diff(abs(z), z)  returns abs(1, z)  and the latter, for a numeric z, is defined only for a nonzero real z.
However,  functions such as abs(I+sin(t)) are (real) differentiable for a real t and diff should work. It usually does, but not always.

restart
f:= t -> abs(GAMMA(2*t+I)):  # We want D(f)(1)
D(f)(1): 
evalf(%);  # Error, (in simpl/abs) abs is not differentiable at non-real arguments
D(f)(1); simplify(%); 
evalf(%);   # 0.3808979508 + 1.161104935*I,  wrong

The same wrong results are obtained with diff instead of D

diff(f(t),t):   # or  diff(f(t),t) assuming t::real; 
eval(%,t=1);
simplify(%); evalf(%);   # wrong, should be real

To obtain the correct result, we could use the definition of the derivative:

limit((f(t)-f(1))/(t-1), t=1); evalf(%); # OK 
fdiff(f(t), t=1);    # numeric, OK

       0.8772316598
       0.8772316599

Note that abs(1, GAMMA(2 + I)); returns 1; this is also wrong, it should produce an error because  GAMMA(2+I) is not real;
 

Let's redefine now `diff/abs`  and redo the computations.

restart
`diff/abs` := proc(u, z)   # implements d/dx |f(x+i*y|) when f is analytic and f(...)<>0
local u1:=diff(u,z);
1/abs(u)*( Re(u)*Re(u1) + Im(u)*Im(u1) )
end:
f:= t -> abs(GAMMA(2*t+I));
D(f)(1); evalf(%);   # OK now

         0.8772316597

Now diff works too.

diff(f(t),t);
eval(%,t=1);
simplify(%); evalf(%);   # it works

This is a probably a very old bug which may make diff(u,x)  fail for expressions having subespressions abs(...) depending on x.

However it works  using assuming x::real, but only if evalc simplifies u.
The problem is actually more serious because diff/ for Re, Im, conjugate should be revized too. Plus some other related things. 

The Putnam 2020 Competition (the 81st) was postponed to February 20, 2021 due to the COVID-19 pandemic, and held in an unofficial mode with no prizes or official results.

Four of the problems have surprisingly short Maple solutions.
Here they are.

A1.  How many positive integers N satisfy all of the following three conditions?
(i) N is divisible by 2020.
(ii) N has at most 2020 decimal digits.
(iii) The decimal digits of N are a string of consecutive ones followed by a string of consecutive zeros.

add(add(`if`( (10&^m-1)*10&^(n-m) mod 2020 = 0, 1, 0), 
n=m+1..2020), m=1..2020);

       508536

A2.  Let k be a nonnegative integer.  Evaluate  

sum(2^(k-j)*binomial(k+j,j), j=0..k);

        4^k

A3.  Let a(0) = π/2, and let a(n) = sin(a(n-1)) for n ≥ 1. 
Determine whether the series   converges.

asympt('rsolve'({a(n) = sin(a(n-1)),a(0)=Pi/2}, a(n)), n, 4);

a(n) ^2 being equivalent to 3/n,  the series diverges.

B1.  For a positive integer n, define d(n) to be the sum of the digits of n when written in binary
 (for example, d(13) = 1+1+0+1 = 3). 

Let   S =  
Determine S modulo 2020.

d := n -> add(convert(n, base,2)):
add( (-1)^d(k) * k^3, k=1..2020 ) mod 2020; 

        1990

Here is a very nice (but not easy) elementary problem.
The equality
ceil(2/(2^(1/n)-1)) = floor(2*n/ln(2));

is not an identity, it does not hold for each positive integer n.
How to find such a number?

[This is a re-post, because the original vanished when trying a conversion Question-->Post]

The problem appears in the recent book:
Richard P. Stanley - Conversational Problem Solving. AMS, 2020. 

The problem is related to a n-dimensional tic-tac-toe game. The first counterexample (2000) was wrong due to a multiprecision arithmetic error.
The  author of the book writes 
"To my knowledge, only eight values of n are known for which the equation fails,
and it is not known whether there are infinitely many such values",
but using Maple it will be easy to find more.

A brute-force solution is problematic because the smallest counterexample is > 7*10^14.

restart;
a := 2/(2^(1/n)-1): b := 2*n/ln(2):
asympt(b-a, n);

It results:  b - a → 1 (for n →oo);
So, to have a counterexample, b must be close to an integer
m ≈ 2*n/ln(2)  ==> n/m ≈ ln(2)/2

The candidates for n/m will be obviously the convergents of the continued fraction of the irrational number ln(2)/2.
 

convert(ln(2)/2, confrac, 200, 'L'):
Digits:=500:
for n in numer~(L[3..]) do
  if not evalf(ceil(a)=floor(b)) then printf("n=%d\n", n) fi;
od:

n=777451915729368
n=140894092055857794
n=1526223088619171207
n=54545811706258836911039145
n=624965662836733496131286135873807507
n=1667672249427111806462471627630318921648499
n=36465374036664559522628534720215805439659141
n=2424113537313479652351566323080535902276508627
n=123447463532804139472316739803506251988903644272
n=97841697218028095572510076719589816668243339678931971
n=5630139432241886550932967438485653485900841911029964871
n=678285039039320287244063811222441860326049085269592368999
n=312248823968901304612135523777926467950572570270886324722782642817828920779530446911
n=5126378297284476009502432193466392279080801593096986305822277185206388903158084832387
n=1868266384496708840693682923003493054768730136715216748598418855972395912786276854715767
n=726011811269636138610858097839553470902342131901683076550627061487326331082639308139922553824778693815
 

So, we have obtained 16 counterexamples. The question whether there are an infinity of such n's remains open.

Download imc2020-1-4.mw

Can you guess what P() produces, without executing it?

P:=proc(N:=infinity) local q,r,t,k,n,l,h, f;
q,r,t,k,n,l,h := 1,0,1,1,3,3,0:
while h<N do 
   if 4*q+r-t < n*t
   then f:=`if`(++h mod 50=0,"\n",`if`(h mod 10=0," ","")); printf("%d"||f,n);   
        q,r,t,k,n,l := 10*q,10*(r-n*t),t,k,iquo(10*(3*q+r),t)-10*n,l
   else q,r,t,k,n,l := q*k,(2*q+r)*l,t*l,k+1,iquo(q*(7*k+2)+r*l,t*l),l+2
   fi
od: NULL
end:

I hope you will like it (maybe after execution).