1998, Part 1
We've been in this year for quite awhile now, but this collection of
puzzles supplied by Philippe Fondanaiche (Paris, France) is one of the best
I've seen. Some of these are easy and some are quite involved. Feel
free to send partial solutions. Part 2 will be posted next week.

With the four digits 1, 9, 9, 8 used in this order and considered
separately or juxtaposed (i.e. 19 or 998) and the symbols
+ ,  , x , / , sqrt, ! , ^ , (...),
build up the longest possible list of integer numbers starting from 1,
without interruption.
As an example:
1 = (1^9)*(98)
2 = 19  9  8
3 = 1*sqrt(9) * (98)
etc..

With the 9 digits 1, 2, 3, 4, 5, 6, 7, 8, 9 used in this order and
considered separately and the symbols above mentioned, we can express
1998 as follows:
1998 = (1+2) * 3/4 *(5*6 + 7)*8*sqrt(9).
Find with the minimum of symbols the 9 equations expressing 1998
by the successive cancellation of one digit from 1 to 9, the other 8
digits being used in the ascending order.

With the above mentioned symbols, express 1998 by using one digit,
that means 9 different formulas to be built up.
As an example, we can calculate 2048 with the digit 4 as follows:
2048 = sqrt(4)*4*4^4
Try to use the smallest number of symbols needed (i.e. not 1+1+...+1=1998).

Find a, b, c, d positive integers (d>c>b>a), all < 10000 such that:
1/a + 1/b + 1/c + 1/d = 1/1998 in the following cases:
 (da) maximum
 (da) minimum
 a+b+c+d maximum
 a+b+c+d minimum
 a*b*c*d maximum
 a*b*c*d minimum

Is it true that 11111^99999 + 99999^88888 is divisible by 1998?
Same question with 111111^999999 + 999999^888888 ?

1998 cards are numbered from 1 to 1998 and are placed in a circle in
this order. Beginning with card 2, we eliminate the cards placed in
an even position, continuing around the circle
until there is one card. What is the number of this card?
Let's consider now a pack of N cards. With the same process above
described, the remaining card is numbered 1998. What is (are) the
possible value(s) of N? What if we choose every other card, starting
with the first card?

1998 cards are numbered from 1 to 1998 and are placed in a line in this
order. The cards are face down (i.e. the numbers are invisible).
In a first step, we turn over all the cards, that means all the numbers become
visible. In a second step, we turn over 1 card out of 2 starting with
the second card. In a third step, we turn over 1 card out of 3 starting
with the third card, and so on......
In the 1998th step, we turn over the 1998th card.
What are the visible numbers at the end this process?
Sources:
Pierre Tougne (Pour la Science) (puzzles 1,2,3,8,13),
Philippe Fondanaiche (puzzles 4,5,11,12), or derived
of many collections of mathematical puzzles
Solutions:
 From Alexander Doskey, the numbers from 1 to 100:
1 = 1*SQRT(9)*SQRT(9)8
2 = 1+SQRT(9)*SQRT(9)8
3 = 1*SQRT(9)*(98)
4 = 1+SQRT(9)*(98)
5 = 1+SQRT(9)+(98)
6 = 19/9+8
7 = 1*9/9+8
8 = 1*(9/9)*8
9 = 1*9/9+8
10 = 1*9+98
11 = 1+9+98
12 = 1+9/SQRT(9)+8
13 = 1+SQRT(9)+SQRT(9)+8
14 = 19+SQRT(9)8
15 = 1*(9+SQRT(9)*8)
16 = 19+SQRT(9)*8
17 = 1*SQRT(9)*SQRT(9)+8
18 = 199+8
19 = 19*(98)
20 = 19+98
21 = 1*(SQRT(9)+SQRT(9)*8)
22 = 1+(SQRT(9)+SQRT(9)*8)
23 = 1+(9/SQRT(9))*8
24 = 1*(9/SQRT(9))*8
25 = 1+9+9+8
26 = 1*9+9+8
27 = 1+9+9+8
28 = 1+SQRT(9)+SQRT(9)*8
29 = FACT(1+SQRT(9))SQRT(9)+8
30 = 19+SQRT(9)+8
31 = 1+FACT(FACT(SQRT(9)))/(SQRT(9)*8)
32 = (1SQRT(9)+FACT(SQRT(9)))*8
33 = 19+FACT(SQRT(9))+8
34 = 1+9+SQRT(9)*8
35 = 1*9*SQRT(9)+8
36 = 19+9+8
37 = 1+FACT(SQRT(9))*FACT(SQRT(FACT(9)/FACT(8)))
38 = (1+9)*SQRT(9)+8
39 = 1*9+FACT(SQRT(9))*8
40 = 19+FACT(SQRT(9))*8
41 = 1FACT(SQRT(9))+FACT(SQRT(9))*8
42 = 1*FACT(SQRT(9))+FACT(SQRT(9))*8
43 = 19+SQRT(9)*8
44 = 1+9*(SQRT(9)+8)
45 = 1*9*(SQRT(9)+8)
46 = 1+9*(SQRT(9)+8)
47 = 1+(9SQRT(9))*8
48 = 1*(9SQRT(9))*8
49 = 19*SQRT(9)8
50 = 1+SQRT(9)+FACT(SQRT(9))*8
51 = 1*SQRT(9)+FACT(SQRT(9))*8
52 = 1+SQRT(9)+FACT(SQRT(9))*8
53 = 19+9*8
54 = 1*FACT(SQRT(9))+FACT(SQRT(9))*8
55 = 1+FACT(SQRT(9))+FACT(SQRT(9))*8
56 = 1+9+FACT(SQRT(9))*8
57 = 1*9+FACT(SQRT(9))*8
58 = 1+9+FACT(SQRT(9))*8
59 = 1 + !(!sr(9)) / !sr(9) / cr(8) (Denis Borris, using cube root)
60 = (1+9)*(FACT(SQRT(FACT(9)/FACT(8))))
61 = 1+FACT(SQRT(9))*9+8
62 = 19+9*8
63 = 1*9+9*8
64 = 19+9*8
65 = 19*SQRT(9)+8
66 = 1*FACT(SQRT(9))*(SQRT(9)+8)
67 = 1+FACT(SQRT(9))*(SQRT(9)+8)
68 = 1SQRT(9)+9*8
69 = 1*(SQRT(9))+9*8
70 = 1SQRT(9)+9*8
71 = 1+SQRT(9)*SQRT(9)*8
72 = 1+9*98
73 = 1*9*98
74 = 1+9*98
75 = 1*SQRT(9)+9*8
76 = 1+SQRT(9)+9*8
77 = 1+FACT(SQRT(9))+9*8
78 = 1*FACT(SQRT(9))+9*8
79 = 19+98
80 = 1+9+9*8
81 = 1*9+9*8
82 = 1+9+9*8
83 = 1+FACT(SQRT(9))*(FACT(SQRT(9))+8)
84 = 1*FACT(SQRT(9))*(FACT(SQRT(9))+8)
85 = 1+FACT(SQRT(9))*(FACT(SQRT(9))+8)
86 = 1  sr(9) + !(!sr(9)) / 8 (Denis Borris)
87 = 1 + !(!sr(9)) / 9 + 8 (Denis Borris)
88 = 1+9*9+8
89 = 1*9*9+8
90 = 1+9*9+8
91 = 19+9*8
92 = 1*FACT(SQRT(9))+98
93 = 1FACT(SQRT(9))+98
94 = 1SQRT(9)+98
95 = 1+(9+SQRT(9))*8
96 = 1*(9+SQRT(9))*8
97 = 1+(9+SQRT(9))*8
98 = 1+9*(SQRT(9)+8)
99 = 1*9*(SQRT(9)+8)
100 = 1+9*(SQRT(9)+8)
 From Alexander Doskey:
no 1's 1998 = 2*3*FACT(4)*(5*6+7)/8*SQRT(9)
no 2's 1998 = 1*3/4*(5*6+7)*8*9
no 3's 1998 = 1*2*FACT(4)*(5*6+7)/8*9
no 4's 1998 = 1*2*3*(5*6+7)/FACT(8)*FACT(9)
no 5's 1998 = 1*2*3*(FACT(4)+6+7)/FACT(8)*FACT(9)
no 6's 1998 = FACT(1+2)*(FACT(3)+FACT(4)+7)/FACT(8)*FACT(9)
no 7's 1998 = (1+2)*(3+4+5*6)*CBRT(8)*9
Oops, cuberoot not allowed...
no 7's 1998 = FACT(1+2)*(34+5*6+8)*9
no 8's 1998 = 1/2*3*4*(5*6+7)*9
no 9's 1998 = (1+2)*FACT(3)*FACT(4)*(5*6+7)/8
 From Alexander Doskey (he says this can be optimized):
all 1's 1998 = FACT(1+1+1)*(1+1+1)*(1+1+1)*(FACT(1+1+1)*FACT(1+1+1)+1) 161's 18ops
all 2's 1998 = FACT(2+2)/(2+2)*(2*2*2+2/2)*((2*(2*2)^2)+2*2+2/2) 172's 17ops
all 3's 1998 = FACT(3)*3*3*(FACT(3)*FACT(3)+3/3) 73's 9ops
all 4's 1998 = FACT(4)/4*(4+4+4/4)*(4*4*SQRT(4)+4+4/4) 124's 13ops
all 5's 1998 = (5*(5+5+5/5)5/5)*(5*(5+5/5)+5+5/5+5/5) 165's 15ops
all 6's 1998 = (6*(6+6/6+6/6+6/6))*(6*6+6/6) 126's 11ops
all 7's 1998 = (7*7+77/77/7)*((77/77/7)*(77/7)+7) 167's 15ops
all 8's 1998 = ((88/8)*88/88/8)*(8+8+8+8+88/88/88/8) 198's 18ops
all 9's 1998 = (9+9+9+9+9+9)*(9+9+9+9+9/9) 129's 11ops

From Paul Botham 4/24/03:
1/a + 1/b + 1/c + 1/d = 1/1998
all a,b,c,d < 10,000

The only solutions in the required range are
a b
c d
A 6660 6993 9324 9990
B 6660 7992 8140 9768
C 6438 7992 8584 9657
(Added by Dave Peck 05/13/03)
D 6660 7884 8103 9990
E 7425 7474 7575 9999
F 6831 7326 8510 9990
G 6435 7722 8658 9990
H 6045 8370 8658 9990
I 5957 8510 8694 9990
J 5850 8658 8775 9990
K 5920 8370 8928 9990
L 6290 7548 9180 9990
M 5920 8160 9180 9990
N 5920 7920 9504 9990
O 5538 8658 9585 9990
P 5535 8658 9594 9990
which gives
P=max(da)=max(a+b+c+d)=max(abcd)
E=min(da)=min(a+b+c+d)=min(abcd)
 From Bert Sevenhant:
11111^99999 + 99999^88888 is not divisble by 1998 since it is obviously
not divisble by 9 (11111 isn't).
111111^999999 + 999999^888888 is divisble by 1998 since it is divisble by
111/3 (obvious) 9 (111111 is divisble by 3, so after multiplying it with
itself more then 3 times it gets divisble by 27, and finally the whole
thing is even, as sum of two odd numbers.
 From Bert Sevenhant (with a slight edit by
jrhowell@ix.netcom.com, 6/17/99):
If there were 1024 cards in the circle and you
start with taking away the cards you get as answer 1.
After taking away 974 cards we are in this situation. The next
card in the
circle is 1949, and since we take 1950 away as following '1949' is the
answer! (Birdsinger@aol.com concurs it will be 1949.)
[Check out the
Josephus Problem Solution for a method for
solving this.  KD]
In general you can always find a power of 2 between N and N/2, and repeat
this. Of course starting with 1 only changes the answer 1.
For any N the 'even' process never will give an 'even' answer, so the next
question should be answered there is not such N.
For the odd process we get the even proces, minus 1 so we get that 1998 in
the even proces would be the last to take away, when coming to a power of
2. So N999 has to be a power of 2. Thje answer seems to be all N such
that N999 is a power of 2. But that is only true if N999 is the greatest
power of 2 smaller then N so that 2023 is the smallest answer. (this is
logical since 1998 wouldn't exists for smaller numbers such that N999 is
a power of 2.
 Perfect squares less than 1998 will be visible. (They're the only
numbers with an odd number of divisors.)
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