GATE 2023 EE GA Q.9: solution and answer

GATE 2023 · EEGA Q.9 · 2 marks · MCQ

The digit in the unit's place of the product $3^{999} \times 7^{1000}$ is ______.

(A)
7
(B)
1
(C)
3
(D)
9

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Answer and solution

ANSWER

The unit digit of $3^{999} \times 7^{1000}$ is $7$ . Option (A).

Official answer: ASource: Official answer key, GATE 2023 EE (IIT Kanpur)

Only unit digits matter: the unit digit of a product equals the unit digit of the product of the factors' unit digits.
Pair the factors: $3^{999} \times 7^{1000} = 3^{999} \times 7^{999} \times 7 = (3 \times 7)^{999} \times 7 = 2 1^{999} \times 7$ .
Any power of a number ending in 1 also ends in 1, so $2 1^{999}$ has unit digit $1$ .
Hence the unit digit of the product is $1 \times 7 = 7$ .
Check with cyclicity (unit digits of powers repeat with period 4). Powers of 3 end in $3, 9, 7, 1$ ; since $999 = 4 \times 249 + 3$ , $3^{999}$ ends in the 3rd term, $7$ .
Powers of 7 end in $7, 9, 3, 1$ ; since $1000 = 4 \times 250$ (remainder 0, i.e. the 4th term), $7^{1000}$ ends in $1$ .
Product of the unit digits: $7 \times 1 = 7$ , which is option (A).

KEY IDEA

Only unit digits matter. Take $7^{999}$ out of $7^{1000}$ and pair it with $3^{999}$ to get $2 1^{999} \times 7$ . A power of a number ending in 1 always ends in 1, so the answer is the unit digit of $7$ . You can check it with the period-4 cycles of 3 ( $3, 9, 7, 1$ ) and 7 ( $7, 9, 3, 1$ ).

WATCH OUT FOR

Getting the cycle position wrong when the remainder is 0. $1000 mod 4 = 0$ means the 4th term of the cycle, which is $1$ . Picking the 1st term ( $7$ ) gives $7 \times 7 = 49$ , unit digit $9$ , which is option (D). Another slip is reading $3^{999}$ off at remainder 3 as $9$ (the 2nd term) and getting $9 \times 1 = 9$ .

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GATE 2023 EE GA Q.9: solution and answer

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