UGC NET Questions (Paper – 1)

For a positive integer N, the number of decimal digits reflects the position of the most significant non-zero digit. The magnitude of N lies between powers of 10, and if 10^(k−1) ≤ N ≤ 10^k − 1, then N has k digits. This inequality rearranges to k = ⌊log₁₀N⌋ + 1. Hence the standard formula used in computer science and mathematics for the digit count is ⌊log₁₀N⌋ + 1.

Option A:
Option A uses the floor function to ensure that fractional logarithm values yield the correct integer digit count. For any N in the interval [10^(k−1),10^k−1], log₁₀N lies in [k−1,k), so ⌊log₁₀N⌋ = k−1 and adding 1 gives k. This matches the definition of decimal digit length.

Option B:
Option B, ⌈log₁₀N⌉, would overestimate the digit count whenever N is exactly a power of 10, such as 100. For N = 100, log₁₀N = 2, and ⌈2⌉ = 2, which ignores that 100 has 3 digits. Hence, the ceiling function alone is not appropriate for this formula.

Option C:
Option C, log₁₀(N−1), lacks both integer rounding and the +1 term, so it does not produce an integer digit count reliably. It also fails for small values like N = 1, where log₁₀(0) is undefined. Thus, this expression is mathematically unsuitable for the purpose.

Option D:
Option D, log₁₀N + 1, can give non-integer, fractional values, which do not correspond directly to a digit count. It also overestimates digits for many values, since there is no rounding mechanism to correct for the logarithm's fractional part.

Converting 1234 to hexadecimal involves dividing repeatedly by 16 or expressing it as a sum of powers of 16. We can write 1234 as 4×16² + 13×16¹ + 2×16⁰, where 13 is represented by D in hexadecimal. This gives the compact form 4D2₁₆. Since this exactly reproduces 1234 when converted back to decimal, option C is correct.

Option A:
Option A has the middle digit B, which represents 11 in hexadecimal, and its expansion gives a smaller decimal value than 1234. The difference in the middle digit means the 16¹ place contributes less to the total. Hence, 4B2₁₆ does not correspond to 1234.

Option B:
Option B uses C (12) as the middle digit and therefore produces a decimal value different from 1234. Its place value expansion does not match the decomposition of 1234 into powers of 16. It is therefore not the correct hexadecimal representation.

Option C:
Option C uses D (13) as the middle digit and 2 in the units place, which precisely fits the decomposition of 1234 into 4×256 + 13×16 + 2. This structure makes it the only option that converts back to 1234. Thus, 4D2₁₆ is the correct hexadecimal form.

Option D:
Option D has E (14) in the middle position, increasing the value beyond 1234. Its decimal equivalent is larger and therefore cannot represent 1234. Hence, 4E2₁₆ is an incorrect choice.