Worked Examples: GRE Arithmetic

Example 1: Remainders (Medium)

When the positive integer $n$ is divided by 5, the remainder is 3. What is the remainder when $4n + 1$ is divided by 5?

Solution. Work in modular arithmetic. $n \equiv 3 \pmod 5$, so $4n + 1 \equiv 4(3) + 1 = 13 \equiv 3 \pmod 5.$ The remainder is 3. No value of $n$ was needed — only its remainder.

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Example 2: Successive Percent Change (Medium)

A laptop's price is increased by 20% and later decreased by 15%. The final price is what percent of the original price?

Solution. Multiply the factors rather than adding the percents: $1.20 \times 0.85 = 1.02.$ The final price is 102% of the original — a net 2% increase, not the 5% a naive subtraction (20 − 15) suggests.

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Example 3: Ratio Word Problem (Medium–Hard)

In a parking lot the ratio of cars to trucks is 5 : 2. If 12 more trucks arrive, the ratio becomes 5 : 4. How many cars are in the lot?

Solution. Let cars $= 5k$ and trucks $= 2k$. Adding 12 trucks: $\frac{5k}{2k + 12} = \frac{5}{4}.$ Cross-multiply: $20k = 5(2k + 12) = 10k + 60$, so $10k = 60$ and $k = 6$. Cars $= 5k = \mathbf{30}$.

Check: trucks start at $2(6) = 12$; after 12 more, $24$; ratio $30 : 24 = 5 : 4$. ✓

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Example 4: Exponents with a Common Base (Hard)

Quantity A: $2^{40}$ Quantity B: $4^{20} + 4^{19}$

Solution. Rewrite Quantity B with base 2: $4^{20} = (2^2)^{20} = 2^{40}$. Therefore $\text{Quantity B} = 2^{40} + 4^{19} > 2^{40} = \text{Quantity A}.$ Quantity B exceeds Quantity A by the positive amount $4^{19}$, so the answer is (B). The trap answer is (C) — students who see $4^{20} = 2^{40}$ but overlook the extra $+\,4^{19}$ term wrongly call them equal.

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Example 5: Simplifying Radicals (Medium)

Simplify $\sqrt{50} + \sqrt{18}$.

Solution. Factor out perfect squares: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}, \qquad \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}.$ $5\sqrt{2} + 3\sqrt{2} = 8\sqrt{2}.$ The trap is $\sqrt{50 + 18} = \sqrt{68}$ — you may never add radicands. Simplify each radical, then combine like terms.

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Example 6: LCM / Divisibility (Hard)

$x$ is a positive integer, and both $\dfrac{x}{6}$ and $\dfrac{x}{9}$ are integers. Which of the following must be a factor of $x$? (A) 12 (B) 18 (C) 27 (D) 54

Solution. If 6 divides $x$ and 9 divides $x$, then $x$ is a multiple of $\text{LCM}(6, 9)$. Factor: $6 = 2 \cdot 3$, $9 = 3^2$, so $\text{LCM} = 2 \cdot 3^2 = 18$. Every valid $x$ is a multiple of 18, so 18 must be a factor. The smallest such $x$ is 18 itself, which is not divisible by 12, 27, or 54 — those need not be factors. Answer: (B).