Number System Questions and Answers

Answer: B

The number is 12111.

\[ \begin{aligned} 11000 + 1100 + 11 &= 12111 \end{aligned} \]

Eleven hundred is 1100, which carries a thousand into the thousands place, so the number has 12 thousands, 1 hundred and 11. 111111 simply writes 'eleven' three times, 11000110011 writes each part's digits side by side, and XIXIXI is not a valid Roman numeral for the number.

Answer: D

The distance is 22000 units.

Between 25000 and 39000 there are 7 equal intervals, so each interval is \( 14000 \div 7 = 2000 \). A is 3 intervals after 25000, and B is 7 intervals after 39000, which fits the 4 intervals from B to 61000:

\[ \begin{aligned} A &= 25000 + 3 \times 2000 = 31000 \\ B &= 39000 + 7 \times 2000 = 53000 \\ B - A &= 22000 \end{aligned} \]

22 units counts the intervals between points as if each were 1000, and 29000 and 84000 come from misreading the scale or adding the values.

Answer: C

The true statement is a + (b + c) = c + (a + b).

Addition of rational numbers is both associative and commutative, so the three numbers can be added in any grouping and any order and the sum is the same: both sides equal a + b + c. With the given values,

\[ \begin{aligned} \tfrac{3}{5} + \tfrac{2}{3} - \tfrac{5}{6} &= \tfrac{18 + 20 - 25}{30} \\ &= \tfrac{13}{30} \end{aligned} \]

on both sides. The other statements fail because multiplication, division and subtraction do not let you swap the numbers like this: for example a × (b + c) = (3/5)(−1/6) = −1/10, but b × (a + c) = (2/3)(−7/30) = −7/45.

Answer: B

p is the square root of q.

If q = p², then by definition p is a number whose square is q, that is, p = √q. For example, if p = 7 then q = 49 and 7 is the square root of 49.

p is not the square of q (that would be q² = 2401 in the example). p equals q only for p = 1, and for every natural number above 1 the square q is greater than p, not smaller, so options C and D are wrong.

Answer: A

The statement that is not correct is all prime numbers are odd numbers.

2 is a prime number, since its only factors are 1 and itself, and it is even. It is the only even prime, so the statement fails.

The other statements are true. Euclid proved that there are infinitely many primes. A prime has exactly two factors, 1 and itself. The single-digit primes are 2, 3, 5 and 7, exactly four.

Answer: D

The number is 2520.

A number divisible by every number from 1 to 10 must be a multiple of their LCM:

\[ \begin{aligned} \text{LCM}(1, \dots, 10) &= 2^{3} \times 3^{2} \times 5 \times 7 \\ &= 2520 \end{aligned} \]

2520 is the smallest such number. 10 and 100 are not divisible by 3 or 7, and 604 = 4 × 151 is not divisible by 3, 5 or 7.

Answer: A

The number is 57.

The sum of the digits must be a multiple of both 3 and 4, so a multiple of 12. For a two-digit number the sum is at most 18, so it must be 12. The digits must be consecutive prime numbers: 5 and 7 are consecutive primes and 5 + 7 = 12.

Check the others: 23 has digits summing to 5, 35 to 8 and 13 to 4, none a multiple of 12. (2 and 3 and 3 and 5 are consecutive primes, but their sums fail.)

Answer: D

The number is 60482.

Fill the places one at a time:

• Tens: 8
• Units: one-fourth of 8 = 2
• Thousands: 0
• Hundreds: twice the units = 4
• Ten thousands: thrice the units = 6

Reading from ten thousands to units gives 6, 0, 4, 8, 2, that is 60482. The other options put the 0 or the 4 in the wrong place, for example 64082 swaps the thousands and hundreds digits.

Answer: D

There are 8 pairs of twin primes between 1 and 100.

Twin primes are pairs of primes that differ by 2. Listing them:

(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73).

That is 8 pairs. Other near misses are not twins because one member is composite: 23 and 25 (25 = 5 × 5), 47 and 49 (49 = 7 × 7), 89 and 91 (91 = 7 × 13). Forgetting that 5 belongs to two pairs, (3, 5) and (5, 7), or treating 91 as prime, leads to the wrong counts 7 or 9.

Answer: D

The value of (4a − 5b + c) is 3.

Factorise 21168 by repeated division:

\[ \begin{aligned} 21168 &= 7 \times 3024 \\ &= 7^{2} \times 432 \\ &= 7^{2} \times 16 \times 27 \\ &= 2^{4} \times 3^{3} \times 7^{2} \end{aligned} \]

So a = 4, b = 3 and c = 2.

\[ 4a - 5b + c = 16 - 15 + 2 = 3 \]

Check: \(16 \times 27 \times 49 = 21168\). Miscounting a power changes the result sharply; for example a = 3 would give 12 − 15 + 2 = −1, which is not even an option.