Beatrice had 28 books and Chloe had 16 books at first 3]-:- 8}- = © Singapore Asia Publishers Pte Ltd 222 @~[ 162 1~ 222 [1~ + 3) = ~ x 100 50 25 100 50 25 , • ooo o oooeo0000000000000eeOe00000000000000 0• ooooo ,• ,• ,• ,• , • •I •I o• o• •o •o •o •• •• • •••••••••••••••••••••••••••••••••••• • • • • oo • •• • @~1 200 1~@2]~[§]~ 75 + =480-:- IO'h 16 Assume there is one bead in each I' 111 nl lh 3rd division: + + + (]) ·I I I, G) + G) + G) + G) + I 51 l division: @ + @ + @ + @ + I 21 + 21 + 21 + 21 + = 85 The minimum number of beads I have is 85 ;.~ (;;:rtr~;f~t'"~ -;,~ r- , :::::::: :: ; ::: ::::::::::::::::::::::::::::::::::::::: ::: ::: : ooo o o o oeooooo e eeeeeeeeee•oooooooooo o o oo ::::: ::: ::: ::::::::::::::::::::::::::::::::::::::::::::: :: :! ::: •:! ' 137) Prime numbers smaller than 12 = 2, 3, 5, 7, 11 Since 13 is not divisible by all these prime numbers, 137 is a prime number (c) 119 is not a prime number as it is divisible by (e) 127 is a prime number ~ (j) ® 210 ~ 10~ @ 21 X (b) X ® 33~ @ 330 G) 330 = (c) X (J) X 11 13 G) 385 = X 2006 11 X -~ X 17 © 209 ~ (@ 418 418 = (J) X 11 X 19 14 @ -221 ~ ~ ll05 1105 = X 13 1309 1309=7 X @ 17 Maths Olympiad- Unleash The Maths Olympian In You! (Intermediate) ® (J) 15 Method 1: Making a Table abc Area=axb+bxc X + X = 15 + 35 = 50 II X + X II= 35 + 77 = 112 II 13 7xll+llx]3=77+143=220 (]) 11 X 17 © Singapore Asia Publishers Pte Ltd (J) 1992 ~996 ~332 ~ 166 ~ 1992 = X X X X 83 + + + + 83 = 92 The sum of all its prime factors is 92 X © 187~ ~ ~@ 32 X 32 = 1024 (> 1003) We divide 1003 by all the prime numbers smaller than 32 1003-;- 17 =59 2006 = X 17 X 59 + 17 +59= 78 The sum of all its prime factors is 78 S 13 Chapter 780 = 2 X "- ® X @ I I I • o It I o o o oooooooooooooooooooOOOOOOOOOoooooooooooooo ~ 0 , , ,, ,, , , , o o oooooooooooooooOOOOOOOOOOOOOOOoooooooooooooooo , , , , , , , , , 83 X 13 ® (j]) 10 = + + + 23 37 = + + 13 + 19 37=2+5+11+19 37 = + + 13 + 17 37=2+7+11+17 more prime numbers in 10 ··········································· ···············•··•······················· ::: :::::::::::::::::::::::::::::::::::::::::::::::::::::: ::: • ••••• • •• •• ••••••••••••••••••••••••••••••••••••••• • Yes, but the last digit must be (b) 1+0+0+0+2=3 10 002 is divisible by but not by No, it is not possible (a) 41- 15 = 26 15 041 is divisible by 13 but not by (b) 523- 397 = 126 126-;- = 18 397 523 is divisible by but not by 13 (c) 597-415 = 182 182-;- 13 = 14 182-;- = 26 415 597 is divisible by and by 13 (d) 508-417 = 91 91+13=7 91 -;- = 13 417 508 is divisible by and by 13 (d) 1+0+0+3+2=6 10 032 is divisible by but not by (a) + + + = 18 2367 is divisible by and by (c) + + + + = 18 18 135 is divisible by and by (J) X @ 21 37 = + + 29 37 = + 11 + 23 37=5+13+19 37=7+11+19 37=7+13+17 37 is a sum of or ways X ::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::: :: 20 a+l8=b+l4=c+35 Using guess and check, a=l9,19+18=37, b=23,23+14=37 c = 2, + 35 = 37 19+23+2=44 The value of a+ b + c is 44 2X 3X3X X X X 13 Two groups, x x x 13 = 234 X X X X X X = 432 234 + 432 = 666 The sum of the palindrome pair is 666 X X 1716 = X X X 11 X 13 X X X X 13 X a=2 X X X 11 X 13 X b The smallest values of a and b are 11 and respectively 351 ~ 117~ 39 ~ ® 101 088 = 2 (J) @ ® X 1716 ~ 858 ~ 429 ~ 143 ~ @ 10 ~ @ ® ® G) 1540 ~770 ~ 77 ~ â đ ~ 13 @ 221 221 = 13 (h) X 780 ~ 78 ~ 39 ~ @ ""' 10 ~ @ @ 17 101 088 ~ ~ ~ @ 12 636' @ (J) 12 636 ~ 4212 ~ 1404 ~ ~ @ "- 351 (J) (f) 28 X 28 = 784 (> 741) We divide 741 by all the prime numbers smaller than 28 741-;- = 247 741 is not a prime number (j) 77~ @ 85 (g) 11 ~143~ @ (d) (f) X @ 286 286 = (e) X 19 1540 = X X X X 11 1540 x m = square number 1540 X m = X X X X X X X 11 m = X X 11 = 385 The smallest possible value of m is 385 (e) 23 x 23 = 529 (> 507) We divide 507 by all the prime numbers smaller than 23 507-;- = 169 507 is not a prime number - 10~ (J) rectangles of different sizes can b form d ll·om 36 identical rectangles (j) (J) (d) 21 X 21 = 441 (> 437) Prime numbers smaller than 21 = 2, 3, 5, 7, 11, 13, 17, 19 437+19=23 437 is not a prime number (J) X 1~ (c) 19 x 19 = 361 (> 337) Prime numbers smaller than 19 = 2, 3, 5, 7, 11,13,17 Since 337 is not divisible by all these prime numbers, 337 is a prime number (f) 133 is not a prime number as it is divisible by 210 = a X b + b X C = 220 b x (a+c)=220 b x (a+ c)= 11 x 20 b = 11 a + c = + 13 = 20 X 13 X 11 = 100 The volume of the cuboid is 1001 cm • (b) 17 X 17 = 289 (> 271) Prime numbers smaller than 17 = 2, 3, 5, 7, 11,13 Since 271 is not divisible by all these prime numbers, 271 is a prime number (d) 123 is not a prime number as it is divisible by 11 (a) Method 2: Using Equations + + + = 15 + + + = 15 5154 is divisible by Its palindromic number, 4515, is also divisible by as the sum of all the digits is the same and can be divided by Matl1s Olympiad - Unleash The Maths Olympian In You! (Intermediate) ICI Singapore Asia Publishers Pte Ltd S 14 Chapter I Chapter] • •• • ••• 4+5+6+7+6+5+4=37 It is not divisible by (4 + + + 4)- (5 + + 5) = 20-17 =3 It is not divisible by 11 either 11 (a) 37 625 is divisible by 25 and by 125 It is not divisible by or by (a) (8 + + 3)- (3 + + 8) =17-17 =0 386 683 is divisible by 11 (d) 548 672 is divisible by and by It is not divisible by 25 or by 125 (b) 93 648 is divisible by and by It is not divisible by 25 or by 125 (c) 87 615 is not divisible by 4, by 8, by 25 or by 125 (c) (3 + + + 2)- (2 + + + 3) = 20-20 =0 23 788 732 is divisible by 11 (d) (2+4+3+1)-(1+3+4+2) = 10- 10 =0 12 344 321 is divisible by 11 To conclude, a palindrome of even number of digits is divisible by 11 as the difference oftheir sums will be 3466 - 645 = 2821 821-2 = 819 819-;- 13 = 63 466 645 is divisible by 13 123 (3-digit number) 123 123 (6-digit number) 123 123-;- 13 = 9471 9471 -;- 11 = 861 861 -;- = 123 I will get back the original3-digit number Reason: x 11 x 13 = 1001 1001 x any 3-digit number = 6-digit number with the 3-digit number repeated 10 (a) 1+2+7+8=18 1278 is divisible by and by (b) 4+6+3+2=15 4632 is divisible by but not by (c) 5+4+6+8+4=27 54 684 is divisible by and by (d) + + + + + = 26 119 375 is not divisible by or by Maths Olympiad - Unleash The Maths Olympian In You! (Intermediate) © Singapore Asia Publishers Pte Ltd 13 5+0+1+5=11 m = 12-11 = m=15-11 =4 m=18-11=7 The possible values ofm are 1, and 14 Since 15 = x 5, b can only be or When b = 0, a + + + + + = a+ 32 For a + 32 to be divided by 3, a = 1,4, Therefore, the possible values of the number are 178 890, 478 890 and 778 890 When b = 5, a+ + + + + =a+ 37 For a+ 37 to be divided by 3, a= 2, 5, Therefore, the possible values of the number can also be 278 895, 578 895 and 878 895 15 333 333 333-333 = Since is divisible by 7, 333 333 is divisible by 888 888 888 - 888 = 888 888 is divisible by The possible values of a must either be or 16 The last digits of 006 D D D have to be 00 so as to be divisible by and by 25 We have 006 D 00 2+0+0+6+ +0+0=8 D = 1,4, The possible values of this number are 006 100, 006 400 and 006 700 515 Chapter 10 21 The ones digit can only be or When it is 0, we have D 340 + + + = 12 D = 18-12 = When it is 5, we have D 345 5+3+4+5=17 =18-17=1 The possible values of D are and The possible values of the number are 56 340 and 51 345 18 The ones digit of the number must be or in order to be divisible by When it is 0, we have 25 D 40 2+5+4+0=11 =1,4, Therefore, the possible values are 25 140, 25 440 and 25 740 When it is 5, we have 25 D 45 2+5+4+5=16 =2 58 Therefore, the possible values ;re '25 245, 25 545 and 25 845 12 517 699 699 - 517 = 182 182-;- 13 = 14 182-;- = 26 517 699 is divisible by and by 13 (b) (5+6+1)-(1+6+5) = 12-12 =0 156 651 is divisible by 11 17 Since it is the largest possible number, the first digit must be (9 + + 9) - (8 + + D) = 11 27- (12 + D) = 11 D =4 The largest possible number is 989 494 19 (100- 1) + = 99 + = 33 multiples of3 (100- 1) + 11 = 99 + 11 = multiples of 11 33, 66 and 99 are multiples of both and 11 100-33-9 + = 61 61 whole numbers between and 100 are not divisible by or by 11 22 The last two digits must be 00 in order to be divisible by and by 25 We have 368 D 00 3+6+8+0+0=17 =18-17=1 The smallest possible number is 368 100 23 We have DO DOD We choose the smallest digits, 0, 1, and 3, to fill in 80123 + + + = 14 =18-14=4 The smallest possible value ofthis number is 801234 20 112- 35 = 77 35 112 is divisible by and by 11 but not by 13 11 (a) -+ = x 3- + = 15-3 = 12 (b) 6-$- = X 3-8 + =18-4=14 (c) 12-$- 14 = 12 x 3- 14 + = 36-7 = 29 @ @ (b) 288 -;- = 96 288 = 96 + = 96 96 -;- = 12 96 = 12 + = 12 (288 3) = 12 = ( + 1) X =5x5 =25 25 = (3 + 1) X 25 =4 X 25 = 100 (a) ~ = + (5 + 1) + (5 + 2) +(5 -1- 3) -1- (5 -1- 4) + (5 + 5) + (5 + 6) + (5 + 7) -1- (5 + ) =5+6+7+8 +9 + 10 + 11 + 12 + =81 • (a) ~ = X 4- X = 28-24 =4 (b) ~ = X 4-6 X = 20-18 =2 (c) ~ = X 4-2 X =16-6=10 (b) m~10 = m + (m + l) + + (m + l0) = 11m + 55 11m + 55 = 99 11m = 99 - 55 = 44 111 = 44 -;- Jl =4 Miilhs Olympiad- Unleash The Maths Olympian In You! (Intermediate) ' Singapore Asia Publishers Pte Ltd (a) 2007 + = 250 R7 2007 = 250 + =257 S 16 Chapter 10 Chapter jj 13 *6 = m x 6- 11 =6m-6 6m-6= 12 6m= 12 + = 18 m=1876=3 12 m m ~ = x m- x = 4m- 18 4m-18 = 4m=20 m= 20 74=5 (b) ~ = X 5-3 X = 20- 12 = 8 ~ m = x - x m = 32-3m 32-3m= 23 3m=9 m=973=3 13 (a) From the last clue, we can rule out 1, 4, 5, A'ff O % ff Z ff (2 numbers are in the correct position) The answers could be 3870 or 0873 Since 0873 is a 3-digit number, the number was -> -> -> The four couples are Herman and Floi"Cil · , Charles and Alicia, David and Beatrice and Graham and Elaine Andy Kevin Policeman c B Matthew X X X / Amy X X Tony / Amanda Beatrice Melissa X Amy X TownB Towne Melissa X X X X Bloomsberry X X X Melissa Tony Basketball X X X Volleyball X X Amy Windschill X X Swimming 10 23 X + 19 X = 130 Dave scores sets of 23 and sets of 19 X X Melissa comes from Windschill and likes basketball Amy comes from Greenville and likes swimming Tony comes from Bloomsberry and likes volleyball 11 If Colin did it, he would be lying Jason and David would also be lying Only Melvin told the truth Hence, Colin was the culprit of this act ingapore Asia Publishers Pte Ltd Volleyball Greenville Amy X Jodie is older than the lady from Town B Jodie is younger than Beatrice Beatrice is not from Town B Beatrice comes from Town A I Basketball OR Jodie Mnths Olympiad- Unleash The Maths Olympian In You! (Intermediate) Chapter 11 I Chapter 12 Windschill Melissa comes from Bloomsberry and likes volleyball Amy comes from Windschill and likes basketball Tony comes from Greenville and likes swimming Box C contains blue marbles © Singapore Asia Publishers Pte Ltd Melissa Town A incorrect statement 517 16th lOth 8th 4th Bloomsberry Tony When Ashley is 10 years old, Stella is 13 years old This fits the statement where each girl makes one - Greenville Swimming IfLeon told the truth, then Peter also told the truth Hence, Teddy lied Leon and Peter both told the truth @, 111, and Melanie is years old Chinese Egyptian German Mexican X Matthew, the doctor, is older than the policeman Since the doctor is younger than Kevin, Kevin is not the policeman Kevin is the teacher Last number in the 14' row = + + + + + 13 + 14 The S'h number in the 15'h row is 110 white marbles white marbles blue marbles Maths Olympiad- Unleash The Maths Olympian In You! (Intermediate) - 13 A D Teacher 11 = 105 106, 107, 108, 109, If Box A has the right label, Box B and Box C will have the wrong labels Box A Box B Box C "There were dry afternoons." Morning oo oo oooooooo oo Afternoon o o o o o o o o "It rained 10 times " eeoooooooooooooo oooooooo•••••••• Tony 3870 If Box A has the wrong label, it contains blue marbles This indicates that Box B has the wrong label too, because only one box has blue marbles It also means that Box C has the wrong label We will then have a scenario of three wrong labels, so Box A cannot be the one having the wrong label 48 = X X 16 = + + 63 = X X The largest number among Jolene's cards is tlll!l Charles' wife cannot be Beatrice or lain , David's wife cannot be Elaine i 12- - l "There were 14 dry mornings." Morning ooo oo oo ooooooo 14 ·::: :: :: ::::::::::::::::::::::::::::::::::::::::::::: ::::: ::: :: ~'Q;i"i liJ'f;~:J~~-~:I~i'ill : ~~h.: · '"' ~·t;n ~."~ lil~J,~t " 12 Herman's wife cannot be Alicia 11 11'1 Elaine Method 2: Solve by Reasoning (10+8+14)72=16 The meteorologist recorded 16 days 14 E9 = + 99 + 999 + 9999 + 99 999 + 999 999 =1111110-6 = 1111104 1E99=1+11+111+1111+11111+111111 + 1111111 + 11111111 + 111111111 = 123 456 789 ·····::::::::::::::::::::::::: ························::·::: ·::::: : ••••••••••••••• .··· ···· Method 1: Solve by Drawings o is dry weather • is wet weather Doctor = X - =28-7=21 22 = X - 22 = 42-22 = 20 30 = X - 30 =56- 30 = 26 20 = X - 20 = 63-20 = 43 II 24 180 m E9 = (m + 1) + (m + 2) + (m + 3) + (m + 4) + + (m + + 1) = 9m + 45 (a) @ = X X X X = 120 (b) @ = X X X X X = 60 480 3@ = X X X X X = 20 160 60 480-20 160 = 40 320 13 + 11 1s x 9m + 45 = 81 9m = 81-45 = 36 m = 36 = 5@ = X 5-3 X = 25-18 = 7®n=5x7-3 x n = 35- 3n 35- 3n = 11 3n=24 n = 247 = 8 11 15 e 12 = 518 Chapter 12 · ·::· ···························································· ::::::::::::::::::mmmmmmm::::::::::::::::::::;:: · ·· ··········••••••••••••••••••••••••••••••••••••······ ···· · ··· r,- ,- :;J I J r-_r /\ - J.-J,·.· _J ~ rJ J_ I )- Let the first number be a a+ (a+ 1) +(a+ 2) +(a+ 3) +(a+ 4) = 465 Sa= 465-10 Let the number of students be p 4p+48 = 6p-8 48 + =6p-4p 2p =56 p =56-;- =28 She gives the sweets to 28 students 4f + 4v-> $240 3f + 5v -> $234 12/ + 12v-> x $240 = $720 12/ + 20v-> x $234 = $936 8v-> $936-$720 = $216 v->$216+8=$27 3/-> $234- X $27 = $99 f-> $99 -;- = $33 The cost of a football is $33 The cost of a volleyball is $27 c+ 60=145 c =145-60 = 85 The farmer has 85 chickens and 60 rabbits 12 X b+5 = 99 12b + 5s = 99 12b = 99- 5s X S b = 99- 5s 12 One possible answer: Whens = 3, b = 99-5 12 =7 X Another possible answer: Whens = 15, 5m + n = 62 5m = 62- 3n 62-3n m= Maths Olympiad - Unleash The Maths Olympian In You! (Intermediate) Let the numbers of big boxes and small boxes be b and s respectively Let the numbers of long tubes and short tubes be m and n respectively x m + x n = 62 © Singapore Asia Publishers Pte Ltd = 60 Letfand v be football and volleyball respectively b = 99-5 X 15 12 =2 S 19 Chapter 13 lA + lB + IC-> $31.50 - $21 = $10 d = 213 The price of item A + item B + ite 111 d= 213 -31m 12 A+B +C =31 C+D+E=47 A+B+D+E=64 (1) + (2) A + B + 2C + D + E = 78 (4)- (3) a+ b + c =20 (1) (2) (3) (1) 20a + 25b + 30c = 515 (1) X 20 (2) 20a + 20b + 20c = 400 (2)- (3) (3) 5b+ lOc= 115 5b=ll5-10c (4) b = 115 -lOc One possible answer: Whenc= 8, b -_ 115-5}0 X =7 2C = 14 C=7 10 Let f, b and d be fries, burger and drinks respectively Another possible answer: When c = 10, 3f+ 2b + ld-> $10.95 If+ 2b + 3d - $11.25 4f+ 4b + 4d- $10.95 + $11.25 = $22.20 j+ b + d-> $22.20-;- = $5.55 b = 115- 10 X 10 =3 14 Le_t a, band c be the number of roosters, hens and chicks respectively The price of a set is $5.55 11 !.!] + :: + (!) + (!) = 52 !.!] + :: + :: + (!) = 47 !.!] + :: + :: + (!) = 49 Compare (1) and (2) • $l O, IS, 13 Leta, b andc be the numbers of years the differ nt types of papers were set We have Jonathan's birthday is on 10'h of March Substituting r = 60 into (1), that day X d=213-3Ix3_120 12 - 12 = 10 2r= 120 r= 120+2 29 000 Thai Baht -> US$1000 29 000 Thai Baht could buy 1000 US dollars on m + }2 Whenm =3, Let the number of chickens and rabbits be c and r respectively c+r=145 - (1) A chicken has legs and a rabbit has legs 2c+4r=410 -(2) (1) X S$1.00-> 20 Thai Baht S$1.45 -> 29 Thai Baht X Again by comparison, 12d=213 -31m 2c + 2r = 145 x = 290 - (3) (2)-(3), 2c+ 4r-2c -2r=410-290 US$1.00-> S$1.45 S$50 -> 1000 Thai Baht L_et t~e month of his birthday be m and the day of his birthday be d 3} = 10 Another possible answer: Whenn = 9, 62-3 X m= =7 = 91 The five consecutive numbers are 91, 92, 93, 94 and 95 One possible answer: When n = 4, 62-3 X m= =455 a= 455 + ,, a+ b + c = 100 (1) (2) (3) 5a+3b+f= 100 (2) 15a + 9b + c = 300 (3)- (1) (!) = :: + Compare (2) and (3) 14a + 8b = 200 !.!] = :: + Replac~ (!) and !.!] in (2) by (:: + 5) and (:: + 2) respectively :: + + :: + :: + :: + = 47 4::= 40 :: = 10 (!) = 15 !.!l=12 14a = 200- 8b a= 200-8b 14 One possible answer: Whenb=4, a_ 200-8 x = 12 14 c=I00-4-12=84 Another possible answer: Whenb = 11, 12 3A + 7B + IC _ $31.50 4A + I OB + 1C-> $42 By comparison, a= 200-8 x 11 14 =8 lA + 3B-> $42-$31.50 = $10.50 2A + 6B-> $2 X $10.50 = $21 Mnths Olympiad- Unleash Th M h e at s lymp1an In You! (Intermediate) mgapore Asia Publishers Pte Ltd X c = 100- 11-8 = 81 520 Chapter 13 (1) (2) (3) ·::::: :::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ••••••• • • •• • •••••••••••••••••••••••••••••••••••••• • } ••• •• X + = 48 7X X + = 32 7X 7X X + = 16 numbers have the same when divided by Sun Mon Tue Wed 0 • • • • • • • c~-~~s J~ ~?_~;~ -~ ~ ) ;-; ? _) - :: / ~ J ~ + = 40 + = 24 1+ 1= quotient and remainder Thu Rl Fri R2 Sat R3 Sun Tue Wed Thu Multiples of7 = 7, 14, 21 , 28, 35, 42, (Multiples of7) + = 12, 19, 26, 33 , 40, 47, 54, ®J, 68, 75 , Multiples of = 9, 18, 27 , 36, 45 , 54, 63 , 72, (Multiples of9) + = 16, 25 , 34, 43 , 52, ®J, 70, black: Rl , R2, R3, R4, R5 blue: R6, R7, R8, R9 white: RIO, Rll , RO + + = 12 150 12 = 12 R The colour of the 150th bead is blue X 12 + = 49 There are 49 blue beads in the first 150 beads Maths Olympiad - Unleash The Maths Olympian In You! (Intennediate) B R2 There is no remainder and the last digit of quotient is 24 digits white th 16 111 111 7 = 15 873 ~ 7 = 15 873 015 873 R4 R5 R6 ~ R8R9 RI OR II Rl red 23 digits There is no remainder and the last digit of the quotient is black 13 X + = 69 There are 69 red beads 1'S page I to page : digits page 10 to page 99 : 90 x = 180 digits page 100 to page 168 : 69 x = 207 digits 9+ 180+207=396 396 digits are used to print the page numbers of the book page to page : digits page 10 to page 99 : 90 x = 180 digits page 100 to page 250 : 151 x = 453 digits + 180 + 453 = 642 642 digits are in the page numbers of the book page to page : digits page 10 to page 99 : 90 x = 180 digits page 100 to page 999 : 900 x = 2700 digits page 1000 to page 1002: x = 12 digits 9+ 180+2700+ 12=2901 2901 digits are used to print the page numbers of the dictionary is 60 12 (Multiples of 4) + = 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, ~, 54, 58 , 62 (Multiples of3) + = 5, 8, 11 , 14, 17, 20, 23 , 26,29, 32, 35, 38, 41 , 44, 47, ~, 53 , 56 (Multiples of7) + = 8, 15, 22, 29, 36, 43, ~ 57, 64, The smallest possible value of this number is 50 521 ~~ The last bead is red 11 The least common multiple of 2, 3, 4, and Chapter 14 ' -y 12 digits ~~~ C D E F G R3 R4 R5 R6 RO 1986 7 = 283 R Region E received the 1986th rescue package 1976 7 = 282 R 282 + = 283 Region B had received 283 rescue packages by the time the 1976t" package was handed out A Rl 15 Since the string of numbers is a mLJ ILi ll ol' '• look at 333 333 alone 333333 77=47619 333 333 7 = 47 619 047 619 14 + + = 12 160712=13R4 10 A B = R 20 A = 7B +20 A + B =340 7B + 20 + B = 340 8B = 340 - 20 = 320 B = 320 =40 A+40= 340 A= 340-40 =300 The two numbers are 300 and 40 + + + + + = 20 122 = 20 R2 20 X 20 + + = 410 The sum of the first 122 numbers is 410 Rl R2 R3 R4 R5 RO 138 = 23 RO The 138thnumber is © Singapore Asia Publishers Pte Ltd 120718=6R12 144718=8 The smallest possible value of the 2-digit number is 18 49, 56, 63 , 70, The 2-digit number is 61 30 + 31+30=91 Sun Mon Tue Wed Thu Fri Sat R3 R4 R5 R6 RO Rl R2 91 7 = 13 R 30th of August was a Thursday in that year Rl Fri Sat Today R3 R4 R5 R6 RO Rl R2 90 7 = 12 R6 It will be a Wednesday 90 days later Mon RO R4 R5 R6 31 + 28 + 31 + =91 917 = 13 R 1" of April was a Wednesday in that year 13 Divisor+ Dvidend = 171- 12-8 = 151 Let A be the divisor and B be the dividend A+B=l51 B=A x l2 + A+B=l51 A+ A X 12 + = 151 13A + = 151 13A=l51-8=143 A= 143 13 = 11 171- 12- 8- 11 = 140 The dividend is 140 and the divisor is 11 _j - page to page : digits page 10 to page 99 : 180 digits 300- 9- I80 = 11I digits for 3-digit page numbers 111 = pages 99 + 37 = 136 The last page number of the book is 136 Muths Olympiad- Unleash The Maths Olympian [n You! {Intennediate) t ' lngnpore Asia Publishers Pte Ltd page I to page : digits page I to page 99 : 180 digits 672- 9- I80 = 483 digits for 3-digit page numbers 483 = I61 pages 99 + 16I = 260 The book has 260 pages page to page : digits page 10 to page 99 : I80 digits page 100 to page 999 : 2700 digits + 180 + 2700 = 2889 2953 - 2889 = 64 digits for 4-digit page numbers 64 = 16 pages 999 + 16 = 10I5 The dictionary has 1015 pages digit '2 ' intheonesplace : 2, 12, 22, ., I42= 15times digit'2' inthetensplace : 20,2I,22, , 129=20times digit '2' in the hundreds place : 15+20=35 The digit '2 ' appears 35 times in a book that is 150 pages thick digit ' I' in the ones place : 1, 1I, 21 , , 191 =20times digit ' 1' in the tens place : 10, 11, I2, , 119=20times digit ' 1' in ~he hundreds place: 100, 101, 102, , 199 = 100 times 20 + 20 + 100 = 140 The digit '1' appears 140 times in a book that is 200 pages thick 522 Chapter 14 Chapter 15 digit '5' in the ones place: 5, 15,25, ,255 =26times digit '5' in the tens place: 50, 51, 52, 00.,255 =26times digit '5' in the hundreds place : 26 + 26 =52 The digit '5' appears 52 times in a book that has 255 pages For 20 to 29, there are 10 '2's = 10 X + 45 11 For40 to 49, = 10 X + 45 45+10+45+20+45+30+45+40+45 = 225 + 100 =325 The sum of the first 89 digits is 325 325 + + + + + + + + + + + = 225 + 100 + 40 = 365 The sum of the first 100 digits is 365 10 digit ' 8' intheonesplace: 8, 18,28, oo ,288=29times digit'8' inthetensplace: 80, 81,82, oo.,288=29times digit '8' in the hundreds place : 29 + 29 =58 The digit '8 ' appears 58 times in a book that has 288 pages Method 1· Solve by Reasoning 6~kg !" person 2"' person L 3'' person L 14 J., digit ' 3' in the ones place 3, 13, 23, oo , 93 : 10 times 103,113,000,193 : lOtimes : 10 times 203, 213, 000 '293 digit '3' in the tens place 30, 31, 000,39 : 10 times : 10 times 130, 131, 000' 139 : 10 times 230,231,00.,239 87 - 60 = 27 digit '3' in the hundreds place : 10 times 300, 301, 000 , 309 310,311, 000,319 : 10 times 320, 321, 322, 323 : times digit '3' in the ones place 303, 313, 323 : times The book has 323 pages ~ 000 = 81 Sum of digits in even-digit blocks The block of digits containing 10 has 20 digits The block of digits containing 11 has 24 digits 20 ) 25 intervals 25 + = 26 numbers Using the formula, (20 + 120) X 26 1820 1820 + 81 = 1901 digits 2000 - 1901 = 99 digits The 2000'h digit is in the block of digits containing 124 digits 124-99 = 25 By counting backwards, 00 1716151413121110987654321 The 2000'h digit is ( 13 For to 9, 1+2+3+00.+9 =45 For 10-19, there are 10 'l's = 10 X + 45 523 Chapter 15 120 : I I ' A= + B tl 00 "' B = 11 + C I 00 00 '" From (2), C = B - 11 oo oo , ,, I ll A+B +C=3 X 70 + B + B + B- 11 = 210 3B=210+JI I 3B = 219 B=219+3 =7 A= 73 + = 75 c = 73-11 = 62 The values of A, B and C are 75, respectively 1111 ,j 11 FI:I:EI:I~83~ + :.:66~+~74~+~73!"""tan:li:I:EI:IDDJ