## mathmath333 one year ago Reasoning question

1. mathmath333

\large \color{black}{\begin{align} & \normalsize \text{Find the odd man out}\hspace{.33em}\\~\\ & a.)\ 324 \hspace{.33em}\\~\\ & b.)\ 244 \hspace{.33em}\\~\\ & c.)\ 514 \hspace{.33em}\\~\\ & d.)\ 136 \hspace{.33em}\\~\\ \end{align}}

2. mathmath333

$$324=2^{2}\times 3^{4}\\ 224=2^{2}\times 61\\ 514=2^{3}\times 257\\ 136=2^{3}\times 3^{4}$$

3. ganeshie8

Hint : sum of digits

4. mathmath333

sum of digits $$324\rightarrow 9\\ 224\rightarrow 8\\ 514\rightarrow 10\\ 136\rightarrow 10$$

5. anonymous

Using sum of digits, a) is the odd man out. Also, d) is odd man out considering the one's digit.

6. mathmath333

d.) correct ?

7. anonymous

@mathmath333 , there's something wrong with your prime factorizations. And the second number is 244, not 224.

8. amilapsn

b) too @ospreytriple because its first digit isn't odd!

9. mathmath333

this one $$324=2^{2}\times 3^{4}\\ 244=2^{2}\times 61\\ 514=2\times 257\\ 136=2^{3}\times 3^{4}$$

10. anonymous

@mathmath333 , $$136 = 2^3 \times 17$$

11. mathmath333

ok

12. thomas5267

It could be argued that 514 is the odd one as its prime factorisation contains only two primes. Or even the answer is none of above as all of the options are numbers not man.