Number System Questions and Answers

Answer: A

Such a prime can be expressed as a sum of squares of two natural numbers.

This is Fermat's theorem on sums of two squares: an odd prime p can be written as \(a^{2} + b^{2}\) exactly when \(p \equiv 1 \pmod 4\). For example \(5 = 1^{2} + 2^{2}\), \(13 = 2^{2} + 3^{2}\), \(17 = 1^{2} + 4^{2}\), \(29 = 2^{2} + 5^{2}\).

Sums of cubes fail at once: 5 and 13 are not sums of two positive cubes (1, 8, 27 ...). Sums of square roots or cube roots of natural numbers are generally irrational and have no link with the remainder on division by 4.

Answer: D

The correct statement is that 1 is neither prime nor a composite number.

A prime number has exactly two distinct factors, 1 and itself, and a composite number has more than two factors. The number 1 has only one factor, itself, so it fits neither definition.

Treating 1 as prime would also break the uniqueness of prime factorisation, since 6 could then be written as 2 × 3 or 1 × 2 × 3. The smallest prime is 2 and the smallest composite is 4.

Answer: B

The statement that holds for all primes is that \(p_{1}p_{2}\) is not a prime number.

A prime has exactly two divisors, 1 and itself. The product \(p_{1}p_{2}\) of two primes, both at least 2, is divisible by \(p_{1}\), which is neither 1 nor the whole product. So \(p_{1}p_{2}\) is always composite, even when \(p_{1} = p_{2}\).

The others fail for some primes. \(p_{1} + p_{2}\) can be prime: 2 + 3 = 5. \(p_{1} + p_{2} + 1\) can be composite: 3 + 5 + 1 = 9. \(p_{1}p_{2} + 1\) can be composite: 3 × 5 + 1 = 16.

Answer: D

The correct statement is that a is odd and b is even.

If a were even, both a and \(a^{2}b^{3}\) would be even, and their sum would be even. So a must be odd. Then \(a + a^{2}b^{3}\) is odd only if \(a^{2}b^{3}\) is even, and since \(a^{2}\) is odd, \(b^{3}\), and so b, must be even.

Check: a = 1, b = 2 gives 1 + 8 = 9, odd. With both odd, the sum is odd + odd = even; with both even it is even; and with a even the sum is even whatever b is.

Answer: C

The smallest such number is 15873.

The product must be a number made only of 8s, so it is 8 times a repunit (111...1). Since \(56 = 8 \times 7\), we need the smallest repunit divisible by 7, and the number asked for is that repunit divided by 7.

\[ \begin{aligned} 111111 &= 7 \times 15873 \\ 15873 \times 56 &= 888888 \end{aligned} \]

No shorter repunit (1, 11, 111, 1111, 11111) is divisible by 7, so 888888 is the first all-8 product and 15873 the smallest multiplier. A quick check of the others: 15093, 15738 and 16283 times 56 end in 8 but contain other digits, for example \(15738 \times 56 = 881328\).

Answer: B

The fraction that changes least is 5/8.

Adding 2 to both parts of a/b changes it by

\[ \begin{aligned} \frac{a + 2}{b + 2} - \frac{a}{b} \\ = \frac{2(b - a)}{b(b + 2)} \end{aligned} \]

\[ \begin{aligned} 1/5 &: \tfrac{8}{35} \approx 0.229 \\ 5/8 &: \tfrac{6}{80} = 0.075 \\ 1/2 &: \tfrac{2}{8} = 0.25 \\ 3/4 &: \tfrac{2}{24} \approx 0.083 \end{aligned} \]

5/8 moves to 7/10, a change of 0.075, the smallest. 3/4 is the near miss: it goes to 5/6, a change of about 0.083. The rule of thumb is that the change is small when the gap \(b - a\) is small relative to \(b^2\).

Answer: A

There are 17 such integers.

First list the numbers between 1 and 100 with a 5 as a digit: 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85 and 95, which is 19 numbers (55 counted once).

Of these, the multiples of 7 are 35 and 56. Removing them leaves \(19 - 2 = 17\).

18 comes from removing only one multiple of 7 (usually 35, missing 56), and 14 or 15 from undercounting the fifties. Listing is safer than a formula here.

Answer: C

All three statements are correct.

Statement 1: an even number is 2k, and \((2k)^3 = 8k^3\), which is even.

Statement 2: an odd number is \(2k + 1\), and \((2k + 1)^2 = 4k^2 + 4k + 1\), which is odd.

Statement 3: \((2a + 1)(2b + 1) = 4ab + 2a + 2b + 1\), which is odd.

The general rule behind all three: a product is even if and only if at least one factor is even, so powers of even numbers stay even and products of odd numbers stay odd.

Answer: B

Both Statement I and Statement II are false.

A prime number has exactly two distinct positive factors, 1 and itself.

1 has only one positive factor, so it is not prime; Statement I is false. 2 has exactly two factors, 1 and 2, so it is prime (the only even prime); Statement II, which says 2 is not prime, is also false.

Both statements reverse a basic fact, so every option calling either of them true is wrong.

Answer: C

The difference between the hundreds digit and the units digit is 1.

Write \(x = 100a + 10b + c\), so its reverse is \(100c + 10b + a\).

\[ \begin{aligned} x - \text{reverse} &= 99a - 99c \\ &= 99(a - c) = 99 \\ a - c &= 1 \end{aligned} \]

The tens digit cancels out, so it can be anything. For example, \(201 - 102 = 99\) and \(2 - 1 = 1\). A difference of 0 would make \(x\) equal its reverse, and 2 or 3 would give differences of 198 or 297.