If s and t are integers greater than 1 and each is a factor of the integer n...

Question

If s and t are integers greater than 1 and each is a factor of the integer n, which of the following must be a factor of n^s t ? I. s^t II. (s t)^2 III. s+t

GMAT Panda · Accepted Answer

I and II (E) Express n in terms of s since s divides n. n = s k Raise both sides to the power st to relate n^st to s^st. n^st = (s k)^st = s^st k^st Since st t, the factor s^t appears in s^st, so s^t divides n^st. s^st = s^t s^st - t Because s>1 and t>1, we have st 2. Hence both s^2 and t^2 divide n^st, and their product (st)^2 divides n^st. (st)^2 = s^2 t^2 Provide a counterexample to show s+t need not divide n^st. s = 2, t = 3, n = 6 6^6 will end in 6, so s+t=5 does not divide 6^6.

Explanation	Calculations	Help
$Express n in terms of s since s divides n .$	$n = s \times k$
$Raise both sides to the power s t to relate n^{s t} to s^{s t} .$	$n^{s t} = (s \times k)^{s t} = s^{s t} \times k^{s t}$
$Since s t \geq t, the factor s^{t} appears in s^{s t}, so s^{t} divides n^{s t} .$	$s^{s t} = s^{t} \times s^{s t - t}$
$Because s > 1 and t > 1, we have s t \geq 2 . Hence both s^{2} and t^{2} divide n^{s t}, and their product (s t)^{2} divides n^{s t} .$	$(s t)^{2} = s^{2} \times t^{2}$
$Provide a counterexample to show s + t need not divide n^{s t} .$	$s = 2, t = 3, n = 6 6^{6} will end in 6, so s + t = 5 does not divide 6^{6} .$

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