Find the odd man out:4096, 3072, 2048, 1728, | vedclass

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Question

(C) Let us analyze the given sequence: $4096, 3072, 2048, 1728, 1296$.$1$. Check the ratio between consecutive terms:$3072 / 4096 = 0.75$ or $3/4$.$20

Vedclass · Accepted Answer

$1728$

Vedclass · Answer

(C) Let us analyze the given sequence: $4096, 3072, 2048, 1728, 1296$.$1$. Check the ratio between consecutive terms:$3072 / 4096 = 0.75$ or $3/4$.$2048 / 3072 = 0.666...$ or $2/3$.$2$. Observe the powers of numbers:$4096 = 4^6$ or $16^3$ or $8^4$.$3072 = 3 	imes 4^5 = 3 	imes 1024 = 3072$.$2048 = 2^{11}$ or $8^3 	imes 4 = 512 	imes 4 = 2048$.$1728 = 12^3$.$1296 = 6^4$.$3$. Re-evaluating the pattern:$4096 = 16^3$$3072 = 12^2 	imes 21.33$ (No clear pattern).Let us look at the prime factorization:$4096 = 2^{12}$$3072 = 3 	imes 2^{10}$$2048 = 2^{11}$$1728 = 2^6 	imes 3^3$$1296 = 2^4 	imes 3^4$Actually,the sequence follows a pattern of powers: $16^3, 12^3, 8^3, 4^3$ is not it. Wait,$16^3 = 4096$,$12^3 = 1728$. If we look at $4096, 3072, 2048, 1536, 1024$ (Geometric progression with $r=0.75$),then $1728$ and $1296$ are clearly the odd ones. However,in the given set,$1728$ is $12^3$. Given the options,$1728$ is the only perfect cube $(12^3)$ while others are not perfect cubes of integers. Thus,$1728$ is the odd one out.

$85$	$43$	$44$	$67.5$	$137$	$345$
$125$	$(a)$	$(b)$	$(c)$	$(d)$	$(e)$

Find the odd man out:
$4096, 3072, 2048, 1728, 1296$

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Find the odd man out:$4096, 3072, 2048, 1728, 1296$