MCQ
Which among $43^2, 67^2, 52^2, 59^2$would end with digit $1?$
- A$43^2$
- B$67^2$
- C$52^2$
- ✓$59^2$
✓
Answer
Correct option: D.
$59^2$
D. $59^2$
Solution:
We know that, the unit's digit of the square of a natural number is the unit's digit of the square of the digit at unit's place of the given natural number.
$\therefore$ Unit's digit of $43^2 = 9$ $\big[\because3^2=9\big]$
Unit's digit of $67^2 = 9$ $\big[\because$ unit's digit of $7^2$ is 9$\big]$
Unit's digit of $52^2 = 4$ $\big[\because2^2=4\big]$
Unit's digit of $59^2 = 1$ $\big[\because$ unit's digit of $9^2$ is 1$\big]$
Clearly, the square of $59$ end with digit $1.$
Solution:
We know that, the unit's digit of the square of a natural number is the unit's digit of the square of the digit at unit's place of the given natural number.
$\therefore$ Unit's digit of $43^2 = 9$ $\big[\because3^2=9\big]$
Unit's digit of $67^2 = 9$ $\big[\because$ unit's digit of $7^2$ is 9$\big]$
Unit's digit of $52^2 = 4$ $\big[\because2^2=4\big]$
Unit's digit of $59^2 = 1$ $\big[\because$ unit's digit of $9^2$ is 1$\big]$
Clearly, the square of $59$ end with digit $1.$
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