Huge Numbers
1 7 1 6 or 1 6 1 7
Which one of the two numbers above is greater?
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6 solutions
Moderator note:
Nice approach to remember.
It can be approached by binomial theorem.we see (16+1)^16 and (16)^17 in form of (1+x)^n and then expanding (16+1)^16 to prove that it is....
We can also find the number of digits in each number.
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True. As 1 6 lo g 1 0 ( 1 7 ) = 1 9 . 6 8 7 . . , (so 1 7 1 6 has 2 0 digits), and 1 7 lo g 1 0 ( 1 6 ) = 2 0 . 4 7 0 . . . , (so 1 6 1 7 has 2 1 digits), we know that 1 6 1 7 > 1 7 1 6 . I just wanted to use a method where a calculator wasn't necessary.
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One could also prove that,
f
(
x
)
=
x
x
1
is a decreasing function for
x
≥
e
1
7
1
7
1
≤
1
6
1
6
1
Raising to
1
6
⋅
1
7
on both sides get our results.
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@A Former Brilliant Member – Great observation! Indeed, f ′ ( x ) = x x 1 − 2 ( 1 − ln ( x ) ) , so f ( x ) peaks at x = e and then diminishes for x ≥ e .
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@Brian Charlesworth – I really like this function from the time I used it to prove e π > π e .
I thought that 2
I compared 3x3x3x3 with 4x4x4 mentally and then my intuition recognised that the smaller number wins.
Or you could just multiply..
For ( x , y ) > e and x > y , y x > x y . Therefore, 1 6 1 7 is greater.
f(x)=
x
l
n
x
is monotonically decreasing x∈[e,+∞ )
16<17
f(16)>f(17)
1
6
l
n
1
6
>
1
7
l
n
1
7
17ln16>16ln17
16^17>17^16
Using log base transform : 17^16 is equal to 16^16.34985136 Hence: 16^17 > 16^16.34985236
We will look at the ratio
R = 1 6 1 7 1 7 1 6 = 1 6 1 ∗ ( 1 6 1 7 ) 1 6 = 1 6 1 ∗ ( 1 + 1 6 1 ) 1 6 = 1 6 f ( 1 6 )
where f ( x ) = ( 1 + x 1 ) x . Now we recognize that f ( x ) is monotonically increasing over the positive reals and that x → ∞ lim f ( x ) = e = 2 . 7 1 8 2 8 . . . . , and so f ( 1 6 ) < e < 1 6 . This in turn implies that R < 1 , i.e., that 1 6 1 7 is the greater of the given quantities.