GATE 2026 EC Q.45: solution and answer

GATE 2026 · ECQ.45 · 2 marks · MCQ

What is the $1 0^{'} s$ complement of $(47)_{10}$ ?

(A)
52
(B)
53
(C)
54
(D)
55

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Answer and solution

ANSWER

The 10's complement of $(47)_{10}$ is $1 0^{2} - 47 = 53$ , option (B).

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

For an $n$ -digit number $N$ in base $r$ , the $r$ 's complement is $r^{n} - N$ . Here $r = 10$ , and $N = (47)_{10}$ has $n = 2$ digits.
Apply the formula directly: $1 0^{2} - 47 = 100 - 47 = 53$ .
Check it with the 9's complement. The $(r - 1)$ 's complement is $(r^{n} - 1) - N = 99 - 47 = 52$ , which just replaces each digit $d$ with $9 - d$ (so $4 \to 5$ and $7 \to 2$ ). Adding 1 gives the 10's complement: $52 + 1 = 53$ .
Check the sum: $47 + 53 = 100$ . Drop the end carry and $00$ is left, which is what a number plus its $r$ 's complement must give. That is why the complement can stand for $- N$ in subtraction.
So the 10's complement of $(47)_{10}$ is $53$ , option (B).

KEY IDEA

In base $r$ , the $r$ 's complement of an $n$ -digit number $N$ is $r^{n} - N$ . It equals the $(r - 1)$ 's complement, $(r^{n} - 1) - N$ , plus 1. For $N = 47$ : $100 - 47 = 53$ , or $99 - 47 = 52$ and then $52 + 1 = 53$ .

WATCH OUT FOR

Stopping at the 9's complement (52, option A) and forgetting to add 1 is the usual slip. A second one is using the wrong power of the base, for example $1 0^{3} - 47 = 953$ . With $n = 2$ digits the power must be $1 0^{2}$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 5 Number systems questions, with notes

GATE 2026 EC Q.45: solution and answer

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