Porisma

Worked solution

Fill increasing bags without exceeding the supply

Alice has 48004800 sweets to divide into bags. She puts 2525 sweets in the first bag and 55 more in each subsequent bag than in the preceding bag. She stops when there are not enough sweets to fill the next bag in the same manner. Find the number of sweets left behind.

[2008/AJC/Promo/5(a)]

CJC · Sequences and Series · Example 21 (Self-Reading)

Answer

120120 sweets.

Compare your reasoning

Model the number used, not the size of one bag

Sn=n2[2(25)+(n−1)5]≤4800\begin{aligned}&S_n\\ & =\frac n2[2(25)+(n-1)5]\le4800\end{aligned}

The source identifies a=25,d=5. Its illustrated guidance emphasises formulating and solving the inequality.

Method 1: solve the quadratic inequality

n2+9n−1920≤0,n2+9n−1920=0⟹n=−9±92−4(1)(−1920)2(1),−9−77612≤n≤−9+77612.\begin{aligned}n^2+9n-1920&\le0,\\n^2+9n-1920&=0\quad\Longrightarrow\quad n=\frac{-9\pm\sqrt{9^2-4(1)(-1920)}}{2(1)},\\\frac{-9-\sqrt{7761}}2&\le n\le\frac{-9+\sqrt{7761}}2.\end{aligned}

The source uses the quadratic equation and an upward-opening parabola: the quadratic is nonpositive between its two roots. Keep exact roots until choosing the positive integer count.

Restrict the real solution to a bag count

n∈Z+,n≤39.548268978…⟹nmax⁡=39n\in\mathbb Z^+,\qquad n\le39.548268978\ldots\quad\Longrightarrow\quad n_{\max}=39

The roots are approximately −48.548 and 39.548, as labelled in the source sketch. A filled-bag count is a positive integer, so the maximum is 39.

Method 2: use the calculator table

nSncomparison with 4800384465<394680<404900>\begin{array}{c|r|c}n&S_n&\text{comparison with }4800\\\hline38&4465&<\\39&4680&<\\40&4900&>\end{array}

The source calls this graphical. Enter Y1=(X/2)(50+5(X−1)) and inspect integer X. Its adjacent values show that 39 bags fit and 40 do not.

Find the unused sweets

S39=392[2(25)+38(5)]=4680,4800−4680=120S_{39}=\frac{39}{2}[2(25)+38(5)]=4680,\qquad4800-4680=120

Both printed methods end with the same S39 calculation and subtraction.

How did this question go?

Common mistakes

Use the greatest feasible whole bag count, then subtract the total already used. The leftover 120 is less than the next bag’s required 220.

Solution notes

The source writes the rounded interval −48.54≤n≤39.54 after showing roots near −48.548 and 39.548. Those truncated endpoints do not describe the exact real solution. The exact interval is retained in the working; the positive integer maximum 39 and the printed answer 120 are unchanged.