iGCSE Mathematics (0580) - C1.1 Types of number- Exam Style Questions Paper 1- New Syllabus
▶️ Answer/Explanation
(a)
A prime number has exactly two factors: 1 and itself.
Checking between 20 and 30: \(21 = 3 \times 7\) (not prime), \(22\) (even), \(23\) — no factors other than 1 and 23 (prime ✓), \(24,25,26,27,28\) (not prime), \(29\) — prime ✓.
✅ Answer: \(23\) or \(29\)
(b)
\(\sqrt{9} = 3\), then \(3^2 = 9\). The square root and squaring operations cancel each other for non-negative numbers.
\[\left(\sqrt{9}\right)^2 = 9\]
✅ Answer: \(9\)
▶️ Answer/Explanation
Let’s scan our list to find matching numbers for each condition:
(a) An even number must end in an even digit; $112$ is the only even number here.
(b) A square number is the result of multiplying an integer by itself. Out of these, $49$ fits because $7 \times 7 = 49$.
(c) A factor of $56$ divides $56$ perfectly with no remainder. Testing our numbers, $56 \div 7 = 8$, so $7$ is a factor.
(d) A prime number has only two distinct factors: $1$ and itself. $7$ is prime, while numbers like $27$ ($3\times9$) or $39$ ($3\times13$) are composite.
✅ Answer:
(a) $112$
(b) $49$
(c) $7$
(d) $7$
(a) An even number must end in an even digit; $112$ is the only even number here.
(b) A square number is the result of multiplying an integer by itself. Out of these, $49$ fits because $7 \times 7 = 49$.
(c) A factor of $56$ divides $56$ perfectly with no remainder. Testing our numbers, $56 \div 7 = 8$, so $7$ is a factor.
(d) A prime number has only two distinct factors: $1$ and itself. $7$ is prime, while numbers like $27$ ($3\times9$) or $39$ ($3\times13$) are composite.
✅ Answer:
(a) $112$
(b) $49$
(c) $7$
(d) $7$
