Chapter 2: Power Play

This is a hands-on activity. Take an ordinary sheet of paper (like A4) and fold it in half repeatedly. After each fold the thickness doubles, so the folded stack grows very fast in thickness while its size shrinks.

You will find that after about $$7$$ folds it becomes almost impossible to fold further — the stack is too thick and too short. With thinner paper (tissue paper or newspaper), you may manage $$8$$ or $$9$$ folds, but no more. So Estu's claim is essentially correct for a normal sheet.

This is a guessing question — we are asked to make an instinctive estimate before calculating. Most people guess only a few centimetres, or maybe a metre or two, because we underestimate how fast doubling adds up.

Starting from an initial thickness of $$0.001$$ cm and doubling once for every fold, the thickness after $$30$$ folds is $$0.001 \times 2^{30}$$ cm. Later we shall see that this works out to about $$10.7$$ km — roughly the altitude at which passenger planes cruise!

The rule behind the table is simple: the initial thickness is $$0.001$$ cm, and each fold doubles it. So the thickness after $$n$$ folds is

Checking a couple of rows: after $$10$$ folds we get $$0.001 \times 2^{10} = 0.001 \times 1024 = 1.024$$ cm, and after $$17$$ folds we get $$0.001 \times 2^{17} = 0.001 \times 1{,}31{,}072 = 131.072$$ cm $$\approx 131$$ cm — both match the table.

So even though the numbers in the table look small at first, the fact that they double at every step means they grow explosively.

This is another guess, made before calculating. Guesses can vary from a few centimetres to a few metres. Let us compute the actual answers using $$0.001 \times 2^{n}$$ cm.

After 30 folds: $$0.001 \times 2^{30}$$ cm $$= 0.001 \times 1{,}07{,}37{,}41{,}824$$ cm $$= 10{,}73{,}741.824$$ cm $$\approx 10.74$$ km. That is close to the altitude at which passenger airplanes fly.

After 45 folds: $$0.001 \times 2^{45}$$ cm $$= 0.001 \times 3.5184 \times 10^{13}$$ cm $$\approx 3.52 \times 10^{10}$$ cm $$\approx 3{,}51{,}844$$ km. This is almost the distance between the Earth and the Moon (about $$3{,}84{,}400$$ km)!

Use the rule $$\text{Thickness after } n \text{ folds} = 0.001 \times 2^{n}$$ cm, doubling from one fold to the next. Converting to metres or kilometres as convenient:

Sample calculation for fold $$26$$: $$0.001 \times 2^{26} = 0.001 \times 6{,}71{,}08{,}864$$ cm $$= 67{,}108.864$$ cm $$\approx 671$$ m, taller than the Burj Khalifa's $$830$$ m. For fold $$45$$: $$0.001 \times 2^{45}$$ cm $$\approx 3{,}51{,}844$$ km, close to the Earth-Moon distance of $$3{,}84{,}400$$ km.

Every fold doubles the thickness. So starting from $$v$$, after each fold we multiply the current thickness by $$2$$:

After $$1$$ fold: $$v \times 2 = 2v$$.
After $$2$$ folds: $$v \times 2 \times 2 = 2^{2} v$$.
After $$3$$ folds: $$v \times 2 \times 2 \times 2 = 2^{3} v$$.

Continuing this pattern, after $$10$$ folds the thickness is $$v \times 2 \times 2 \times \cdots \times 2$$ ($$10$$ times) $$= 2^{10} v$$.

So the correct choice is (v) $$2^{10} v$$.

The other choices are wrong because $$10v$$, $$10 + v$$, $$2 \times 10 \times v$$ and $$10^{2} v$$ describe linear (additive) or square growth, while doubling is exponential; and $$2^{10}$$ alone forgets the starting thickness $$v$$.

Divide $$32400$$ successively by the smallest prime that divides it:

$$32400 \div 2 = 16200$$
$$16200 \div 2 = 8100$$
$$8100 \div 2 = 4050$$
$$4050 \div 2 = 2025$$
$$2025 \div 3 = 675$$
$$675 \div 3 = 225$$
$$225 \div 3 = 75$$
$$75 \div 3 = 25$$
$$25 \div 5 = 5$$
$$5 \div 5 = 1$$

So $$32400 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5$$. Collecting equal factors:

Multiplying $$(-1)$$ by itself, the sign flips at every step. In fact,

$$(-1) \times (-1) = 1$$ (two negatives make a positive), and $$1 \times (-1) = -1$$.

So an odd number of $$-1$$s multiplied together gives $$-1$$, and an even number gives $$+1$$.

$$(-1)^{5} = (-1)(-1)(-1)(-1)(-1) = -1$$ (since $$5$$ is odd) — negative.

$$(-1)^{56} = +1$$ (since $$56$$ is even) — positive.

Multiply $$-2$$ by itself four times, pairing the negatives:

$$(-2)^{4} = (-2) \times (-2) \times (-2) \times (-2)$$
$$= \big[(-2) \times (-2)\big] \times \big[(-2) \times (-2)\big]$$
$$= 4 \times 4 = 16.$$

So yes, $$(-2)^{4} = 16$$. In general, an even power of a negative number is positive.

Raising $$0$$ to a positive power means multiplying zero by itself that many times, and any product that contains a $$0$$ factor is $$0$$.

$$0^{2} = 0 \times 0 = 0.$$
$$0^{5} = 0 \times 0 \times 0 \times 0 \times 0 = 0.$$

So for any counting number $$n$$, $$0^{n} = 0$$.

Multiply the count at each level by $$3$$:

Daughters: $$3$$.
Baskets: $$3 \times 3 = 3^{2} = 9$$.
Keys: $$9 \times 3 = 3^{3} = 27$$.
Rooms (each key opens $$3$$ rooms): $$27 \times 3 = 3^{4} = 81$$.
Tables ($$3$$ per room): $$81 \times 3 = 3^{5} = 243$$.
Necklaces ($$3$$ per table): $$243 \times 3 = 3^{6} = 729$$.
Diamonds ($$3$$ per necklace): $$729 \times 3 = 3^{7} = 2187$$.

So there are $$3^{7} = 2187$$ diamonds. The chain of seven multiplications by $$3$$ is exactly why the answer is a power of three.

Follow the chain from the poem, multiplying by $$3$$ each time we go one level deeper:

$$3$$ daughters $$\times \ 3$$ baskets each $$\times \ 3$$ keys each $$\times \ 3$$ rooms each.

So there are $$81$$ rooms in total.

From the poem, each room has $$3$$ tables, each table $$3$$ necklaces, and each necklace $$3$$ diamonds. So the number of diamonds per room is $$3 \times 3 \times 3 = 3^{3} = 27$$.

We already found $$81$$ rooms, i.e., $$3^{4}$$. So the total number of diamonds is

Yes — one multiplication ($$81 \times 27$$) is enough, once we know that rooms give $$3^{4}$$ and diamonds per room give $$3^{3}$$.

Expand each side using the definition of a power (a product of the same factor):

$$3^{7} = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$$ (seven $$3$$s).

Group these seven $$3$$s as $$2$$ and $$5$$:

$$3^{7} = (3 \times 3) \times (3 \times 3 \times 3 \times 3 \times 3) = 3^{2} \times 3^{5}.$$

Multiplying two $$3$$s together with five $$3$$s gives $$2 + 5 = 7$$ threes in all, so the product is $$3^{7}$$.

This is a special case of the general rule $$n^{a} \times n^{b} = n^{a+b}$$.

By the rule $$n^{a} \times n^{b} = n^{a+b}$$,

Or, expanding: $$(p \times p \times p \times p) \times (p \times p \times p \times p \times p \times p) = p^{10}$$ ($$4 + 6 = 10$$ factors of $$p$$).

Group the ten $$2$$s into two blocks of five:

$$2^{10} = (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2) = 2^{5} \times 2^{5} = (2^{5})^{2}.$$

Numerically, $$2^{5} = 32$$, so $$(2^{5})^{2} = 32 \times 32 = 1024 = 2^{10}$$. Yes, they are equal.

This is a special case of the general rule $$(n^{a})^{b} = n^{a \times b}$$: here $$a = 5, b = 2$$, so $$(2^{5})^{2} = 2^{5 \times 2} = 2^{10}$$.

The number of lotuses doubles every day. So on any day, the number is exactly half what it will be the very next day. Since the pond is fully covered on day $$30$$, on day $$29$$ it must be exactly half covered — one day away from doubling to full.

This shows how deceptive exponential growth can be: even the day before the pond is fully covered, only half of it looks green!

Step 1 — doubling pond: Start with $$1$$ lotus. After each of the first $$4$$ days, the count doubles:

Damayanti now moves all $$16$$ lotuses to the tripling pond.

Step 2 — tripling pond: Start with $$16$$ lotuses. After each of the next $$4$$ days, the count triples:

Written together, the total is $$2^{4} \times 3^{4} = (2 \times 3)^{4} = 6^{4} = 1296$$.

Swap the two ponds. Start with $$1$$ lotus in the tripling pond.

After $$4$$ days there: $$1 \times 3^{4} = 81$$ lotuses.

Move all $$81$$ to the doubling pond. After $$4$$ more days there: $$81 \times 2^{4} = 81 \times 16 = 1296$$ lotuses.

The count is the same, $$1296$$, because multiplication is commutative:

So the order in which she places the flowers does not matter — the total after $$8$$ days is the same.

When two powers share the same exponent, we can pair up the bases:

Applying this with $$m = 2, n = 5, a = 5$$:

So the product is $$10^{5} = 1$$ lakh.

When two powers with the same exponent are divided, the bases can be divided:

So

Each outfit is a choice of one dress and one cap. For every one of the $$4$$ dresses, there are $$3$$ caps to pair it with, so the total number of dress-and-cap combinations is

(We could count the other way round too: for each of the $$3$$ caps, there are $$4$$ dresses, so $$3 \times 4 = 12$$. Multiplication does not care about order.)

An outfit is now a choice of one dress and one hat and one pair of shoes. So we multiply the three counts:

You could imagine a tree: each of $$7$$ dresses branches into $$2$$ hats, and each of those into $$3$$ pairs of shoes, giving $$7 \times 2 \times 3 = 42$$ paths from the root to a leaf — one for each outfit.

Each of the $$5$$ slots can hold any of the $$10$$ digits $$0, 1, 2, \ldots, 9$$, independently of the others. So the number of possible passwords is

(This is exactly the same as listing every 5-digit combination from $$00000$$ up to $$99999$$, i.e., $$1$$ lakh strings.) Since the lock opened only on the last try, they checked all $$1{,}00{,}000$$ passwords.

There are $$26$$ letters ($$A$$ to $$Z$$), and each of the $$6$$ slots can independently be any letter. So the count is

Computing: $$26^{2} = 676$$, $$26^{3} = 26 \times 676 = 17{,}576$$, and

So Estu's lock has about $$3.09 \times 10^{8}$$ possible passwords — indeed far safer than the $$1$$ lakh 5-digit passwords of the earlier lock.

When dividing powers of the same base, exponents subtract: $$n^{a} \div n^{b} = n^{a - b}$$ (for $$a > b, n \ne 0$$).

Why? Expanding, $$\dfrac{2^{100}}{2^{25}}$$ has $$100$$ twos on top and $$25$$ on the bottom. Each of the $$25$$ twos on the bottom cancels one from the top, leaving $$100 - 25 = 75$$ twos on top.

The rule $$n^{a} \div n^{b} = n^{a - b}$$ comes from cancellation in the fraction $$\dfrac{n^{a}}{n^{b}}$$. But if $$n = 0$$, then $$n^{b} = 0$$ in the denominator, and division by zero is not defined.

For example, $$0^{4} \div 0^{2}$$ would mean $$\dfrac{0}{0}$$, which has no meaningful value. So we must require $$n \ne 0$$ whenever we divide powers.

Extend the rule $$2^{a} \div 2^{b} = 2^{a - b}$$ to the case $$a = b$$. On the left, we get $$\dfrac{2^{a}}{2^{a}} = 1$$; on the right, we get $$2^{a - a} = 2^{0}$$. For both to agree, we must define

Concretely, $$2^{0} = 2^{4 - 4} = \dfrac{2^{4}}{2^{4}} = \dfrac{16}{16} = 1$$.

The same reasoning gives $$x^{0} = 1$$ for every nonzero number $$x$$.

Yes. We know $$10^{-3} = \dfrac{1}{10^{3}}$$ (that is how negative exponents are defined). So

More generally, $$n^{a} = \dfrac{1}{n^{-a}}$$ for every $$n \ne 0$$ — negative exponents "flip" a fraction, and flipping twice returns the original.

Yes. Once we have defined $$n^{0} = 1$$ and $$n^{-a} = \dfrac{1}{n^{a}}$$ (with $$n \ne 0$$), all three exponent rules extend to any integer exponents:

$$n^{a} \times n^{b} = n^{a + b}$$
$$(n^{a})^{b} = n^{a \times b}$$
$$n^{a} \div n^{b} = n^{a - b}$$.

Quick sanity check with negative exponents:

$$2^{3} \times 2^{-5} = 2^{3 + (-5)} = 2^{-2} = \dfrac{1}{4}.$$
Also directly: $$2^{3} \times 2^{-5} = 8 \times \dfrac{1}{32} = \dfrac{8}{32} = \dfrac{1}{4}$$. ✓

Or: $$2^{4} \div 2^{7} = 2^{4 - 7} = 2^{-3} = \dfrac{1}{8}$$. Directly, $$\dfrac{16}{128} = \dfrac{1}{8}$$. ✓

"How many times larger" is a division question:

So $$4^{7} = 16 \times 4^{5}$$, i.e., $$16384 = 16 \times 1024$$. Check: $$16 \times 1024 = 16384$$. ✓

Yes — $$16384$$ is exactly $$16$$ times larger than $$1024$$.

Divide the larger quantity by the smaller:

Check: $$4^{2} = 16$$ and $$4^{-2} = \dfrac{1}{16}$$, so $$\dfrac{16}{1/16} = 16 \times 16 = 256$$. ✓

So $$4^{2}$$ is $$256$$ times larger than $$4^{-2}$$.

In scientific notation we write each number as $$x \times 10^{y}$$, where $$1 \le x < 10$$ and $$y$$ counts the number of places we shifted the decimal point.

(i) $$30{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}000$$ m has $$23$$ digits. Move the decimal point $$22$$ places to the left to bring it after the first non-zero digit:

(ii) $$1{,}00{,}00{,}00{,}00{,}000$$ has $$12$$ digits (a $$1$$ followed by $$11$$ zeros), so it equals

(iii) $$59{,}76{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}00{,}000$$ kg has $$25$$ digits. The significant part is $$5.976$$, and shifting the decimal $$24$$ places gives

In scientific notation, the exponent tells us the order of magnitude. Comparing:

Sun-Saturn: $$1.4335 \times 10^{12}$$ m
Saturn-Uranus: $$1.439 \times 10^{12}$$ m
Sun-Earth: $$1.496 \times 10^{11}$$ m

The first two both have exponent $$12$$; the third has exponent $$11$$. Since $$10^{11}$$ is $$10$$ times smaller than $$10^{12}$$, the Sun-Earth distance is clearly the smallest. (In fact, the Sun-Earth distance is roughly $$\dfrac{1.496 \times 10^{11}}{1.4335 \times 10^{12}} \approx 0.10$$, i.e., about a tenth of the Sun-Saturn distance.)

Compute the fraction of the total distance:

So Earth sits at roughly $$10.4\%$$ of the way from the Sun to Saturn — very close to the Sun end.

On a number line drawn from Sun (left, at $$0$$) to Saturn (right, at $$1.4335 \times 10^{12}$$ m), mark Earth's position at about one-tenth of the total length from the Sun.

This question is really asking us to set up the relationships before plugging in any numbers. There are four quantities in the problem: Roxie's weight, Estu's weight, the cost per kg of jaggery, and the cost per kg of wheat.

The worth of a donation is just the amount donated multiplied by the cost per unit:

The actual number requires us to assume reasonable values for the weights and prices, which is what the next question does.

Assumptions. A typical Indian 13-year-old weighs about $$45$$ kg, and a typical 11-year-old about $$40$$ kg. Jaggery costs roughly $$₹70$$ per kg, and wheat about $$₹50$$ per kg (retail prices in 2026 can vary; these are round-number estimates).

Worth of jaggery (Roxie's weight $$\approx 45$$ kg):

Worth of wheat (Estu's weight $$\approx 40$$ kg):

Your answer may vary a little depending on the assumed weights and prices — that is expected in an estimation problem.

To match Roxie's weight with coins, we need to know two things:

1. Roxie's weight (in the same units as the coin weight, say grams).
2. The weight of a single 1-rupee coin.

Then the required number of coins is

The weight of one coin can be measured with a small kitchen scale (weigh a small stack and divide), or looked up — the current $$1$$-rupee coin weighs about $$4$$ g.

A rough head-check: a $$1$$-rupee coin is small and light — noticeably lighter than a rupee note but not featherweight. It feels like a few grams. Roxie weighs a few tens of kilograms, i.e., tens of thousands of grams.

So the ratio $$\dfrac{\text{tens of thousands of grams}}{\text{a few grams}}$$ should be in the thousands, not lakhs.

Indeed, with the reasonable numbers used in the next question, we get about $$11{,}000$$ coins — squarely in the thousands.

Assumptions. Take Roxie's weight $$\approx 45$$ kg $$= 45{,}000$$ g, and the weight of one 1-rupee coin $$\approx 4$$ g.

Compute.

Since each coin is worth $$₹1$$, this is $$₹11{,}250$$ in coins — a mountain of change, but the count is about $$11$$ thousand, matching our instinctive guess.

Guess first. A $$5$$-rupee coin is heavier than a $$1$$-rupee coin (fewer coins), and a $$10$$-rupee note is very light (many notes). So we expect the number of $$5$$-rupee coins to be somewhat less than $$11{,}250$$, but the value larger; and the $$10$$-rupee notes to be a huge count but very valuable.

Assumptions. A $$5$$-rupee coin weighs about $$6$$ g; a $$10$$-rupee note weighs about $$1$$ g. Take Roxie's weight as $$45$$ kg $$= 45{,}000$$ g.

5-rupee coins.

10-rupee notes.

So switching to $$5$$-rupee coins gives about $$₹37{,}500$$, and switching to $$10$$-rupee notes about $$₹4.5$$ lakh — a huge jump, driven mostly by how light the notes are.

Assumptions. As adults, Estu weighs $$\approx 60$$ kg and Roxie $$\approx 55$$ kg. A single notebook ($$100$$ pages) weighs about $$120$$ g, and each student needs, say, $$5$$ notebooks. One filling meal uses about $$300$$ g of rice/grain (uncooked weight, roughly $$150$$ g dry rice with dal, vegetables etc., averaged out).

Estu — notebooks. Notebooks in $$60$$ kg $$= 60{,}000$$ g at $$120$$ g each:

At $$5$$ notebooks per student, this helps $$\dfrac{500}{5} = 100$$ students each year.

Roxie — annadāna. Meals in $$55$$ kg $$= 55{,}000$$ g at $$300$$ g of grain each:

So roughly $$180$$ people can be fed one filling meal each year from Roxie's offering.

Numbers can shift depending on the assumed weights, notebook size, or meal size — the useful takeaway is that both offerings can benefit around a hundred to a couple of hundred people every year.

Assumptions. A comfortable walking pace on a pilgrimage (with rests, load and heat) is about $$4$$ km per hour. In a day, a walker can manage roughly $$8$$ hours of actual walking, so about $$4 \times 8 = 32$$ km per day.

Total walking hours:

Days: At $$8$$ hours per day,

So they most likely started about two weeks ago. Answers may differ a little depending on the assumed pace and daily walking hours, but $$10$$-$$15$$ days is a good estimate.

This is a general instruction that goes with the estimation questions that follow (walking around the world, building a ladder to the moon, etc.). The recipe is:

1. Guess a value before starting any calculation. It doesn't have to be right — the goal is to build intuition.
2. Model the situation: what quantities matter, and how are they related?
3. Assume reasonable values for the unknown quantities. Round to easy numbers where you can.
4. Compute the estimate.
5. Compare the computed answer with your guess. How far off were you?

Following these steps for a few problems trains you to sense whether a number is 'in the thousands' or 'in the crores' without doing the full computation.

Assumptions. Adult walking pace $$\approx 5$$ km/h. If a person walks $$8$$ hours a day (a realistic maximum, with time for sleep, rest and meals), the daily distance is $$5 \times 8 = 40$$ km. A human lifetime is about $$70$$ years — but a person cannot walk from birth or in extreme old age, so let us assume they can effectively walk for $$50$$ productive years.

Total distance walked in a lifetime:

Number of circumnavigations:

So roughly $$15$$-$$20$$ circumnavigations in a lifetime is a good estimate. Answers may vary with pace, hours walked per day, and lifetime length.

If instead we take a stricter 'non-stop but realistic' pace, only walking a few hours daily and covering $$20$$ km/day, we get about $$\dfrac{20 \times 365 \times 50}{40{,}000} \approx 9$$ circumnavigations. Either way, the answer is tens, not hundreds.

Guess before calculating! The Moon is about $$3{,}84{,}400$$ km away. Depending on how far apart the steps are, guesses might range from a few thousand ('like a very tall staircase') to a few crore ('like tons and tons of steps').

The actual answer, worked out in later questions, is about $$1{,}92{,}20{,}00{,}000$$ steps (nearly $$2$$ billion) if the steps are $$20$$ cm apart — most people badly underestimate it. This is a nice illustration of linear growth: adding just $$20$$ cm at a time, we still need a huge number of steps for a very long distance.

Quick order-of-magnitude check. Assume each step is $$20$$ cm apart. The Moon is at $$3{,}84{,}400$$ km $$= 3.844 \times 10^{5}$$ km $$= 3.844 \times 10^{10}$$ cm.

$$10^{9}$$ is one arab (a billion), also called $$100$$ crore. So the number of steps is in the crores, specifically about $$192$$ crore steps.

Put everything into the same unit (centimetres):

Now divide by the length of one step ($$20$$ cm):

In Indian numerals, $$1{,}92{,}20{,}00{,}000$$ — about $$192$$ crore or $$1$$ arab $$92$$ crore steps.

Linear growth adds the same amount each step, so the running total grows by adding: it is additive. Examples:

• A cyclist riding at a constant speed: after $$1, 2, 3, \ldots$$ hours, distance is $$15, 30, 45, \ldots$$ km.
• Putting $$₹50$$ into a piggy bank every week: after $$n$$ weeks, the total is $$50n$$.
• Height of a ladder as you climb: each step adds a fixed $$20$$ cm.
• Simple interest on a savings deposit.

Exponential growth multiplies by the same factor each step, so it is multiplicative. Examples:

• Paper folding: thickness doubles at every fold ($$0.001, 0.002, 0.004, \ldots$$ cm).
• Bacteria dividing every $$20$$ minutes: $$1, 2, 4, 8, \ldots$$.
• Compound interest in a bank account.
• Number of ancestors going back generations: $$2$$ parents, $$4$$ grandparents, $$8$$ great-grandparents, $$\ldots$$.
• The lotus pond that doubles daily, or the diamonds in the poem (three at each level).
• A viral message: each person forwards it to $$3$$ friends, they each forward to $$3$$ more, and so on.

People per elephant is a division:

So yes — there are about $$20{,}000$$ people for every African elephant. That is one elephant per average town, if we were to distribute them evenly!

One day contains $$24 \times 60 \times 60 = 86{,}400$$ seconds. Convert $$10^{6}$$ seconds to days:

So a million seconds is roughly $$11$$ and a half days — less than a fortnight. That is much shorter than most students imagine. (For contrast, a billion seconds is about $$31.7$$ years.)

Aim for events roughly $$1$$-$$2$$ days (for $$10^{5}$$ s) or roughly $$1$$-$$2$$ weeks (for $$10^{6}$$ s).

(i) Order of $$10^{5}$$ seconds ($$\approx 1$$-$$2$$ days):

• One full day $$= 86{,}400 \text{ s} = 8.64 \times 10^{4}$$ s (just below $$10^{5}$$).
• A long train journey across India (say $$36$$ hours): $$36 \times 3600 = 1.296 \times 10^{5}$$ s.
• A five-day Test cricket match: $$5 \times 86{,}400 = 4.32 \times 10^{5}$$ s.
• Weekend + a day (a school holiday break of $$3$$ days): $$\approx 2.6 \times 10^{5}$$ s.

(ii) Order of $$10^{6}$$ seconds ($$\approx 12$$ days):

• A two-week summer holiday: $$14 \times 86{,}400 = 1.21 \times 10^{6}$$ s.
• The Cricket World Cup pool stage ($$\approx 15$$ days): $$\approx 1.3 \times 10^{6}$$ s.
• The lunar month (time between full moons, $$29.5$$ days): $$29.5 \times 86{,}400 \approx 2.55 \times 10^{6}$$ s.
• A pilgrimage of $$400$$ km on foot (as in the $$p\bar{a}day\bar{a}tra$$ question, $$\approx 12$$-$$13$$ days): $$\approx 1.1 \times 10^{6}$$ s.

The first part of each name is a Latin prefix that stands for a counting number:

If the prefix stands for $$n$$, then the number equals $$1000^{n+1} = 10^{3(n+1)}$$. For example, 'sext' means $$6$$, so sextillion $$= 1000^{7} = 10^{21}$$.

Convert everything to a single base $$2$$. Since $$4 = 2^{2}$$,

So

Now we need the units digit of $$2^{160}$$. Powers of $$2$$ have units digits in a cycle of length $$4$$:

The cycle $$(2, 4, 8, 6)$$ repeats every $$4$$ powers. To find which position $$160$$ lands on, compute the remainder of $$160$$ divided by $$4$$: $$160 = 4 \times 40$$, so the remainder is $$0$$, i.e., $$160$$ is a multiple of $$4$$ (like $$4, 8, 12, \ldots$$), corresponding to units digit $$6$$.

So the units digit of $$2^{160}$$, and hence of $$2^{224} \div 4^{32}$$, is $$\mathbf{6}$$.

One container arrives each day, and each container adds $$5$$ bottles. This is linear growth ($$+5$$ every day), not exponential.

After $$40$$ days, the total number of containers is $$40$$, so the total number of bottles is

Rewrite each expression as a product of powers of the primes $$2$$ and $$3$$.

$$2^{4} \times 3^{6}$$: already in prime form.

$$6^{4} \times 3^{2} = (2 \times 3)^{4} \times 3^{2} = 2^{4} \times 3^{4} \times 3^{2} = 2^{4} \times 3^{6}$$. ✓ same as the first.

$$6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$$.

$$18^{2} \times 6^{2} = (2 \times 3^{2})^{2} \times (2 \times 3)^{2} = 2^{2} \times 3^{4} \times 2^{2} \times 3^{2} = 2^{4} \times 3^{6}$$. ✓ also same as the first.

$$6^{24} = 2^{24} \times 3^{24}$$.

So the three equal quantities are $$2^{4} \times 3^{6}$$, $$6^{4} \times 3^{2}$$, and $$18^{2} \times 6^{2}$$. Their common value is $$16 \times 729 = 11{,}664$$. The other two ($$6^{10}$$ and $$6^{24}$$) are much larger and different from each other.

An $$n$$-digit code from $$10$$ digits allows $$10^{n}$$ distinct codes. We need $$10^{n} \ge 8.5 \times 10^{9}$$.

Check $$n = 9$$: $$10^{9} = 1$$ billion $$< 8.5$$ billion. Not enough.

Check $$n = 10$$: $$10^{10} = 10$$ billion $$> 8.5$$ billion. Enough.

So the code must have at least $$10$$ digits.

A number that is both a square and a cube must have every prime exponent divisible by both $$2$$ and $$3$$ — i.e., divisible by $$\mathrm{lcm}(2, 3) = 6$$. Equivalently, it must be a sixth power: some $$n^{6}$$.

Check: $$n^{6} = (n^{3})^{2}$$ (a square) $$= (n^{2})^{3}$$ (a cube). So every $$n^{6}$$ is both a square and a cube.

Examples:

• $$1^{6} = 1$$.
• $$2^{6} = 64$$ ($$= 8^{2} = 4^{3}$$, our given example).
• $$3^{6} = 729$$ ($$= 27^{2} = 9^{3}$$).
• $$4^{6} = 4096$$ ($$= 64^{2} = 16^{3}$$).
• $$5^{6} = 15{,}625$$ ($$= 125^{2} = 25^{3}$$).

So the general description is: a number is both a square and a cube iff it is a sixth power.

Each slot can be any of $$10$$ digits or $$26$$ letters, giving $$10 + 26 = 36$$ choices per slot. With $$5$$ independent slots, the total number of passcodes is

Compute step by step:

So there are $$36^{5} = 6{,}04{,}66{,}176$$ possible codes — about $$6$$ crore.

The total is a simple sum, not a product:

So the correct choices are (v) $$2 \times 10^{9}$$ and (vi) $$10^{9} + 10^{9}$$ (which are the same number).

Why the others are wrong:

• (i) $$20^{9} = 2^{9} \times 10^{9} = 512 \times 10^{9}$$ — off by a factor of $$256$$.
• (ii) $$10^{11}$$ is $$100$$ times too big.
• (iii) $$10^{10}$$ is $$10$$ times too big.
• (iv) $$10^{18}$$ comes from multiplying ($$10^{9} \times 10^{9}$$), not adding — vastly too big.

Convert $$1$$ arab ($$10^{9}$$) seconds into days:

Convert days to years, using $$365.25$$ days per year (to account for leap years):

So $$1$$ billion seconds is roughly $$31$$ years and $$8$$ months. Counting back that far from today's date 17 July 2026:

$$2026 - 31 = 1995$$, and $$17$$ July $$1995$$ minus about $$8$$ more months lands near November 1994 — around $$8$$-$$10$$ November 1994.

Interestingly, a billion seconds is comparable to a full generation — much longer than most students guess.