Practice Set 1 | Real Numbers

Problem 1
LCM★★★★☆
For prime numbers p and q, determine the LCM of A = $p^{5}q^{6}$ and B = $p^{6}q^{3}$.
Answer
LCM(A, B) = $p^{6}q^{6}$.
Solution
1. For prime-power forms, the LCM takes the larger exponent of each prime that appears.
2. For p, the larger exponent is $\max(5, 6)$ = 6.
3. For q, the larger exponent is $\max(6, 3)$ = 6.
4. So LCM(A, B) = $p^{6}q^{6}$.
Problem 2
Squares and Cubes★★★☆☆
Show that the square of any positive integer cannot be of the form $3q + 2$.
Answer
The possible forms are $3q$ or $3q + 1$.
Solution
1. When any integer is divided by 3, the possible remainders are 0, 1, ..., 2.
2. Squaring numbers with these remainders gives possible remainders 0, 1.
3. So a square cannot leave remainder 2 when divided by 3.
Problem 3
LCM★★★★☆
Determine the smallest positive integer divisible by every integer from $1$ to $9$.
Answer
The required least number is $2520$.
Solution
1. The required number is the LCM of the integers in the given range.
2. Take the highest prime powers needed in that range: $2^{3} \times 3^{2} \times 5 \times 7$.
3. So the LCM is $2520$.
Problem 4
Prime Factorization★★☆☆☆
Find the prime factorization of $147$.
Answer
$147$ = $3 \times 7^{2}$.
Solution
1. Divide by prime numbers until only prime factors remain.
2. The collected prime powers are $3 \times 7^{2}$.
3. Thus $147$ = $3 \times 7^{2}$.
Problem 5
Prime Factorization★★★★☆
Show that $12^n$ cannot have last digit 0 or 5 for any natural number n.
Answer
It cannot end with 0 or 5.
Solution
1. The prime factors of $12$ are 2, 3; no factor 5 appears.
2. Every power $12^n$ therefore has no factor 5.
3. But any number ending in 0 or 5 is divisible by 5.
4. So $12^n$ cannot end with 0 or 5.
Problem 6
HCF★★★☆☆
For the numbers $440$, $380$, the common divisors are $1$, $2$, $4$, $5$, $10$, $20$. Which one is the HCF?
Answer
The HCF is $20$.
Solution
1. The HCF is the largest value in the list of common divisors.
2. The largest listed common divisor is $20$.
3. Hence the HCF is $20$.
Problem 7
Decimal Expansion★☆☆☆☆
When $98.6$ is expressed as $p/q$ with no common factor, what primes can divide q?
Answer
In lowest terms, q can have only the prime factors 2 and 5.
Solution
1. Convert the decimal to a fraction: $98.6$ = $\frac{986}{10}$.
2. After reducing, $\frac{986}{10}$ = $\frac{493}{5}$.
3. The denominator in lowest terms factors as $5$.
4. So no primes other than 2 and 5 can divide q.
Problem 8
Divisibility and Remainders★★★☆☆
Establish that exactly one of n, n + 2, n + 4 is divisible by $3$.
Answer
Exactly one of these terms is divisible by $3$.
Solution
1. Since 2 and 3 have no common factor except 1, adding 2 each time gives every possible remainder when divided by 3.
2. The 3 listed terms therefore take all remainders 0, 1, ..., 2 in some order.
3. Hence exactly one term has remainder 0.
Problem 9
Divisibility and Remainders★★★☆☆
Demonstrate that among any 3 consecutive positive integers, one is divisible by $3$.
Answer
Exactly one of the 3 consecutive integers is divisible by $3$.
Solution
1. When consecutive integers are divided by 3, their remainders are consecutive.
2. Across 3 consecutive integers, the remainders are 0, 1, ..., 2 in some order.
3. Exactly one remainder is 0, so one term is divisible.
Problem 10
HCF★★★★☆
Let A = $x^{4}y$ and B = $x^{6}y^{2}$, where x and y are primes. Find HCF(A, B).
Answer
HCF(A, B) = $x^{4}y$.
Solution
1. For prime-power forms, the HCF takes the smaller exponent of each common prime.
2. For x, the smaller exponent is $\min(4, 6)$ = 4.
3. For y, the smaller exponent is $\min(1, 2)$ = 1.
4. So HCF(A, B) = $x^{4}y$.
Problem 11
Euclid's Division Lemma★★★★☆
If division by $4$ leaves remainder $2$, express the integer using q.
Answer
It is of the form $4q + 2$, where q is an integer.
Solution
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 $4q + 2$.
Problem 12
Euclid's Division Algorithm★★★★☆
Find the largest possible divisor for $38$, $86$, $152$ when the respective remainders are $3$, $2$, $5$.
Answer
The greatest possible divisor is $7$.
Solution
1. Remove each stated remainder from its number; the required divisor must divide $35$, $84$, $147$.
2. So the required divisor is the HCF of $35$, $84$, $147$.
3. That HCF is $7$.
Problem 13
Euclid's Division Lemma★★★★☆
For a = $59$ and b = $6$, write a in the form $a = bq + r$ and give the condition on r.
Answer
$59 = 6 \times 9 + 5$, so q = $9$ and r = $5$.
Solution
1. Euclid's division lemma gives unique integers q and r in a = bq + r.
2. Here the quotient is 9 and the remainder is 5.
3. The remainder satisfies $0 \le r < b$.
Problem 14
Decimal Expansion★★★★☆
State whether the decimal expansion of $\frac{135}{500}$ terminates, without doing long division.
Answer
It has a terminating decimal expansion.
Solution
1. Put the fraction in lowest terms: $\frac{135}{500}$ = $\frac{27}{100}$.
2. Its denominator in lowest terms is $100$ = $2^{2} \times 5^{2}$.
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.
Problem 15
HCF★☆☆☆☆
If $54m - 210$ represents the HCF of $54$ and $210$, find m.
Answer
The value of m is $4$.
Solution
1. The HCF of $54$ and $210$ is $6$.
2. Match the expression to the HCF: $54m - 210 = 6$.
3. Rearranging gives $54m = 216$.
4. Thus m = $4$.