Solutions to the Olympiad Cayley Paper

Transcription

1 Soluions o he Olympiad Cayley Paper C1. How many hree-digi muliples of 9 consis only of odd digis? Soluion We use he fac ha an ineger is a muliple of 9 when he sum of is digis is a muliple of 9, and no oherwise. Consider a hree-digi ineger wih he required properies. Each digi is beween 0 and 9, and none of hem is zero, so he sum of he digis is beween 1 and 27. Since we wan he ineger o be a muliple of 9, he sum of he digis is herefore 9, 18 or 27. However, i is no possible o wrie 18 as a sum of hree odd numbers, and he only way of making he sum of he digis equal o 27 is wih 999, which is hus one possible ineger. Bu he remaining quesion is how can we make he sum of he digis equal o 9? If one of he digis is 1, hen we can make he remaining 8 in wo ways: giving he hree inegers 117, 171 and 711; giving he six inegers 135, 153, 315, 351, 513 and 531. If we do no use a 1, hen he only possible ineger is 333. Hence here are eleven hree-digi muliples of 9 consising only of odd digis.

2 C2. In a 6 6 grid of numbers: (i) all he numbers in he op row and he lefmos column are he same; (ii) each oher number is he sum of he number above i and he number o he lef of i; (iii) he number in he boom righ corner is Wha are he possible numbers in he op lef corner? Soluion Suppose we sar by leing he number in he op lef corner be. Since, by rule (i), he numbers in he op row and lef-hand column are all he same, we can fill hem all in as. Then we can use rule (ii) o fill in he res of he grid, as shown From rule (iii), we now obain 252 = Dividing each side by 252, we ge = 8. So he only possible number in he op lef corner is 8.

3 C3. All he elephone numbers in Georgeown have six digis and each of hem begins wih he digis 81. Kae finds he scrap of paper shown, wih par of Jenny's elephone number on i. How many differen possibiliies are here for Jenny's elephone number? Soluion The scrap of paper could read 1018 or Le us sar by working ou which pars of he phone number Kae could have been seeing. We show he scrap of paper as a recangle d There are en possibiliies for he unknown digi d. There is only one possibiliy cd There are one hundred possibiliies for he pair of unknown digis c and d. There is only one possibiliy. However, all he possibiliies in he firs case d are also couned under he hird case 8101 cd. Hence here are in fac 102 possibiliies for Jenny's elephone number.

4 C4. The diagram shows an equilaeral riangle ABC and wo squares AWXB and AYZC. Prove ha riangle AXZ is equilaeral. Z B Y X W C A Soluion We sar by sudying he angles wihin he square AYZC, each of whose angles is 90. Since he riangle ABC is equilaeral, each of is angles is 60. Hence BAY = CAY CAB = = 30. Also, ZAY = 45 since ZA is a diagonal of he square. Now le us move over o he oher square AWXB. We have BAX = 45 since AX is a diagonal. Therefore ZAX = ZAY BAY + BAX = = 60. Also, he wo squares have he same size, because he riangle ABC is equilaeral and so AB = AC. Thus AZ and AX are diagonals of squares of he same size, and hence AZ = AX. I follows ha he riangle AXZ is an equilaeral riangle i has wo equal sides and he angle beween hem is 60.

5 C5. Dean wishes o place he posiive inegers 1, 2, 3,, 9 in he cells of a 3 3 square grid so ha: (i) here is exacly one number in each cell; (ii) he produc of he numbers in each row is a muliple of four; (iii) he produc of he numbers in each column is a muliple of four. Is Dean's ask possible? Prove ha your answer is correc. Soluion We claim ha Dean's ask is impossible, and will prove our claim by conradicion. Suppose Dean has placed he inegers according o he given rules. Now only one row conains 4, and only one row conains 8. Bu here are hree rows, so here is a leas one row R which conains neiher 4 nor 8. From rule (ii), he produc of he numbers in Ris a muliple of 4 and hence Rconains boh 2 and 6. In a similar way, here is a leas one column C ha conains neiher 4 nor 8. From rule (iii), he produc of he numbers in C is a muliple of 4 and hence C conains boh 2 and 6. As a resul, boh 2 and 6 are in he same row, and in he same column, which conradics rule (i). Therefore Dean's ask is no possible.

6 C6. The diagram shows wo regular hepagons ABCDEFG and APQRSTU. The verex P lies on he side AB (and hence U lies on he side GA). Also, Q lies on OB, where O is he cenre of he larger hepagon. C B Q P A U Prove ha AB = 2AP. R O S T G D E F Soluion We will show ha PQ = PB. Then i will follow ha AP = PB, because AP and PQ are wo sides of a regular hepagon, and hence ha AB = 2AP. Draw he lines OA and OC and exend AB o X X, as shown. B P A Now he angles CBX and QPB are equal because each is an exerior angle of a regular hepagon. Thus BC is parallel o PQ Q (corresponding angles converse), and C herefore angle OBC is equal o angle BQP (alernae angles). Consider riangles OAB and OCB. The side O OB is common, he sides OA and OC are equal because O is he cenre of he larger hepagon, and he sides AB and BC are equal because he hepagon is regular. Thus riangles OAB and OCB are congruen (SSS), and herefore angle OBC is equal o angle OBA, which is he same as angle PBQ. Hence he angles PBQ and BQP are equal. Therefore, using sides opposie equal angles are equal in he riangle BQP, we ge PQ = PB as required.