GATE 2024 EC GA Q.5: solution and answer

GATE 2024 · ECGA Q.5 · 1 mark · MCQ

The greatest prime factor of $(3^{199} - 3^{196})$ is

(A)
13
(B)
17
(C)
3
(D)
11

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Answer and solution

ANSWER

The greatest prime factor of $(3^{199} - 3^{196})$ is $13$ , option (A).

Official answer: ASource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Take the smaller power out as a common factor, using the law of exponents $a^{m} = a^{n} \cdot a^{m - n}$ (here $3^{199} = 3^{196} \cdot 3^{3}$ ): $3^{199} - 3^{196} = 3^{196} (3^{3} - 1)$
Evaluate the bracket: $3^{3} - 1 = 27 - 1 = 26$ .
Break 26 into primes: $26 = 2 \times 13$ .
So the number is $3^{199} - 3^{196} = 2 \times 3^{196} \times 13$ and its only prime factors are $2$ , $3$ and $13$ .
The largest of these is $13$ . Prime factorisation is unique, so no other prime divides the number. That rules out $17$ and $11$ , and $3$ is a factor but not the largest one. The answer is (A) 13.

KEY IDEA

Factor out the smaller power: $3^{199} - 3^{196} = 3^{196} (3^{3} - 1) = 3^{196} \times 26 = 2 \times 3^{196} \times 13$ . The huge power of 3 adds only the prime 3. The other primes all come from the small bracket $3^{3} - 1$ .

WATCH OUT FOR

Picking (C) 3 because the expression is 'made of powers of 3'. The subtraction brings in the new primes 2 and 13 through $3^{3} - 1 = 26$ . A second slip is to write $3^{199} - 3^{196} = 3^{3}$ by subtracting the exponents. That rule applies to division, $3^{199} \div 3^{196} = 3^{3}$ , not to subtraction.

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GATE 2024 EC GA Q.5: solution and answer

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