Question

Find the odd one out among these:
42 , 21 , 63 , 36 {Provide the reason}

Answer 1

36 is right answer

Answer:

because all are divisible by 7 but 36 isn't

Answer 2

Answer:

The odd one is 36 among 42 , 21 , 63 , 36.

Explanation:

Given numbers are 42 , 21 , 63 , 36.

We want to find odd one out from among these numbers.

By prime factorisation,

$42 = 2 \times 3 \times 7 \\ 21 = 3 \times 7 \\ 63 = 3 \times 3 \times 7 \\ 36 = 2 \times 2 \times 3 \times 3$

Here clear that 42,21,63 are divisible by 7 but 36 is not divisible by 7.

Therefore, 36 is required odd number.

This is a problem of Mathematics.

Some important Mathematics formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )$

Two more important Mathematics problem:

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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