Number System Aptitude MCQ

This problem uses the algebraic property that the HCF of two numbers in the form (a^m - 1) and (a^n - 1) is given by:

First, find the highest common factor of the powers 24 and 18:

Substituting this value into the formula:

Therefore, the HCF is 63.

A fundamental property of any two numbers is that their LCM multiplied by their HCF equals the product of those two numbers.

Here, using the formula:

Substituting the given values:

Dividing 7560 by 84 gives:

The distractor options represent common mistakes, such as dividing the LCM directly by the known number or ignoring the role of HCF in the formula.

This problem uses the repeating remainder cycle of powers of 3 modulo 8.

The sum of 99 such powers is computed using the geometric series formula.

Reducing the numerator modulo 8 using the known remainder of 3^99 simplifies cleanly to a remainder of 3.

This "remainder cycle" trick is a staple shortcut in UPSC, SSC, and Banking exams whenever a large-exponent sum must be divided.

We use the identity:

Setting this equal to 84:

We look for factor pairs of 84 where the larger factor is greater than the smaller factor and both factors have the same parity. This ensures that the values of a and b are whole numbers.

The valid factor pairs are:

which gives:

and

which gives:

Thus, there are 2 valid pairs of natural numbers.

The last digit of a number raised to a power follows a repeating cycle. For 7, the cycle of last digits is:

The cycle length is 4, so we reduce the exponent modulo 4 to find the effective position in the cycle.

Since every power of 7 is odd:

So, the effective exponent is 3.

The 3rd position in the cycle of 7 is 3. Therefore, the remainder when the number is divided by 10 is: 3

The product of two numbers with a fixed sum is greatest when the two numbers are equal. This follows from the AM-GM inequality.

Here, the sum of the two numbers is 24:

For the maximum product, the numbers should be equal:

The maximum product is:

Any unequal split gives a smaller product. For example:

The incorrect options represent common mistakes, such as choosing adjacent integers, not splitting the sum equally, or applying the sum condition incorrectly.

These special primes are known as emirp primes, where the reversed form of the prime is also a prime.

The key condition is that the reversed number must be different from the original prime. Therefore, palindromic primes such as 11 are excluded because their reverse remains the same.

Here are the 2 digit emirp primes: 13, 17, 31, 37, 71, 73, 79, and 97

We check each two-digit prime, reverse its digits, and verify whether the reversed number is also prime. Only those primes where both the original number and its reversal are distinct primes are counted.

After checking all two-digit primes, the total number of emirp primes is: 8.

This problem uses the cyclicity of powers in modular arithmetic. The remainders of powers of 5 when divided by 6 repeat in a fixed pattern.

The pattern is:

So, odd powers of 5 leave a remainder of 5, while even powers leave a remainder of 1.

Since 84 is even:

Therefore, the remainder is 1.

Recognising repeating cycles is the key technique for solving large-exponent remainder problems in competitive exams.

We split all page numbers into different ranges — 1-digit (1–9), 2-digit (10–99), and 3-digit (100 onwards).

For each range, we count the occurrences of the digit 2 separately in the units, tens, and hundreds places. We keep adding these counts until the total reaches 96.

The key observation is that in the 200s range, the hundreds digit contributes one occurrence of 2 on every page, causing the count to increase rapidly.

Continuing the count carefully, the total reaches 96 at page 242.

Therefore, the book ends at page 242.

First, find the remainder of each number divided by 19:

Next, multiply these remainders together: