Question

odd one out
121 ;169; 225; 289​

Answer 1

Given: Four numbers- 121, 169, 225 and 289

To find: The number which is different (in terms of some property, feature etc.) from the other three.

Solution:

We observe that the given numbers are squares of four integers.

∵ $11^{2} = 121$

∵ $13^{2} = 169$

∵ $15^{2} = 225$ and

∵ $17^{2} = 289$

The four integers which are square roots of given numbers are- 11, 13, 15 and 17.

On carefully observing the integers, we find that 15 is the only composite number. Rest of the integers are prime numbers.

So, 225 is the only number which is the square of a composite number among the given numbers.

Hence, 225 is the odd/unique number among the given numbers.

Answer 2

Given:

$121,169,225,289$

All the given number are square of number

$121=11^2\\169=13^2\\225=15^2\\289=17^2$

$11,13,17$ are prime number whereas $15$ is composite number.

Hence, $225$ is odd one