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Like us on Facebook! O lympiad-Maths-Trainer ~ J andwunti"8 _Committed To Your Learning Complete your learning with these books! The Entrepreneur ~ _ / #', of the year _, ~ ~A ROTARY-ASME AWARD www.sapgrp.com ISBN-13 978-981-4672-13-9 ISBN I S INGAPORE A SIA P UBLISHERS 981-4672-13-0 I I 78 672139 $19 95 A Word by the Author Russian mathematicians have suggested that mathematics is the gymnasium ofthe braif!S 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 M" H OLYMPIAD - UN EASH THE MATHS OLVM IAN IN Y Ul ( Children who exhibit certain traits and penchant for numbers at the age of or years old, or even earlier, have great potential to be the mathematical olympians among their peers - provided they ar groomed via a systematic, rigorous and routinized training INN 11) l'li utl!dlllou ,"llngapor Aslll Publlshct·s Pte Ltd ingapore was ranked 3rct in Mathematics in a recent TIMSS survey, after Hong Kong and Taiwan N tably, China was not among the list of countries surveyed Pu/ill.vh d nnrll lslrlbuted by: 'lngnJloro Asiu Publishers Pte Ltd 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 21! lloudorson Road 1/.10-04 II nd rson Industri al Park ~l n gopo ro !59556 Tol : + ,5 J276 8280 Fox : ·I· 1S 6276 8292 fl mui l: info.sg@sapgrp.com Wubs ilc: www.sapgrp.com l Global Partners' Network: Antigua & Barbuda Australia Bahrain Bangladesh Barbados Bhutan Bostwana Bmzil Brunei umbodia unudn hinu llgypt Fiji lRinnd hunu Grenada Guyana Hong Kong India Indonesia Jamaica Japan Laos Lebanon Macau Malaysia Malawi Maldives Mauritius Myanmar Namibia Nepal New Zealand Nigeria Pakistan Papua New Guinea Philippines Portugal Qatar Rwanda Saudi Arubin Seychelles Singoporo olomon lslunds ri Lonkn Saint Lucia Saint Vincent & the Grenadines South Africa South Korea Syria Taiwan Tanza nia 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: ·The Pigeonhole Principle Values of Ones Digit The Shortest Path Method Defining New Operations Enhancements have also been made to the following: Counting Speed Page Number Problem 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 l~ or inlu m ~tti nal bu ine _ _ _ nquiries, email ibg@sapgrp.com www.sapgrp.com 'hnpt ~~· 18 inding P -rimeter ············· -164 · hapter 19 The Page-Number Problem ~~-~ -172 Chapter 20 Defining New Operations -181 Chapter 21 Value of Ones Digit 188 Chapter 22 Pigeonhole Principle -199 S i UTI0 NS S - S3 In mathematics, there are various patterns: some are relatively straightforward and others are more challenging We, therefore, have to think outside the box and be flexible in our search for answers Besides adding or subtracting the terms in a number pattern, applying multiplication, division or even the use of any two arithmetic skills may help in the solving of the problems In Fibonacci numbers, the third term in the number pattern is the sum of the first and second terms; the fourth term is the sum of the second and third terms; the fifth term is the sum of the third and fourth terms, and so on In essence, each term, after the first two terms, is the sum of the two preceding terms Complete each number pattern (a) 4, 7, 10, 13, ( ), Analysis: The difference between any two consecutive terms in the above number pattern is 3, so the next term must be 13 + = 16 (b) 2, 6, 12, 20, ( ), Analysis: This is more interesting than the number pattern shown in (a) The second term is more than the first one Thereafter, the difference between any two consecutive terms increases by 2+4=6 + + = 12 12 + + + = 20 The next term is, therefore, 20 + + + + = 30 ' Amlly~·i~" Il l In th e ve number pattern it i between a y tw o consecutive num e t·~ Th · tin r ·n · tw een the first and econd terms is The differenc tw n tho second ·:~.nd third terms is 12 Observing the two differences will r· al tl t 12 i three times of Hence the second term is three times the fir t t rt ; t third term is three times the second term and so on (a) 6+2=3 18+6=3 18 X =54 22, 20, 10, 8, ( ), ( ) 1' 2, 3, 5, ( ), ( ), (b) 1, 1, 1, 1, 4, 7, 13, ( ), (c) 5, 9, 15, 1, 3, 6, · 10, (e) 0, 3, 8, 1, 2, (g) 0, 23, 33, ( ), 15, 24, ( 6, 24, 120, 1, 3, 8, (h) 1, 3, 7, 15, (i) 1, 1, 3, 7, (j) 1, 2, 5, 13, (f) Analysis: The above number pattern uses two arithmetic skills: division and subtraction The first term is divided by and the second term is subtracted by2 44-7-2 = 22 20+2=10 8+2=4 1, (d) The next term is 54 (d) 44, Complete each number pattern below 22-2 =20 10-2 = 4-2=2 21, 21, 28, 36, ), 48, 63, ), 5040, ), ( ( 31, ( 34, ), ( 55, 13, ( 45, ), ), 89, ( ), 127, 31, ( ), ), ( The next two terms are and respectively (e) 1, 1, 2, 3, 5, 8, 13, ( ), ( ), 55, 89, Analysis: This number pattern is an example of Fibonacci numbers 13+8=21 21 + 13 = 34 The two terms are 21 and 34 respectively Ah, the magical trick of9! Observe the pattern and write the correct answers in the brackets provided 21 X = 189 321 X = 889 321 X = 38 889 54 321 X = ( ) 654 321 X = ( ) 654 321 X = ( ) 87 654 321 X = ( ) 987 654 321 X = ( ) th Pa cal Triangle and write th c rrect answers in the brackets 1 30 1 What are the missing numbers in the number patterns below? (a) 1 3 40 24 20 28 25 10 20 11 17 54 32 20 14 (b) 14 ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) X 11 = 1111 101 X 22 = 2222 101 X 33 = 3333 101 X 44 = ( ) 101 X 55 = ( ) 101 X 66 = ( ) 101 X 77 = ( ) 101 X 88 = ( ) 101 X 99 = ( ) 12 22 20 The marvel of multiplication of 1011 Observe the pattern and write the correct answers in the brackets provided 101 11 (c) 18 15 14 23 19 (d) 48 32 38 (e) 12 24 44 Fill in each blank with digits from to -~ acb digit may be used only once The number on the left-hand-side of the arrow is add d to 12 to reach the number on the right-hand-side 10 Can you put 32 balls in the boxes below so that there are equal number of balls in every line? l Fill in each blank with digits from to Each digit may be used only once The number on the left-hand-side of the arrow is multiplied by before adding to reach the number on the right-hand-side 11 Find the 81h term of the sequence Fill in each blank with digits from to Each digit may be used only once The number on the left-hand-side of the arrow is divided by before subtracting from it to reach the number on the right-hand-side 3, 5, 9, 15, 23, 33, ( 45, ), 12 Find the 81h term of the sequence 1, 4, 11, 29, 76, 199, ( 521, ), 13 Find the sixth term of the sequence Fill in each blank with digits from to Each digit may be used only once The number on the left-hand-side of the arrow is multiplied by before adding to reach the number on the right-hand-side 1, 4, 9, 16, 25, ), ( 49, 64, 14 Find the seventh and eighth terms of the sequence 1, 2, 4, 7, 11, 16, ( ( ), ), Find the seventh and eighth terms of the sequence 2, 4, 8, 14, 22, 32, ( ), ( ), I J •ind the sixth, seventh and eighth terms of the sequence 4, 9, 10, 15, ( ), ( ), ( ), 17 ind the seventh and eighth terms of the sequence 1, 3, 8, 6, 16, 18, ( ), ( ), 76, 18 Which of the following sequences is different from the rest? 1, 1, 2, 3, 5, 8, 13, (b) 0, 2, 2, 4, 6, 10, 16, 1, 3, 4, 7, 11, 18, 29, (d) 1, 2, 3, 6, 11, 20, 37, (a) (c) 78, 20 lt' ·licia saved $10 in January She saved $20 in February The amount of money Hhe saved in March was the total amount of money she had saved in January and '' bruary The amount of money she saved in April was the total amount of money 1e had saved in February and March If she had saved in this pattern up to October, (a) in which month did she save $210? (b) how much did she save in October? 19 Find the missing number in each number pattern (a) (b) I Write the missing numbers in the brackets provided (c) ( 12 ( ) ) 12 ( ( ) ) 16 ( ( ) ) ( ( ) ) ( ) Find the value of + + + + + + + + + 10 Analysis: We can make pairs of 11 1+10=11 + = 11 3+8=11 4+7=11 5+6=11 Carl Gauss, one of the greatest mathematicians, was born in 1777 in Brunswick, Germany He was gifted in mathematics and showed a talent in that field at an early age Instead of adding up all the numbers, we multiply by 11 to get 55 A popular story about this mathematical genius goes like this One day, his primary school teacher asked all the pupils to find the value of Find the value of + + + + + + + + + 10 + 11 + 12 + 13 + 14 + 15 Analysis: We can make pairs of 16 + + + + + + + + + + 98 + 99 + 100 It was his teacher's hope to use this lengthy addition of integers to quieten down the class Surprisingly, the mathematical prodigy worked out the correct answer almost instantaneously! Well? Don't you want to know how he did it? Gauss added to 100,2 to 99, to 98, ···Each pair added to 101 Since there were 50 such pairs from integers to 100, he multiplied 101 by 50 to get the final answer, 5050 + 15 = 16 2+14=16 + 13 = 16 + 12 = 16 + 11 = 16 6+10=16 + = 16 Now, what about the number that is not in pairs? We just have to add the remaining number to the product of the pairs X 16 = 112 112 + = 120 Find the value of + + + + + + + Analysis: We can make pairs of 1+8=9 2+7 = 3+6=9 4+5=9 Instead of adding up all the numbers, we simply multiply by to get 36 In some situations where you need to find the remaining number, just take the sum of the first and last numbers in the sequence to divide by Find the value of 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30 Find the value of + + + 10 + 12 + 14 Fit d th value of + + + + 11 + 13 + 15 Find the value of + + + + + + ind the value of + + 11 + 14 + 17 + 20 Find the value of + + + + + + + 10 M 11 k ithe value of + 13 + 17 + 21 + 25 + 29 Find the value of + + + + + 10 + 11 10 number of intervals from the first tree to the sixth tree= 6-1 = 10-;- = Benny takes to walk from one tree to another number of intervals from the first tree to the fifteenth tree = 15 - = 14 X 14=28 It takes him 28 to walk to the fifteenth tree At pm, there are only intervals between the first and third chimes 6s+2=3s It takes s to chime once At pm, there are only intervals between the first and sixth chimes X S = 15 S The grandfather clock takes 15 s to chime times at pm 14 number of intervals = 40 + = 20 number of pots of plants = 20 - = 19 There are 19 pots of plants 19 15 16 In the case of circular tracks, number of intervals = number of trees 600-;- = 100 100 trees are planted round the lake From the diagram, , , , , , , X = 16 16+4=20 ' ' 'X ' X X X X ' 'X X = 24 24-4 = 20 38+2=19 There were 19 pupils in each row number of intervals = number of pupils- =19-1=18 18 x 50 em = 900 em The length of the queue was 900 em or m There are 20 flowers on the handkerchief altogether 17 M ethod 1: Solve Using Drawing From the diagram, 00 X = 12 12 + = 15 48-;- = 24 There are 24 pupils in each row 0 0 If the number of coins on each side is multiplied by three, you will have extra three coins as the coins on the three comers are counted twice X = 18 18-3=15 15 coins were used to make the triangle 13 number of intervals = + = 10 100 m + 10 = 10 m The distance between each tree is 10m 18 number of intervals = 79 + = 80 320m+80=4 m The distance between each tree is m Maths Olympiad - Unleash The Maths Olympian In You! (Beginner) © Singapore Asia Publishers Pte Ltd S 11 Chapter (a) 23 + 99 = 23 + 100-1 = 123- = 122 (c) 162-97 = 162-100 + = 62+3 =65 (b) 152- 97 = 152-100 + =52+ =55 (d) 565-298 = 565 - 300 + = 265 + =267 (e) 638 + 397 = 638 + 400-3 = 1038-3 = 1035 (f) 499 + 127 = 500 + 127-1 = 627-1 =626 (g) 645-98 = 645-100 + = 545 + =547 (h) 533- 199 = 533-200 + = 333 + =334 (i) 285 + 297 = 285 + 300-3 = 585-3 =582 (j) 424-196 = 424-200 + = 224 + =228 0 000000 Method 2: Solve by R easoning number of intervals = 100 - = 99 99 x 6m=594m The road is 594 m long X 21 = 105 There are 105 performers altogether Method 1: Solve Using Drawing There are 20 flowers on the handkerchief altogether Method i: Solve by Reasoning If the number of flowers on each side is multiplied by four sides, you will have extra four flowers as the flowers on the four comers are counted twice number of intervals = number of trees - = 95-1 = 94 94 x m = 470 m The road was 470 m long Numberofintervals in each r w =20 m+1 m =20 Number of performers in each row =20+ =21 Number of intervals= 160m+ m = 32 32 + = 33 There are 33 pine trees on this side of the road 15 coins were used to make the triangle 12 20 Length of the road= m x 40 =160m number of intervals = 24 - = 23 23 x 1m= 23m The queue is 23 m long 11 number of intervals = number of plum trees - = 41-1 = 40 number of intervals= 11- = 10 10 + 10 = 30 min+ min= 30 intervals 30 + = 31 Uncle Sam would be at the thirty-first lamp post (o) 323 + 677 + 92 + 108 = (323 + 677) + 92 + 108 = 1000 + 200 = 1200 (p) 167 + 355 + 345 + 133 = 167 + 133 + 355 + 345 = 300 + 700 = 1000 ( q) 9999 + 999 + 99 + = (10000-1) + (1000 -1) + (100 -1) + (10-1) = 10 000 + 1000 + 100 + 10 - = 11106 (r) 18 + 28 + 38 + 48 +58 = (20- 2) + (30- 2) + (40- 2) +(50- 2) + {60- 2) = 200-10 = 1?0 (s) 9997 + + 99 + 998 + + = (9997 + 3) + (1 + 99) + (998 + 2) + (1 + 9) = 10 000 + 100 + 1000 + 10 = 11110 (k) 43 + 99 +58 =40 + (99 + 1) +(58+ 2) = 40 + 100 + 60 = 200 (t) 74 + 75 + 77 + 80 + 82 + 85 = 71 +(3 + 77) + 80 + 80 + + 80 + (5 + 75) = 71 + + 80 + 80 + 80 + 80 + 80 = 73 +400 =473 (I) 56 + 87 + 77 = 50 + (3 + 87) + (3 + 77) =50+ 90 + 80 = 20 + (10 + 90) + (20 + 80) =220 (u) 299 999 + 29 999 + 2999 + 299 + 29 = (300 000- 1) + (30 000- 1) + (3000 - 1) + (300 -1) + (30- 1) = 333 330-5 = 333 325 (m) 74 + 75 + 28 = 74 + + (75 + 25) = 77 + 100 = 177 (n) 27 + 86 + 63 + 14 = 27 + 63 + (86 + 14) = 90 + 100 = 190 (v) 133+135+140+147+145+142 + 13 = 140 + (133 + 7) + 140 + (135 + 5) + 140 + 140 + (2 + 138) = 140 X =980 Maths Olympiad- Unleash The Maths Olympian In You! (Beginner) IC> Singapore Asia Publishers Pte Ltd S 12 Chapter I Chapter -•, (a) 420- 102-98 = 420-(102 + 98) = 420-200 = 220 (b) 240 - 104 - 96 = 240- (104 + 96) = 240 - 200 = 40 (c) 325-10-20-80-90 = 325 - (1 + 90 + 20 + 80) = 325 - 200 = 125 (e) 1897 + 128-597 = 1897 - 597 + 128 = 1300 + 128 = 1428 (d) 400 - 90 - 80 - 20 - 10 - = 400 - (90 + 10 + 80 + 20) - = 400-200-5 = 195 (t) 728 - (28 + 320) = 728 - 28 - 320 = 700 - 320 = 380 (e) 235- 15-25 -75- 85 = 235- (15 + 85 + 25 + 75) = 235 - 200 = 35 (g) 1290 - 164-736 (f) 200 - - - - - = 200- (1 + + + + 5) = 200 - 15 = 185 (h) 1330 - 288 - 342 = 133 - (288 + 342) = 1330 - 630 = 700 (g) 120-117 + 116-113 + 112-109 + - + 100 - 97 120 - 117 = 116 - 113 = 112-109=3 100 - 97 = Every pair of subh·action has a diffe rence of Count the minuends, 100, 104, 108, 112, 11 6, 120 + 3+3 + + +3= 18 The answer is 18 (i) 843 + 78 - 43 = 843 - 43 + 78 = 800 + 78 = 878 (a) 375 + 286 + 125 + 714 = (375 + 125) + (286 + 714) = 500 + 1000 = 1500 © Singapore Asia Publishers Pte Ltd 513 Chapter (z) 583 - 297- 183 = (583- 183)- 297 =400 - 297 = 103 = 537- 543 + 163-57 = (537 + 163)- (543 + 57) = 700 - 600 = 100 (u) 713-(513-229) = 713-513 + 229 = 200 + 229 =429 (I) 847-578 + 398 - 222 = 847- (578 + 222) + 398 = 847 - 800 + (400- 2) = + 400 - = 445 (m) 936- 867 - 99 + 267 = 936- (100- 1)- (867- 267) =936 - 100-600+1 = 236 + = 237 (n) 33 + 87 + 67 + 13 = (33 + 67) + (87 + 13) = 100 + 100 = 200 (o) 96 + 103 + 97 + 104 + 101 + 99 = (96 + 104) + (103 + 97) + (101 + 99) = 200 + 200 + 200 = 600 Count the minuends, 300,306, 312, 318, 324,330, 336,342, 348,354,360 3+3+3+3+3+3+3+3+3 + 3+3=33 The answer is 33 (a) 14 X 25 X = 14 X 100 = 1400 (b) (c) 25 X 32 X 125 = 25 X4 X8 X 125 = 100 X 1000 = 100 000 (d) 125 X 25 X = 125 X 100 = 12 500 (e) 56 x 125 = X X 125 = X 1000 =7000 (t) 125 X 72 = 125 X X = 1000 X =9000 17 X 25 17 X X 25 = 17 X 100 = 1700 X (i) 25 X 16 X 125 X X = 25 X X X 125 X X = (25 X 4) X (8 X 125) X (2 X 5) = 100 X 1000 X 10 = 000 000 = G) 25 X 64 X 125 X 25 = 25 X X X 125 X 25 = (25 X 4) X X (8 X 125) X 25 = 100 X 50 X 1000 = 000 000 (k) 25 X 24 X X 125 = (25 X 4) X X (8 X 125) = 100 X X 1000 = 600 000 (g) 64 X X 25 X 125 (I) 25 X 40 X 125 X = (25 X 4) X 10 X (125 X 8) = 100 X 10 X 1000 = 000 000 = X X X X 25 X 125 = (2 X 5) X (4 X 25) X (8 X 125) = 10 X 100 X 1000 = 000 000 2.1 (a) 24 (h) 56 X 25 X X = 14 X (4 X 25) X (2 X 5) = 14 X 100 X 10 = 14 000 lvluths Olympiad- Unleash The Maths Olympian In You! (Beginner) I ingapore Asia Publishers Pte Ltd 354-351 = Every pair of subtraction has a difference of (w) 46 + 37 + 54 + 63 = (46 +54)+ (37 + 63) = 100 + 100 =200 (k) 564 - (387- 136) = 564 + 136- 387 = 700- 387 = 313 360- 35 + 354- 351 + ; + 300 - 297 360- 357 = 300-297 = (v) 850-486- 114 = 850-(486+ 114) = 850-600 =250 ( q) + 99 + 999 + 9999 = (10- 1) + (100- 1) + (1000- 1) + (10 000- 1) = 10 + 100 + 1000 + 10 000 - =11110-4=11106 (c) 37+97+3997+99 = (40 - 3) + (100- 3) + (4000- 3) +(100-1) = 4240- 10 = 4230 (y) 989-675- 189 = (989 -189)- 675 =800-675 = 125 (t) 537- (543- 163)- 57 (p) 23 + 56 + 77 + 18 + 44 + 82 = (23 + 77) +(56+ 44) + (18 + 82) = 100 + 100 + 100 = 300 (b) 728 + 140 + 272 = (728 + 272) + 140 = 1000 + 140 = 1140 (x) 947 + (372 - 447) - 572 = 947 + 372-447 - 572 = (947- 447)- (572 - 372) = 500-200 =300 (s) 464- 548 + 99 + 348 = 464 + (100- 1)- (548- 348) = 564-1-200 = 364-1 = 363 G) 528 - (186 + 328) = 528 - 328- 186 = 200 - 186 = 14 (i) 800-10-20 - 30-40-50-60-70-80-90 = 800-(1 0+90 + 20 + 80 + 30 + 70 +40 + 60 +50) = 800- (100 + 100 + 100 + 100 +50) = 800 - 450 =350 Maths Olympiad - Unleash The Maths Olympian In You! (Beginner) (r) + 98 + 998 + 9998 = ( 10 - 2) + ( 100 - 2) + ( 1000 - 2) + (10 000- 2) = 10 + 100 + 1000 + 10 000 - = 11 110- = 11102 = 1290 - (164 + 736) = 1290 - 900 = 390 (h) 145 - 141 + 140 - 136 + 135 - 131 + - + 95-91 145 - 141=4 140-136=4 135 - 131=4 95-91=4 Every pair of subtraction has a difference of Count the minuends, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145 4+4+4+4+4+4+4+4 + 4+4+4=44 The answer is 44 (d) 71 + 69 + 68 + 73 + 74 = 70 + (1 + 69) + (68 + 2) + 70 + + 70 + = 70 + 70 + 70 + 70 + 70 + = 350 + = 355 514 Chap ter I Chapter (b) 36 =6 X6 = XJ X2X = X2 X3 X ~ (c) 27 =3 x9 =3 x x (e) 42 (f) 56 =7 x =2x2 x x =6 x =2 x x (g) 33 =3 X (e) 58-:- 12 + 86-:- 12 =(58 + 86)-:- 12 = 144-:- 12 =12 (h) 64 =8 x =2 x x x x x 11 X X X (g) 35 X 202 = 35 X (200 + 2) = (35 X 200) + (35 = 7000 + 70 = 7070 //.JO~ 90~ ~ (b) 175 = 175/ X X 354 ~ (c) 180 = X X X X 104 180/ ~18~ (d) 225 = X X ~ 225 ~ X ~ 9~ (j) 38 X 99 =38 X (100-1) = (38 X 100)- 38 = 3800 - 38 =3762 (a) X 14 + X 86 = X (14 + 86) = X 100 =700 (I) 83 = 24 X (23 + 77) = 24 X 100 =2400 Maths Olympiad - Unleash The Maths Olympian In You! (Beginner) © Singapore Asia Publishers Pte Ltd 2) = = = = = = 2) + (b) + (d) (e) + (f) + 3 (g) + 6 (h) + 6 (c) X (98-48+74) X (50 + 74) X 124 X (125 - 1) (8 X 125)- (8 X 1) 1000-8 992 X (28 X + 72 X 4) = 25 X 28 X + 25 X 72 = 100 X 28 + 100 X 72 = 100 X (28 + 72) = 100 X 100 = 10 000 1) X X X 1) = = = = X 810 + 49 X 81 81 + X 10 X 81 + 49 81 X (50+ 49 + 1) 81 X 100 8100 (k) 23 X 101 =2323 (I) 45 (m) 82 x 101 = 8282 (n) 98 x 101 =9898 (o) 63 x 101 =6363 (p) 76 (a) x (4+1)=20 X = 21 (b) X 52 X 47 = 2021 = 16 X 58= 3016 (e) X (7 + 1)=56 X 73 X 73 X (g) X X 92 = 21 X 77 = 5621 (9 + 1) = 90 = 16 X 98 = 9016 (i) x (4+1)=20 X = 16 42 X 48 = 2016 (k) 4x(4+1)=20 X = 24 81 Maths Olympiad- Unleash The Maths Olympian In You! (Beginner) O Singapore Asia Publishers Pte Ltd i + 7 (j) 67 X 101 =6767 (t) 81 + X () I) ~ (i) 14 X 101 = 1414 (c) x (5 + 1) = 30 (s) 74 + X 740 + 49 X 74 = 74 + X 10 X 74 + 49 X 74 = 74 +50 X 74 + 49 X 74 = 74 X (50+ 49 + 1) = 74 X 100 =7400 X I " X 101 =4545 X 101 =7676 X X 54 (5 + 1)=30 = 24 X 56= 3024 X (r) 25 X 73 + 50 X 73 + 25 = 73 X (25 + 50 + 25) = 73 X 1QQ =7300 3) II ·I X (q) 25 515 Chapter (a) 28 43 99 =83 x (100-1) = (83 X 100) - (83 = 8300-83 =8217 (c) 24 X 23 + 24 X 77 li X (k) 33 X 97 = 33 X (100- 3) = (33 X 100)- (33 = 3300-99 =3201 (b) 65-:- 11 +56 -:- 11 = (65 +56) -:- 11 = 121 -:- 11 =11 (0) 125 X (84 X + 16 X 8) = 125 X 84 X + 125 X 16 = 1000 X 84 + 1000 X 16 = 1000 X (84 + 16) = 1000 X 100 = 100 000 (p) (i) 72 X 101 = 72 X (100 + 1) = (72 X 100) + (72 = 7200 + 72 =7272 45 < X (h) 45 X 98 = 45 X (1 00 - 2) = (45 X 100) - (45 = 4500-90 =4410 X (n) (110 + 25 - 9) x = (110 + 16) X = 126 X = (125 + 1) X = (125 X 8) + (1 X 8) = 1000 + = 1008 (f) 100-:- 14 + 96-:- 14 = (1 00 + 96) -:- 14 = 196-:- 14 = 14 G) 99 (i) 48 = X 11 =8 x =2 x x x x =3x3 x 11 2.2 (a) 90 = (m) X (72 X 25 + 28 X 25) = (4 X 25) X 72 + (4 X 25) = 100 X 72 + 100 X 28 = 100 X (72 + 28) = 100 X 1QQ = 10 000 (d) 31 X 84 + 31 X 16 =31 X (84+ 16) = 31 X 100 =3100 (d) 44 =4 X 11 =2 x x 11 44 516 Chapter X 46 = 2024 (d) 6x(6+1)=42 X = 24 66 X 64 = 4224 (f) X (8 + 1) = 72 x 1=9 89 (h) X X 57 (j) X X 85 (I) X X 74 X 81 = 7209 (5 + 1) = 30 = 21 X 53= 3021 (8 + 1) = 72 = 25 X 85 = 7225 (7 + I) Cl = 24 X 76 • 24 .::;; ;;; ;;;;;;;;;;;;:;::!!!!!!!!!!!::;::::::::::::::::::::::::: • •••••••• • ••••••• • •• • ••••::::: ••• : •••••••••••••••• · · · : : ·::::::::::::::::::::::::::::: · r~ I J I J J-j I\ c r- _1 _l r) J : Use a table Accountant Bus driver Manager ;::j _J Zachary Nicholas Philip )( )( I Since Philip is older than the accountant but younger than Nicholas, Nicholas must be the manager Zachary is the accountant Hence, the total weight of the three girls is 75 kg Mr Wood's chi ldren Royston (8 years old) Jack (2 years old) Sabrina is 30 kg, Kim is 25 kg and Bernice is 20 kg Since all the children are below years old, we assume that the oldest child, Royston, must be years old Sean must be Mr Smith's child and should be = years old Mike is - = year old 75 kg- 45 kg= 30 kg 75 kg- 55 kg= 20 kg 75 kg- 50 kg= 25 kg I If the 41h witness is telling the truth, it means that the 3'd witness is telling a lie 8x0=64 0=6478=8 = X =56 If Sean took the chocolate cake, Truth 5 X 0 X X Sean Cindy Jack Ll=5x0 0=45 0=45 =9 0=3 f:, = X = 15 Leon Doctor Teacher Army officer Sean Cindy Jack Henry I I 10 c E +0 + + 13 Truth I I I I I E F I I G H Lie I Sun I I 13 I I 20 27 17 If Kate took the blueberry pie, Jacelyn Kate Polly Truth Jac~yn Lie Kate Polly I I Truth I Man Lie 14 21 28 15 22 29 I Thu Fri Sat 16 23 30 10 17 24 31 12 19 26 11 18 25 If Sean is lying, Karen could be the tallest among them If Kelvin is lying, no one is the shortest This should not be the case Hence, Sean must be lying Lie tallest Karen I Sean Susan shortest Kelvin I 18 Use a drawing A E15?B + ·:· = 50- 36 = 14 + + •!• + •!• + •!• 36 D 19 = 14-8 = Truth Lie I I I IfKaty broke the window, Truth Bob Katy Joe _b_io I I I If Joe broke the window, Bob Katy Joe =7 = 12-7 =5 Maths Olympiad- Unleash The Maths Olympian In You! (Beginner) If Bob broke the window, Bob Katy Joe + = 55 - 43 = 12 + + + + + + = 43 12 + 12 + 12 + = 43 = 43- 12- 12- 12