A) 0
B) 1
C) 2
D) 4
Correct Answer: D
Solution :
[d] Let the number be x and let \[x=5q+3.\] Then, \[{{x}^{2}}={{(5\,q+3)}^{2}}=(25\,{{q}^{2}}+30\,q+9)\] \[=5\,(5{{q}^{2}}+6q+1)+4\] \[\therefore \] On dividing \[{{x}^{2}}\] by 5, the remainder is 4. |
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