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Neokaal Study

Real Numbers

Competency Practice

Prime power ending digits, repeat-interval LCM applications, and rigorous irrationality proofs.

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  1. Prove that $36^n$ cannot have last digit 0 or 5 for any natural number n.

  2. For step lengths $30$, $42$, $50$ cm, determine the least distance that is an exact number of steps for all three walkers.

  3. Prove that $\sqrt{2}$ is irrational.

  4. Explain why $\frac{1}{5} - \sqrt{5}$ cannot be rational.

  5. Use factorization to show that $13 \times 2 \times 3 + 3$ is composite.

  6. Use prime factorisation to find the HCF and LCM of $98$, $245$, $6$.

  7. The HCF of $65$ and $40$ is $5$. Use the relation between HCF, LCM, and the two numbers to find their LCM.

  8. Break $245$ into its prime factors.

Neokaal Study

Answer Key

  1. It cannot end with 0 or 5.

    1. The prime factors of $36$ are 2, 3; no factor 5 appears.
    2. Every power $36^n$ therefore has no factor 5.
    3. But any number ending in 0 or 5 is divisible by 5.
    4. So $36^n$ cannot end with 0 or 5.

  2. The shortest common distance is $1050$ cm.

    1. The distance must be a common multiple of the three step lengths.
    2. The shortest possible distance is therefore their LCM.
    3. LCM(30, 42, 50) = 1050.

  3. $\sqrt{2}$ is irrational.

    1. Assume $\sqrt{2}$ = $\frac{a}{b}$ for coprime positive integers $a$ and $b$.
    2. Squaring gives $a^2 = 2b^2$, so 2 divides $a^2$. By unique prime factorisation, 2 divides $a$.
    3. Write $a = 2c$. Substitution gives $b^2 = 2c^2$, so 2 also divides $b$.
    4. This contradicts that $a$ and $b$ are coprime. Therefore the square root is irrational.

  4. It is irrational by the operation rule for rational and irrational numbers.

    1. The number $\frac{1}{5}$ is rational and $\sqrt{5}$ is irrational.
    2. If $\frac{1}{5} - \sqrt{5}$ were rational, subtracting it from $\frac{1}{5}$ would make $\sqrt{5}$ rational.
    3. That is impossible; hence the result is irrational.

  5. It is composite because it equals $3 \times 27$.

    1. Factor out the common factor $3$.
    2. This gives $13 \times 2 \times 3 + 3$ = $3(13 \times 2 + 1)$.
    3. Both factors $3$ and $27$ are greater than 1.
    4. So the number has a non-trivial factorization and is composite.

  6. HCF = $1$ and LCM = $1470$.

    1. Write each number as a product of prime powers: $98$ = $2 \times 7^{2}$; $245$ = $5 \times 7^{2}$; $6$ = $2 \times 3$.
    2. Take the smallest common exponents for the HCF: $1$ = $1$.
    3. Take the greatest exponents present for the LCM: $2 \times 3 \times 5 \times 7^{2}$ = $1470$.

  7. The LCM is $520$.

    1. For two positive integers, HCF x LCM equals the product of the integers.
    2. So LCM = ($65$ x $40$) / $5$.
    3. Therefore the LCM is $520$.

  8. $245$ = $5 \times 7^{2}$.

    1. Divide by prime numbers until only prime factors remain.
    2. The collected prime powers are $5 \times 7^{2}$.
    3. Thus $245$ = $5 \times 7^{2}$.