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1.P — Worked Problems: Numbers & Arithmetic
Each problem below is stated, then solved line by line with nothing skipped. Try it first with the solution folded away; the value is in the attempt, not in the reading. Every step names the formula it uses, and every formula is derived in 1.4 — powers roots logarithms.
Problem 1 — Fractions with different denominators
Work out \dfrac{3}{4} + \dfrac{5}{6} - \dfrac{2}{3} and give the answer in lowest terms.
Solution
Step 1 — find a common denominator. The denominators are 4, 6 and 3. Rather than multiplying them all together (which gives 72 and extra work), find the smallest number all three divide into. Their prime factors are 4 = 2^2, 6 = 2\cdot3, 3 = 3. The lowest common multiple takes the highest power of each prime present: 2^2 \cdot 3 = 12.
Step 2 — rewrite each fraction over 12. Multiply top and bottom by whatever turns the denominator into 12. This changes how the fraction is written, not what it is worth, because you are multiplying by a disguised 1.
\frac{3}{4} = \frac{3 \times 3}{4 \times 3} = \frac{9}{12}, \qquad \frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}, \qquad \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12}
Step 3 — now that the pieces are the same size, count them.
\frac{9}{12} + \frac{10}{12} - \frac{8}{12} = \frac{9 + 10 - 8}{12} = \frac{11}{12}
Step 4 — reduce if possible. 11 is prime and does not divide 12, so \frac{11}{12} is already in lowest terms.
Answer: \dfrac{11}{12}. Sanity check in decimals: 0.75 + 0.833 - 0.667 = 0.916, and 11/12 = 0.9167. Good.
Problem 2 — A repeating decimal to a fraction
Write 0.4\overline{27} (that is, 0.4272727\ldots, where only the "27" repeats) as a fraction.
Solution
The complication here is the non-repeating 4 sitting in front of the repeating block. The fix is to shift the number twice and subtract, choosing the shifts so the repeating tails line up exactly.
Step 1 — name it. Let x = 0.4272727\ldots
Step 2 — shift past the non-repeating part. One digit does not repeat, so multiply by 10:
10x = 4.272727\ldots
Step 3 — shift past one full repeating block as well. The block is two digits long, so multiply the original by 10^{1+2} = 1000:
1000x = 427.272727\ldots
Step 4 — subtract. Both lines now end in the identical infinite tail .272727\ldots, so the tails cancel completely:
1000x - 10x = 427.2727\ldots - 4.2727\ldots
990x = 423
Step 5 — solve and reduce.
x = \frac{423}{990}
Both are divisible by 9 (digit sums 4+2+3=9 and 9+9+0=18):
\frac{423 \div 9}{990 \div 9} = \frac{47}{110}
47 is prime and does not divide 110, so this is lowest terms.
Answer: \dfrac{47}{110}. Check: 47 \div 110 = 0.42727\ldots ✓
What to notice. The general recipe: multiply by 10^{(\text{non-repeating digits}) + (\text{block length})} and by 10^{(\text{non-repeating digits})}, then subtract.
Problem 3 — Simplifying a mess of powers
Simplify \dfrac{(2x^3y^{-2})^3 \cdot (4x^{-1}y^4)}{8x^2y^{-3}}, leaving no negative exponents.
Solution
Step 1 — deal with the bracket first, using (ab)^n = a^n b^n and (a^m)^n = a^{mn}:
(2x^3y^{-2})^3 = 2^3 \cdot x^{3\times3} \cdot y^{-2\times3} = 8x^9y^{-6}
Step 2 — multiply the top together, adding exponents of like bases (a^m a^n = a^{m+n}):
8x^9y^{-6} \cdot 4x^{-1}y^4 = (8\cdot4)\,x^{9+(-1)}\,y^{-6+4} = 32x^8y^{-2}
Step 3 — divide by the bottom, subtracting exponents (a^m / a^n = a^{m-n}):
\frac{32x^8y^{-2}}{8x^2y^{-3}} = \frac{32}{8}\,x^{8-2}\,y^{-2-(-3)} = 4x^6y^{1}
Watch the y exponent: -2 - (-3) = -2 + 3 = +1. Subtracting a negative is the commonest slip here.
Answer: 4x^6y.
Check with numbers. Put x=1, y=2 into the original: top is (2 \cdot 1 \cdot \tfrac14)^3 \cdot (4 \cdot 1 \cdot 16) = (0.5)^3 \cdot 64 = 8; bottom is 8 \cdot 1 \cdot \tfrac18 = 1; so the original is 8. And 4 \cdot 1 \cdot 2 = 8. ✓ Substituting easy numbers is the fastest way to catch an algebra error.
Problem 4 — Solving for an exponent
A radioactive sample loses 3% of its mass every year. How long until half of it is gone?
Solution
Step 1 — write what happens in one year. Losing 3% means keeping 97%, so the mass is multiplied by 0.97 each year. After t years:
M(t) = M_0 \times 0.97^t
Step 2 — write the question as an equation. Half gone means M(t) = 0.5M_0:
M_0 \times 0.97^t = 0.5 M_0
The starting mass cancels — the half-life does not depend on how much you began with, which is the whole reason half-life is a useful number:
0.97^t = 0.5
Step 3 — the unknown is in the exponent, so take logarithms. Any base works; use the natural log:
\ln(0.97^t) = \ln(0.5)
Step 4 — bring the exponent down with the power law \log(x^p) = p\log x:
t \ln(0.97) = \ln(0.5)
Step 5 — divide.
t = \frac{\ln 0.5}{\ln 0.97} = \frac{-0.693147}{-0.030459} = 22.76
Both logs are negative, since both numbers are below 1, and the two minus signs cancel to give a sensible positive time.
Answer: about 22.8 years.
Cross-check with the rule of 72. Halving at 3% per year should take roughly 72/3 = 24 years. The rule is tuned for growth rather than decay, so being 5% out is expected — it confirms the answer is the right size, which is what a cross-check is for.
Problem 5 — Compound interest, and the cost of paying monthly
You borrow £5,000 at 18% a year. Compare the debt after 3 years if interest is added (a) once a year, (b) every month, (c) continuously.
Solution
The formula is A = P(1 + r/n)^{nt} with P = 5000, r = 0.18, t = 3.
(a) Yearly, n = 1:
A = 5000(1 + 0.18)^{3} = 5000 \times 1.18^3
1.18^2 = 1.3924, and 1.3924 \times 1.18 = 1.643032.
A = 5000 \times 1.643032 = £8215.16
(b) Monthly, n = 12: the rate per month is 0.18/12 = 0.015, and there are 12 \times 3 = 36 months.
A = 5000(1.015)^{36}
1.015^{36} = e^{36\ln 1.015} = e^{36 \times 0.0148886} = e^{0.535989} = 1.70914.
A = 5000 \times 1.70914 = £8545.70
(c) Continuously: use A = Pe^{rt}.
A = 5000\,e^{0.18 \times 3} = 5000\,e^{0.54} = 5000 \times 1.716007 = £8580.04
Answer. £8,215.16 / £8,545.70 / £8,580.04.
What to notice. Two things. First, more frequent compounding always costs the borrower more, because interest starts earning interest sooner. Second, the increase is not unlimited: going from monthly to infinitely often only adds £34, because the sequence closes in on the continuous limit e^{rt}. That ceiling is exactly the limit that defines e.
Problem 6 — Which average is the right one?
A fund grows 40% in year one, then falls 30% in year two. What is its average annual growth rate?
Solution
The wrong answer first, because it is the tempting one. (40\% + (-30\%))/2 = +5\% a year. This is wrong, and here is the proof: £100 growing at 5% for two years becomes £110.25, but the actual fund did this:
100 \xrightarrow{\times 1.40} 140 \xrightarrow{\times 0.70} 98
The fund is down, at £98. An "average" that turns a loss into a gain is not an average of anything useful.
Why the arithmetic mean fails. Growth multiplies. The two years combine as 1.40 \times 0.70, not as 1.40 + 0.70. An average that is added cannot describe a process that multiplies.
The right tool — the geometric mean of the growth factors. We want the single factor g that, applied twice, gives the same result:
g^2 = 1.40 \times 0.70 = 0.98
g = \sqrt{0.98} = 0.98995
Answer: about -1.0\% a year. Check: 100 \times 0.98995^2 = 98.0. ✓
The general formula. For growth factors f_1, f_2, \ldots, f_n over n periods, the average factor is
g = \sqrt[n]{f_1 f_2 \cdots f_n}
and the average percentage rate is (g - 1) \times 100.
What to notice. The gap between +5\% and -1\% is exactly the AM–GM inequality of 1.3 — fractions decimals ratios §5: the arithmetic mean is always at least the geometric mean, so quoting the arithmetic average of returns always flatters the fund. This is why regulators require published fund performance to use the geometric figure.
Problem 7 — Divisibility without dividing
Without doing the division, decide whether 1,234,566 is divisible by 3, by 4, by 9 and by 11.
Solution
By 3 and 9 — add the digits. 1+2+3+4+5+6+6 = 27.
- 27 is divisible by 3, so the number is divisible by 3. ✓
- 27 is also divisible by 9, so the number is divisible by 9. ✓
This works because every power of ten is one more than a string of nines, so the whole number equals (a multiple of 9) plus (the digit sum).
By 4 — look at the last two digits. They are 66. Since 66 = 4 \times 16 + 2, it is not divisible by 4. Only the last two digits matter because 100 is itself a multiple of 4, so everything above the tens column contributes a multiple of 4.
By 11 — alternate the signs from the right.
6 - 6 + 5 - 4 + 3 - 2 + 1 = 3
3 is not a multiple of 11, so the number is not divisible by 11.
Answer: divisible by 3 and by 9; not by 4; not by 11.
Confirm the 9: 1234566 \div 9 = 137174 exactly. ✓
Problem 8 — Euclid's algorithm, and the lcm that comes free
Find \gcd(1071, 462) and then \operatorname{lcm}(1071, 462).
Solution
Step 1 — run the algorithm. At each line, replace the pair (a, b) with (b,\ a \bmod b), which never changes the set of common divisors.
1071 = 2 \times 462 + 147 \quad \Rightarrow \quad \gcd(1071,462) = \gcd(462,147)
462 = 3 \times 147 + 21 \quad \Rightarrow \quad \gcd(462,147) = \gcd(147,21)
147 = 7 \times 21 + 0 \quad \Rightarrow \quad \gcd(147,21) = \gcd(21,0) = 21
A remainder of zero means the smaller number divides the larger exactly, and that number is the answer.
Step 1 result: \gcd = 21. Three steps, no factorising — this is why the algorithm has survived since about 300 BC and still runs inside every public-key handshake.
Step 2 — use the identity \gcd \times \operatorname{lcm} = a \times b:
\operatorname{lcm}(1071, 462) = \frac{1071 \times 462}{21}
Divide before multiplying to keep the numbers small: 1071 / 21 = 51, so
\operatorname{lcm} = 51 \times 462 = 23{,}562
Answer: gcd = 21, lcm = 23,562.
Verify by factorising (slower, which is the point): 1071 = 3 \times 357 = 3 \times 3 \times 119 = 3^2 \times 7 \times 17, and 462 = 2 \times 3 \times 7 \times 11. Shared: one 3 and one 7, giving \gcd = 21 ✓. All primes at their highest power: 2 \times 3^2 \times 7 \times 11 \times 17 = 23{,}562 ✓.
Problem 9 — Counting divisors, and the odd fact about squares
How many divisors does 720 have? And prove that a number has an odd number of divisors exactly when it is a perfect square.
Solution
Step 1 — factorise 720. Halve repeatedly: 720 = 2 \times 360 = 4 \times 180 = 8 \times 90 = 16 \times 45. And 45 = 9 \times 5. So
720 = 2^4 \times 3^2 \times 5^1
Step 2 — apply the divisor-count formula, which multiplies one-more-than-each-exponent:
d(720) = (4+1)(2+1)(1+1) = 5 \times 3 \times 2 = 30
Answer: 30 divisors. The formula works because building a divisor means choosing between 0 and 4 twos, between 0 and 2 threes, and between 0 and 1 fives, independently.
Step 3 — the perfect-square claim. Divisors come in pairs: if d divides n then so does n/d, and the two multiply to n. So you can walk through every divisor pairing it with its partner. That accounts for all of them two at a time, giving an even count — unless some divisor is its own partner. That happens exactly when
d = \frac{n}{d} \quad \Rightarrow \quad d^2 = n
which is possible only if n is a perfect square, and then only for the single value d = \sqrt{n}. So exactly one divisor is unpaired, exactly when n is a square, and the count is odd exactly in that case.
Check. 36 = 2^2 3^2 gives (2+1)(2+1) = 9 divisors: 1, 2, 3, 4, 6, 9, 12, 18, 36. Nine of them, and 6 is the lonely one pairing with itself. ✓
Where you have met this. It is the locker problem: 100 lockers, 100 students, student k flips every k-th locker. A locker ends open if it was flipped an odd number of times, which happens exactly for the perfect squares — so lockers 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 stay open.
Problem 10 — A huge power, modulo a small number
Find the last digit of 7^{2024}.
Solution
The last digit of a number is its remainder when divided by 10, so the question is really: what is 7^{2024} \bmod 10?
Step 1 — compute the first few powers and look for the cycle. The rule that makes this legal is ac \equiv bd \pmod m: you may reduce at every step instead of carrying giant numbers.
7^1 \equiv 7, \quad 7^2 = 49 \equiv 9, \quad 7^3 \equiv 9 \times 7 = 63 \equiv 3, \quad 7^4 \equiv 3 \times 7 = 21 \equiv 1 \pmod{10}
Step 2 — notice the cycle closes. Because 7^4 \equiv 1, multiplying by another four sevens returns you to where you were. The last digits repeat with period 4: 7, 9, 3, 1, 7, 9, 3, 1, \ldots
Step 3 — find where 2024 lands in the cycle.
2024 = 4 \times 506 + 0
so 2024 is an exact multiple of 4. Then
7^{2024} = \left(7^{4}\right)^{506} \equiv 1^{506} = 1 \pmod{10}
Answer: the last digit is 1.
What to notice. The period had to divide 4, and this is not luck: \phi(10) = 4 (the numbers 1, 3, 7, 9 are the ones sharing no factor with 10), so Euler's theorem guarantees 7^{4} \equiv 1 \pmod{10} before you calculate anything. The same method finds the last digit of any power: find the cycle length, then reduce the exponent modulo that length.
Problem 11 — Fermat's little theorem doing real work
Compute 3^{100} \bmod 7.
Solution
Step 1 — check the theorem applies. 7 is prime and 3 is not a multiple of 7, so Fermat's little theorem gives
3^{6} \equiv 1 \pmod 7
Step 2 — split the exponent to expose that block. Divide 100 by 6:
100 = 6 \times 16 + 4
Step 3 — rewrite the power using a^{m+n} = a^m a^n and (a^m)^n = a^{mn}:
3^{100} = 3^{6\times16 + 4} = \left(3^{6}\right)^{16} \times 3^{4}
Step 4 — replace the whole first factor with 1, which is the entire payoff of the theorem:
3^{100} \equiv 1^{16} \times 3^4 = 3^4 \pmod 7
Step 5 — finish by hand. 3^4 = 81, and 81 = 7 \times 11 + 4.
Answer: 3^{100} \equiv 4 \pmod 7.
What to notice. A number with 48 digits was reduced to 3^4 by one theorem and one division. This is precisely the trick that makes RSA decryption fast enough to run on a phone: the same reduction, with \phi(n) in place of p-1, is what turns an impossible exponentiation into a few dozen multiplications.
Problem 12 — Scientific notation and the size of things
A water molecule has a mass of about 3.0 \times 10^{-26} kg. How many are in a 250 ml glass of water, given that 250 ml has a mass of 0.25 kg?
Solution
Step 1 — set up the division. Number of molecules = total mass ÷ mass of one molecule:
N = \frac{0.25}{3.0 \times 10^{-26}}
Step 2 — put the top into scientific notation too, so both sides are in the same form:
0.25 = 2.5 \times 10^{-1}
Step 3 — divide the fronts and subtract the exponents, using \frac{a^m}{a^n} = a^{m-n}:
N = \frac{2.5}{3.0} \times 10^{-1 - (-26)} = 0.8333 \times 10^{25}
Step 4 — tidy into standard form (one digit before the point):
N = 8.3 \times 10^{24}
Answer: about 8.3 \times 10^{24} molecules — over eight million billion billion.
Sanity check against a known constant. A mole of water is 18 g and contains Avogadro's number, 6.02 \times 10^{23}, molecules. 250 g is 250/18 = 13.9 moles, giving 13.9 \times 6.02 \times 10^{23} = 8.4 \times 10^{24}. ✓ The two routes agree, which is the strongest kind of check: two independent methods landing on the same number.
Problem 13 — A percentage trap worth meeting once
A shop raises a price by 20%, then advertises "20% off". Is the customer back where they started? And what single percentage change would undo a 20% rise?
Solution
Step 1 — track a concrete price. Start at £100.
100 \xrightarrow{+20\%} 100 \times 1.20 = 120 \xrightarrow{-20\%} 120 \times 0.80 = 96
Answer to the first question: no — the customer pays £96, which is 4% below the original.
Why. The two percentages are taken of different bases. The rise is 20% of 100, which is £20. The fall is 20% of 120, which is £24. Bigger base, bigger cut.
Step 2 — algebra, so it is clear this is not about the number 100. Any starting price P ends at
P \times 1.2 \times 0.8 = 0.96P
The 4% loss is universal, and the two factors multiplying to 0.96 rather than 1 is the whole story. In general (1+x)(1-x) = 1 - x^2, so the shortfall is always x^2 — for a 50% rise and 50% fall it would be a brutal 25%.
Step 3 — what does undo a 20% rise? Find f with
1.2 \times f = 1 \quad \Rightarrow \quad f = \frac{1}{1.2} = 0.8333
Answer: a 16.67% cut. Check: 120 \times 0.8333 = 100 ✓.
What to notice. Percentage rises and falls do not cancel; multiplicative factors do. Whenever you are chaining percentage changes, convert each to a factor first and multiply the factors. This is the same insight as Problem 6, and it is the single most useful habit in everyday numeracy.
Problem 14 — Proving something is irrational
Prove that \sqrt{3} is irrational.
Solution
This is Euclid's argument from 1.7, run on 3 instead of 2. The method is proof by contradiction: assume the opposite, follow it faithfully, and reach something impossible.
Step 1 — assume the opposite. Suppose \sqrt3 is rational. Then it can be written as a fraction
\sqrt3 = \frac{a}{b}
where a and b are whole numbers with no common factor — because any fraction can be reduced to lowest terms first, so we may as well assume it already is.
Step 2 — square both sides.
3 = \frac{a^2}{b^2} \quad \Rightarrow \quad a^2 = 3b^2
Step 3 — conclude that 3 divides a. The right-hand side is a multiple of 3, so a^2 is a multiple of 3. Now use the fact that 3 is prime: if a prime divides a product, it divides one of the factors. Since a^2 = a \times a, the prime 3 must divide a.
Step 4 — write that down and substitute. If 3 divides a, then a = 3k for some whole number k. Put it back:
(3k)^2 = 3b^2 \quad \Rightarrow \quad 9k^2 = 3b^2 \quad \Rightarrow \quad 3k^2 = b^2
Step 5 — the same argument now applies to b. The right-hand side b^2 is 3 times something, so 3 divides b^2, so 3 divides b.
Step 6 — the contradiction. We have shown 3 divides a and 3 divides b. But Step 1 said the fraction was in lowest terms, with no common factor. Both cannot be true.
The assumption was the only thing we chose, so the assumption is what must be wrong.
Conclusion: \sqrt3 cannot be written as a fraction. It is irrational. \blacksquare
What to notice. Step 3 is where the primeness is essential, and it is why this proof does not "show" that \sqrt4 is irrational: 4 is not prime, so "4 divides a^2" does not force "4 divides a" — take a = 2. The proof works for the square root of any prime, and with more care for any non-square whole number.
Problem 15 — Putting several tools together
A bacterial culture doubles every 20 minutes. Starting from 500 cells at 9:00 am, how many are there at 1:30 pm, and at what time does it first pass one million?
Solution
Step 1 — build the growth model. The population multiplies by 2 every 20 minutes. If t is measured in minutes, the number of doublings by time t is t/20, so
N(t) = 500 \times 2^{t/20}
The exponent is t/20 rather than t because one doubling costs 20 minutes; get this wrong and every later answer is wrong.
Step 2 — first question. From 9:00 to 13:30 is 4.5 hours, which is 270 minutes.
N(270) = 500 \times 2^{270/20} = 500 \times 2^{13.5}
Split the exponent: 2^{13.5} = 2^{13} \times 2^{0.5} = 8192 \times 1.41421 = 11{,}585.2.
N = 500 \times 11{,}585.2 = 5{,}792{,}619
Answer to the first question: about 5.79 million cells.
Step 3 — second question, where the unknown is in the exponent. Set N(t) = 10^6:
500 \times 2^{t/20} = 1{,}000{,}000
Divide both sides by 500:
2^{t/20} = 2000
Step 4 — take logarithms and bring the exponent down.
\frac{t}{20}\ln 2 = \ln 2000
\frac{t}{20} = \frac{\ln 2000}{\ln 2} = \frac{7.6009}{0.6931} = 10.966
That number is worth reading: it says the culture needs just under 11 doublings, which makes sense, since 11 doublings multiply by 2^{11} = 2048 and we need a factor of 2000.
Step 5 — solve for t.
t = 20 \times 10.966 = 219.3 \text{ minutes} = 3 \text{ h } 39 \text{ min}
Answer: it passes one million at about 12:39 pm.
What to notice. Two questions, one model. Going forwards (given time, find size) needs only powers; going backwards (given size, find time) needs logarithms. That is the practical reason logarithms exist, and it is why they appear the moment any real problem asks "how long until…".
Next: Part 2 — Algebra, where letters replace numbers and the identities of 2.2 — polynomials — including the cube factorisations most people never see derived — take over the work.