Practice Set 2

It cannot end with 0 or 5.

1. The prime factors of $18$ are 2, 3; no factor 5 appears.
2. Every power $18^n$ therefore has no factor 5.
3. But any number ending in 0 or 5 is divisible by 5.
4. So $18^n$ cannot end with 0 or 5.

It has a terminating decimal expansion.

1. Put the fraction in lowest terms: $\frac{136}{160}$ = $\frac{17}{20}$.
2. Its denominator in lowest terms is $20$ = $2^{2} \times 5$.
3. A rational number terminates in decimal form exactly when the lowest-term denominator uses only the primes 2 and 5.
4. Only the primes 2 and 5 appear in the denominator, so the decimal terminates.

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

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

It is composite because it equals $5 \times 23$.

1. Factor out the common factor $5$.
2. This gives $11 \times 2 \times 5 + 5$ = $5(11 \times 2 + 1)$.
3. Both factors $5$ and $23$ are greater than 1.
4. So the number has a non-trivial factorization and is composite.

HCF(A, B) = $x^{3}y$.

1. For prime-power forms, the HCF takes the smaller exponent of each common prime.
2. For x, the smaller exponent is $\min(3, 6)$ = 3.
3. For y, the smaller exponent is $\min(1, 3)$ = 1.
4. So HCF(A, B) = $x^{3}y$.

The greatest possible divisor is $9$.

1. Remove each stated remainder from its number; the required divisor must divide $234$, $117$, $333$.
2. So the required divisor is the HCF of $234$, $117$, $333$.
3. That HCF is $9$.

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

1. When any integer is divided by 5, the possible remainders are 0, 1, ..., 4.
2. Squaring numbers with these remainders gives possible remainders 0, 1, 4.
3. So a square cannot leave remainder 2, 3 when divided by 5.

The value of m is $6$.

1. The HCF of $36$ and $210$ is $6$.
2. Match the expression to the HCF: $36m - 210 = 6$.
3. Rearranging gives $36m = 216$.
4. Thus m = $6$.

It is of the form $3q + 1$, 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 = 3 and r = 1, so n has the form $3q + 1$.

The HCF is $8$.

1. Start with $152$ and $104$: $152 = 104 \times 1 + 48$; $104 = 48 \times 2 + 8$; $48 = 8 \times 6 + 0$.
2. This gives HCF(152, 104) = $8$.
3. Now combine this HCF with $72$: $72 = 8 \times 9 + 0$.
4. Hence the HCF of all three numbers is $8$.

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 $224$ and $182$, whose HCF is $14$ and LCM is $2912$.

The shortest common distance is $5040$ 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(42, 45, 48) = 5040.

The decimal expansion ends after 3 decimal place(s).

1. Reduce the fraction first: $\frac{246}{500}$ = $\frac{123}{250}$.
2. In lowest terms, the denominator is $250$ = $2 \times 5^{3}$.
3. A denominator of the form $2^{m} \times 5^{n}$ gives a terminating decimal with $\max(m, n)$ decimal places.
4. Here $\max(m, n)$ = 3; hence the expansion ends after 3 decimal place(s).

The least such number is $2520$.

1. The smallest number divisible by every term in the range is their LCM.
2. The needed prime powers are $2^{3} \times 3^{2} \times 5 \times 7$.
3. Hence the required number is $2520$.

$a^2 + b^2$ is even and leaves remainder $2$ when divided by $4$.

1. Let an odd integer be 2k + 1.
2. Then $(2k + 1)^2 = 4k^2 + 4k + 1$ $= 4k(k + 1) + 1$.
3. So an odd square leaves remainder 1 when divided by 4.
4. So $a^2$ and $b^2$ each leave remainder 1 when divided by 4.
5. Adding gives $a^2 + b^2$, which leaves remainder 2 when divided by 4.
6. A number that leaves remainder 2 when divided by 4 is even but not divisible by 4.

Exactly one of the 4 consecutive integers is divisible by $4$.

1. When consecutive integers are divided by 4, their remainders are consecutive.
2. Across 4 consecutive integers, the remainders are 0, 1, ..., 3 in some order.
3. Exactly one remainder is 0, so one term is divisible.