Three-Two-One Gooo
If
a
b
c
=
6
5
6
1
, then find the smallest possible value of
(
a
×
b
×
c
)
such that
a
,
b
,
c
are all integers.
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3 solutions
Moderator note:
Right. We can abuse the fact that 1 x = 1 for all x . Bonus question: what would the answer be if we add the constraint b = ± 1 ?
@Sandeep Bhardwaj Arghh.... you got me! Tricky question; I had automatically, but mistakingly, assumed that we were dealing only with positive integers. (Sorry for the earlier fuss with my report.) Nice catch, Abhishek. :)
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Same with me. A very nice question at all.
Re Bonus question: I'm thinking ( 3 8 ∗ 8 n ) 8 − n , making a b c = − 8 n ∗ 3 8 n + 1 for any positive integer n , and thus there is no least possible value as we can make it arbitrarily large (and negative).
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a = 3 8 × 8 n , b = 8 , c = − n .
Remember that tower of exponents are evaluated top down, and not bottom up, IE a ( b c ) instead of ( a b ) c .
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@Calvin Lin – Is the minimum 0 in that case? 6 5 6 1 5 0
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@Pranjal Jain – As Brian pointed out, a b c = − 8 n 3 8 n + 1 . Let n → ∞ and a b c → − ∞ , so there is no infimum / minimum.
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@Calvin Lin – How unfortunate that the question says "smallest" instead of "least" or "lowest".
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@Erick Wong – How is that unfortunate? What is the difference between "smallest" and "least" and "lowest"?
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@Calvin Lin – "Lowest" unequivocally means "closest to -infinity", whereas "smallest" can very reasonably be used to mean "closest to 0" (e.g. sin x is close to x for small x). What makes it unfortunate is that the question is ambiguous when phrased using "smallest", when it could very easily have been written with clearer intent.
Level 5 :O
a = -81, b = 2, c = 1 gives -162 with b^2 not 1.
for the values a=6561 b=1 and c=-1000000...... this wouldn't be equal to 6561 when [(a^{b})^{c}] is calculated although the product will be quite small
this is not 6561. a^b^c = a^bc. In your case it would be 6561^(-100000000), that would be a very small number.
Randomly press the calculator and found out that 3^(2^3) = 6561 where a = 3, b = 2 and c = 3. I got the answer wrong because I didn't realize that integers can be negative in this case.
If b <> +/- 1 then (-81)^2^1 = -162 shall be the one.
6561^(-1)^(Positive even) was what I thought most negative.
6561^1^(Negative) is also another extreme provided we are not talking about magnitude such as zero.
This has already been solved before where I did not get correct solution.
Oh, I don't remember seeing it before. At least you didn't get "caught" this time around like I did. :)
No finite least possible value exists.
Example - a = 6 5 6 1 , b = 1 and c = − 1 0 0 0 0 0 0 0 0 ( a very small negative number that is large in terms of magnitude).