Maths Olympiad Prep

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, 2017

Number theory Difficulty 2.0 Junior Prove it Canada

A store sells packages of red pens and packages of blue pens. Red pens are sold only in packages of 6 pens. Blue pens are sold only in packages of 9 pens.

Figure 0 Igor bought 5 packages of red pens and 3 packages of blue pens. How many pens did he buy altogether?
Figure 1 Robin bought 369 pens. She bought 21 packages of red pens. How many packages of blue pens did she buy?
Figure 2 Explain why it is not possible for Susan to buy exactly 31 pens.

Solution

Expressing 15\dfrac15 and 14\dfrac14 with a common denominator of 40, we get 15=840\dfrac15=\dfrac{8}{40} and 14=1040\dfrac14=\dfrac{10}{40}. We require that n40>840\dfrac{n}{40}>\dfrac{8}{40} and n40<1040\dfrac{n}{40}<\dfrac{10}{40}, thus n>8n>8 and n<10n<10. The only integer nn that satisfies both of these inequalities is n=9n=9. Expressing m8\dfrac m8 and 13\dfrac13 with a common denominator of 24, we require 3m24>824\dfrac{3m}{24}>\dfrac{8}{24} and so 3m>83m>8 or m>83m>\dfrac83. Since 83=223\dfrac83=2\dfrac23 and mm is an integer, then m3m\geq3. Expressing m+18\dfrac {m+1}{8} and 23\dfrac23 with a common denominator of 24, we require 3(m+1)24<1624\dfrac{3(m+1)}{24}<\dfrac{16}{24} or 3m+3<163m+3<16 or 3m<133m<13, and so m<133m<\dfrac{13}{3}. Since 133=413\dfrac{13}{3}=4\dfrac13 and mm is an integer, then m4m\leq4. The integer values of mm which satisfy m3m\geq3 and m4m\leq4 are m=3m=3 and m=4m=4. At the start of the weekend, Fiona has played 30 games and has ww wins, so her win ratio is w30\dfrac{w}{30}. Fiona’s win ratio at the start of the weekend is greater than 0.5=120.5=\dfrac12, and so w30>12\dfrac{w}{30}>\dfrac12. Since 12=1530\dfrac12=\dfrac{15}{30}, then we get w30>1530\dfrac{w}{30}>\dfrac{15}{30}, and so w>15w>15. During the weekend Fiona plays five games giving her a total of 30+5=3530+5=35 games played. Since she wins three of these games, she now has w+3w+3 wins, and so her win ratio is w+335\dfrac{w+3}{35}. Fiona’s win ratio at the end of the weekend is less than 0.7=7100.7=\dfrac{7}{10}, and so w+335<710\dfrac{w+3}{35}<\dfrac{7}{10}. Rewriting this inequality with a common denominator of 70, we get 2(w+3)70<4970\dfrac{2(w+3)}{70}<\dfrac{49}{70} or 2(w+3)<492(w+3)<49 or 2w+6<492w+6<49 or 2w<432w<43, and so w<432w<\dfrac{43}{2}. Since 432=2112\dfrac{43}{2}=21\dfrac12 and ww is an integer, then w21w\leq21. The integer values of ww which satisfy w>15w>15 and w21w\leq21 are w=16,17,18,19,20,21w=16,17,18,19,20,21.

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