Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
The digit in the unit's place of the product $(2464)^{1793}$ $\times(615)^{317} \times(131)^{491}$ is
(a)0
(b)2
(c)3
(d)5
Answer
Answer (as printed): A
Explanation
Unit digit in the given product = Unit digit in $\left(4^{1793} \times 5^{317} \times 1^{491}\right)$ Unit digit in $4^{2}$ is 6 and so the unit digit in $\left(4^{2}\right)^{896}$ is 6. ∴ Unit digit of $4^{1793}=$ Unit digit in $(6 \times 4)=$ Unit digit in 24 , which is 4 . Unit digit in $5^{317}$ is 5 and the unit digit in $1^{491}$ is 1. ∴ Unit digit in the given product = Unit digit in (4 × 5 $\times 1$ ), which is 0 .
Explanation as extracted from the printed page; notation may be imperfect.