Revision Set

The HCF is $9$.

1. Start with $216$ and $207$: $216 = 207 \times 1 + 9$; $207 = 9 \times 23 + 0$.
2. This gives HCF(216, 207) = $9$.
3. Now combine this HCF with $126$: $126 = 9 \times 14 + 0$.
4. Hence the HCF of all three numbers is $9$.

The possible forms are $4q$ or $4q + 1$.

1. When any integer is divided by 4, the possible remainders are 0, 1, ..., 3.
2. Squaring numbers with these remainders gives possible remainders 0, 1.
3. So a square cannot leave remainder 2, 3 when divided by 4.

No. The possible remainders are 0, 1, 2, 3, 4.

1. Euclid's division lemma lets us write a positive integer a as $a = bq + r$.
2. For division by 5, the remainder must satisfy 0 <= r < 5.
3. Thus r can be 0, 1, 2, 3, 4, so the proposed list is incomplete.

The least such number is $60$.

1. The smallest number divisible by every term in the range is their LCM.
2. The needed prime powers are $2^{2} \times 3 \times 5$.
3. Hence the required number is $60$.

It has a non-terminating repeating decimal expansion.

1. Put the fraction in lowest terms: $\frac{276}{440}$ = $\frac{69}{110}$.
2. Its denominator in lowest terms is $110$ = $2 \times 5 \times 11$.
3. A rational number terminates in decimal form exactly when the lowest-term denominator uses only the primes 2 and 5.
4. The denominator has a prime factor other than 2 or 5, so the decimal is non-terminating repeating.

The possible forms are $6q$, $6q + 1$, $6q + 2$, $6q + 3$, $6q + 4$, or $6q + 5$.

1. When any integer is divided by 6, the possible remainders are 0, 1, ..., 5.
2. Cubing numbers with these remainders gives possible remainders 0, 1, 2, 3, 4, 5.
3. Hence every cube has one of those remainders when divided by 6.

The shortest common distance is $2400$ 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, 32, 50) = 2400.

$700$ = $2^{2} \times 5^{2} \times 7$.

1. Divide by prime numbers until only prime factors remain.
2. The collected prime powers are $2^{2} \times 5^{2} \times 7$.
3. Thus $700$ = $2^{2} \times 5^{2} \times 7$.

The product is divisible by $6$.

1. Among three consecutive integers, one is divisible by 3 and at least one is even.
2. The product therefore contains the prime factors required for 6.
3. Hence $n(n + 1)(n + 2)$ is divisible by $6$.

$26 = 2 \times 13 + 0$, so q = $13$ and r = $0$.

1. Euclid's division lemma gives unique integers q and r in a = bq + r.
2. Here the quotient is 13 and the remainder is 0.
3. The remainder satisfies $0 \le r < b$.

It is irrational.

1. The number $\frac{3}{5}$ is rational and $\sqrt{3}$ is irrational.
2. If $\frac{3}{5} + \sqrt{3}$ were rational, subtracting the rational number $\frac{3}{5}$ would make $\sqrt{3}$ rational.
3. This contradicts the irrationality of the square root of a prime.

Yes, such a pair can exist.

1. For any two positive integers, the HCF must be a divisor of the LCM.
2. One example is $154$ and $182$, whose HCF is $14$ and LCM is $2002$.

$\sqrt{5} + \sqrt{13}$ is irrational.

1. Suppose, for contradiction, that $\sqrt{5} + \sqrt{13} = r$, where r is a positive rational number.
2. Then $\sqrt{5} = r - \sqrt{13}$.
3. Squaring both sides gives $5 = r^2 + 13 - 2r\sqrt{13}$.
4. So $\sqrt{13} = \frac{r^2 + 13 - 5}{2r}$, which would make $\sqrt{13}$ rational.
5. Since 13 is prime, $\sqrt{13}$ is irrational. The contradiction proves the claim.

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.

Exactly one of these terms is divisible by $5$.

1. Since 4 and 5 have no common factor except 1, adding 4 each time gives every possible remainder when divided by 5.
2. The 5 listed terms therefore take all remainders 0, 1, ..., 4 in some order.
3. Hence exactly one term has remainder 0.

It is of the form $4q + 2$, where q is an integer.

1. Euclid's division lemma writes any positive integer n as $n = bq + r$ with $0 \le r < b$.
2. Here b = 4 and r = 2, so n has the form $4q + 2$.

$n^3 - n$ is divisible by $6$.

1. Factor the expression: $n^3 - n = n(n - 1)(n + 1)$.
2. These are three consecutive integers.
3. One of three consecutive integers is divisible by 3, and one is even.
4. So their product is divisible by 6.

HCF(A, B) = $p^{3}q^{2}$.

1. For prime-power forms, the HCF takes the smaller exponent of each common prime.
2. For p, the smaller exponent is $\min(3, 3)$ = 3.
3. For q, the smaller exponent is $\min(3, 2)$ = 2.
4. So HCF(A, B) = $p^{3}q^{2}$.

In lowest terms, q can have only the prime factors 2 and 5.

1. Convert the decimal to a fraction: $148.51$ = $\frac{14851}{100}$.
2. After reducing, $\frac{14851}{100}$ = $\frac{14851}{100}$.
3. The denominator in lowest terms factors as $2^{2} \times 5^{2}$.
4. So no primes other than 2 and 5 can divide q.

The greatest possible divisor is $15$.

1. Remove each stated remainder from its number; the required divisor must divide $375$, $510$, $120$.
2. So the required divisor is the HCF of $375$, $510$, $120$.
3. That HCF is $15$.