One can show that for a > b >= e , b^a > a^b, so immediately 59^60 > 60^59 This is proven by considering the graph ln(x)/x for x > 0, and showing it has a global maximum at x = e, and monotonically decreasing to 0+ as x->infinity. Hence for a > b >= e: ln(b)/b > ln(a)/a ==> aln(b) > bln(a) ==> b^a > a^b Similarly, ln(x)/x is monotonically increasing from 0 to e, and hence for p < q p^q < q^p This is pretty much the best that we can do, if p < e and q >= e, we have a deadzone of values where we can’t evaluate the inequalities. For example: 2^3 < 3^2 2^4 = 4^2 2^5 > 5^2 Edit: There is actually a way of comparing the inequalities in the "deadzone" for special cases, which is a little trickier. We now consider intersections of the straight line y = mx + c with the curve y = ln(x) for varying m > 0 and c, and x > 0: 1) If there exists a line of the form y = mx for m < 1/e which intersects ln(x) at x = a and x = b (one can show quite easily that there are 2 intersections in this case, and they occur at a < e and b > e), then a^b = b^a. Lets attempt to show this with a = 2 and b = 4; then the gradient of such a straight line is: (ln(4) - ln(2))/(4-2) = ln(2)/2. Hence, it suffices to show that y = xln(2)/2 intersects y = ln(x) at x = 2 and x = 4. This is true as 2ln(2)/2 = ln(2), and 4ln(2)/2 = ln(4). Hence, 2^4 = 4^2 !! 2) If we consider the line y = x/e + c, for some c < 0, and it intersects ln(x) at two points, x = p and x = q, with p < e and q > e. One can show in this case that p^q < q^p. This is a highly specific case however. More details of these can be found on the STEP II 2023 Paper (Question 1).

Instead "tricky" part, even simpler would be to use approximation: (1+1/59)^59~e Then you got the same: e/59

Considering 59 and 60 are almost the same number, and you have 59 being multipled 1 additional time over the 60^59, this alone would implie 59^60>60^59

Just calculate the number of digits by using Logarithm as follow : Num of digits = (characteristic of log) + 1 Now , 59 ^ 60 [ ] 60 ^ 59 Taking (log base 10) on both sides => 60 * log ( 59 ) [ ] 59 * log ( 60 ) => 106.2511 [ ] 104.9109 Thus , number of digits in 59 ^ 60 = 106 + 1 = 107 number of digits in 60 ^ 59 = 104 + 1 = 105 Hence , 59 ^ 60 [ > ] 60 ^ 59

Your videos are awesome! Thank you

I don't get the tricky part. How is (1+1/2)*(1+1/3) etc related to the actual question?

It's much simpler than that: 59^60 = (60-1)^60 which can be factorized as = (60-1) × (60^59 + 60^58 + 60^57 + .... + 1) Notice than the first term in the second bracket is simply 60^59.. and everything else is a positive number which is added to it.. hence 59^60 > 60^59

A calculation using smaller figures will also prove it... 5⁶ and 6⁵ 5x5x5x5x5x5 = 15625 6x6x6x6x6 = 7776 6⁷ and 7⁶ 6x6x6x6x6x6x6 = 279936 7x7x7x7x7x7 = 117649

a^b is ever lager b^a where 1< a

Using the Binomial Theorem approximation for small x (approx 1+nx) then no need to go beyond line two.

I think with these types of problems, the larger exponent and smaller base wins. As in (a-1)^b > a^(b+1). This will work for a bases and exponents greater than 2 I believe.

Just try a few beginning with 2^3 and 3^2 and then a few more. You can see the trend very quickly. The larger exponent will always give the larger number. Is that a "proof"? No. But you can see the two possible results diverging.

59(60/59) v 60. Doesn't take a large exponent to bump 59 above 60. 59^60 > 60^59.

Above some number the size of the exponent always matter more than the size of the base. This is already the case with 3⁴ (=81) > 4³ (=64) but not yet with 2³ (=8) < 3² (=9) So the turnaround is at ~3 (e?).

Brilliant!

Or, just see that x^(1/x) is descending for x>e, so 59^(1/59)>60^(1/60)

Exponents ate a fucking powerful thing. They always win

lol the litle number divided by the big number

60^59 = (59*(60/59))^59=59^59*(1+1/59)^59 < 59^59*e(1+1/n)^n for all positive n)

a >= b >= e a^b

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I don't understand oo

Hypothesis: x^(x+1) > (x+1)^x for any positive number where x>2

this problem makes no sense, it's common knowledge that the number with the bigger exponent is always bigger than the one with the bigger base, if the numbers are inverted if the numbers are bigger than 3.

WRONG! But it did make me watch it more than I should have... so they won anyway - hahahahah