Large Numbers = Worse!

Level pending

If $a, b, c, d, e$ are integers such that

$a = b^{2}$

$b = c^{3}$

$c = d^{4}$

$d = e^{5}$ ,

If $e>1$ , what is the smallest possible whole number that can be the integer $a$ ?

Don't be confused with the choices. The last digit is important.

1329227995784915872903807060280344576 1329227995784915872903807060280344578 1329227995784915872903807060280344572 1329227995784915872903807060280344574

1 solution

Jeremy Bansil
Jul 31, 2014

Since $0$ and $1$ cannot be (which would make it logical), the remaining smallest whole number that is equal to $e$ is $2$ . So $2^{5}$ is $32$ , which would be $d$ . $32^{4}$ is $1048576$ , which would be $c$ . $1048576^{3}$ is equal to $1152921504606846976$ , which would be $b$ . And $a$ would be $1152921504606846976^{2}$ , which would be, well, $\boxed{1329227995784915872903807060280344576}$ .

To simply solve it, go multiply the exponents. $(^{2})(^{3})(^{4})(^{5}) = (^{120})$ So make $2^{120}$ and you'll get the same answer.

The question as previously stated, could have $a = 2$ with $e = \sqrt[120] { 2}$ . I've added in the fact that they are all integers.

Calvin Lin Staff - 6 years, 10 months ago

Oh. Thank you...

Jeremy Bansil - 6 years, 10 months ago

Large Numbers = Worse!

1 solution

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