1. **State the problem:** John bought large and small bottles of drinks. Large bottles are sold 2 for 15, small bottles 3 for 10. He spent 100 more on large bottles than small bottles. We need to find the total number of bottles bought. 2. **Define variables:** Let $x$ be the number of large bottles bought, and $y$ be the number of small bottles bought. 3. **Write the cost equations:** - Cost per large bottle = $\frac{15}{2}$ - Cost per small bottle = $\frac{10}{3}$ Total spent on large bottles = $\frac{15}{2}x$ Total spent on small bottles = $\frac{10}{3}y$ 4. **Use the given condition:** $$\frac{15}{2}x = \frac{10}{3}y + 100$$ 5. **Express $x$ in terms of $y$:** $$\frac{15}{2}x - \frac{10}{3}y = 100$$ Multiply both sides by 6 (LCM of 2 and 3): $$6 \times \frac{15}{2}x - 6 \times \frac{10}{3}y = 6 \times 100$$ $$45x - 20y = 600$$ 6. **Simplify the equation:** $$45x = 20y + 600$$ $$x = \frac{20y + 600}{45}$$ 7. **Since $x$ and $y$ represent number of bottles, they must be integers.** We look for integer $y$ such that $x$ is integer. Rewrite: $$x = \frac{20y + 600}{45} = \frac{20y}{45} + \frac{600}{45} = \frac{4y}{9} + \frac{40}{3}$$ For $x$ to be integer, $\frac{4y}{9}$ must be such that $\frac{4y}{9} + \frac{40}{3}$ is integer. Since $\frac{40}{3} = 13\frac{1}{3}$, $\frac{4y}{9}$ must be of the form $n - 13\frac{1}{3}$ where $n$ is integer. Multiply both sides by 9: $$4y = 9n - 40$$ We want $4y + 40$ divisible by 9. Try values of $y$ that make $4y + 40$ divisible by 9. Try $y=2$: $$4(2)+40=8+40=48$$ divisible by 9? 48/9=5.33 no. Try $y=4$: $$4(4)+40=16+40=56$$ no. Try $y=7$: $$4(7)+40=28+40=68$$ no. Try $y=5$: $$4(5)+40=20+40=60$$ yes, 60/9=6.66 no. Try $y=10$: $$4(10)+40=40+40=80$$ no. Try $y=20$: $$4(20)+40=80+40=120$$ yes, 120/9=13.33 no. Try $y=25$: $$4(25)+40=100+40=140$$ no. Try $y=35$: $$4(35)+40=140+40=180$$ yes, 180/9=20 integer. So $y=35$ works. 8. **Calculate $x$ for $y=35$:** $$x = \frac{20(35) + 600}{45} = \frac{700 + 600}{45} = \frac{1300}{45} = 28.88$$ not integer. Try $y=30$: $$4(30)+40=120+40=160$$ no. Try $y=40$: $$4(40)+40=160+40=200$$ no. Try $y=45$: $$4(45)+40=180+40=220$$ no. Try $y=50$: $$4(50)+40=200+40=240$$ yes, 240/9=26.66 no. Try $y=55$: $$4(55)+40=220+40=260$$ no. Try $y=60$: $$4(60)+40=240+40=280$$ no. Try $y=65$: $$4(65)+40=260+40=300$$ yes, 300/9=33.33 no. Try $y=70$: $$4(70)+40=280+40=320$$ no. Try $y=75$: $$4(75)+40=300+40=340$$ no. Try $y=80$: $$4(80)+40=320+40=360$$ yes, 360/9=40 integer. Calculate $x$ for $y=80$: $$x = \frac{20(80) + 600}{45} = \frac{1600 + 600}{45} = \frac{2200}{45} = 48.88$$ no. Try $y=90$: $$4(90)+40=360+40=400$$ no. Try $y=95$: $$4(95)+40=380+40=420$$ yes, 420/9=46.66 no. Try $y=100$: $$4(100)+40=400+40=440$$ no. Try $y=105$: $$4(105)+40=420+40=460$$ no. Try $y=110$: $$4(110)+40=440+40=480$$ yes, 480/9=53.33 no. Try $y=115$: $$4(115)+40=460+40=500$$ no. Try $y=120$: $$4(120)+40=480+40=520$$ no. Try $y=125$: $$4(125)+40=500+40=540$$ yes, 540/9=60 integer. Calculate $x$ for $y=125$: $$x = \frac{20(125) + 600}{45} = \frac{2500 + 600}{45} = \frac{3100}{45} = 68.88$$ no. Try $y=135$: $$4(135)+40=540+40=580$$ no. Try $y=140$: $$4(140)+40=560+40=600$$ yes, 600/9=66.66 no. Try $y=145$: $$4(145)+40=580+40=620$$ no. Try $y=150$: $$4(150)+40=600+40=640$$ no. Try $y=155$: $$4(155)+40=620+40=660$$ yes, 660/9=73.33 no. Try $y=160$: $$4(160)+40=640+40=680$$ no. Try $y=165$: $$4(165)+40=660+40=700$$ no. Try $y=170$: $$4(170)+40=680+40=720$$ yes, 720/9=80 integer. Calculate $x$ for $y=170$: $$x = \frac{20(170) + 600}{45} = \frac{3400 + 600}{45} = \frac{4000}{45} = 88.88$$ no. Try $y=180$: $$4(180)+40=720+40=760$$ no. Try $y=185$: $$4(185)+40=740+40=780$$ yes, 780/9=86.66 no. Try $y=190$: $$4(190)+40=760+40=800$$ no. Try $y=195$: $$4(195)+40=780+40=820$$ no. Try $y=200$: $$4(200)+40=800+40=840$$ yes, 840/9=93.33 no. Try $y=210$: $$4(210)+40=840+40=880$$ no. Try $y=215$: $$4(215)+40=860+40=900$$ yes, 900/9=100 integer. Calculate $x$ for $y=215$: $$x = \frac{20(215) + 600}{45} = \frac{4300 + 600}{45} = \frac{4900}{45} = 108.88$$ no. 9. **Try a different approach:** Since $x$ and $y$ must be multiples of 2 and 3 respectively (because bottles sold in packs), let: $$x = 2a, \quad y = 3b$$ Rewrite the cost equation: $$\frac{15}{2}x = \frac{10}{3}y + 100$$ Substitute: $$\frac{15}{2} (2a) = \frac{10}{3} (3b) + 100$$ Simplify: $$15a = 10b + 100$$ 10. **Solve for $a$:** $$15a = 10b + 100$$ Divide both sides by 5: $$3a = 2b + 20$$ 11. **Express $a$ in terms of $b$:** $$a = \frac{2b + 20}{3}$$ 12. **For $a$ to be integer, $2b + 20$ must be divisible by 3.** Check values of $b$: - $b=1$: $2(1)+20=22$ no - $b=2$: $24$ yes - $b=3$: $26$ no - $b=4$: $28$ no - $b=5$: $30$ yes So $b=2,5,8,11,...$ (every $b$ where $2b+20$ divisible by 3) 13. **Try $b=2$:** $$a = \frac{2(2)+20}{3} = \frac{24}{3} = 8$$ Calculate total bottles: $$x = 2a = 2 \times 8 = 16$$ $$y = 3b = 3 \times 2 = 6$$ Total bottles = $16 + 6 = 22$ 14. **Verify cost difference:** Cost large = $\frac{15}{2} \times 16 = 15 \times 8 = 120$ Cost small = $\frac{10}{3} \times 6 = 10 \times 2 = 20$ Difference = $120 - 20 = 100$ correct. **Final answer:** John bought **22 bottles** altogether.