The missing number is 13 so that
208 -13 = 15 x 13
A problem of the form
ac ? ab = ca ! ab
where ? and ! are different arithmetic signs was published in The Daily Telegraph (30 October 1993). The only solutions are
27 - 24 = 72 / 24 and 49 - 47 = 94 / 47
Sparky the Electrician
The values are A=D=2, B=3, C=4. The juxtaposed batteries should be B in row 4 column 3, and D in row 2 column 1. Exactly two rows and two columns total 10V, the remaining two rows and two columns being involved in the interchange. We now examine the six combinations where two rows are both 10V and set them equal. For example, for rows 2 and 4, we have: A + C +2D = 2A + C + D so that A = D and use the fact that the total of all batteries is 40V, so that 5(A + D) + 2(2B + C) = 40. By experimenting with possible values we find five solutions : A = D = 1, B = 4, C = 7; A = D = 2, C = 3, D = 4; A = D = 3, B = 2, C = 1; A = B = 2, C = 1, D = 4; or A = B = 3, C = 4, D = 1; only the first three allowing a single juxtaposition. The condition that exactly three batteries total 10V reduces us to the second case ( 4 + 3 + 3).
The path was 5000 paces long. If the path is straightened out into a line, its area is that of the square, 100x100 square paces. Dividing by the path width 2 gives the solution.
Grandma and Trickle
Grandma was 72 years old. Let the two digits be A and B. Then 10A+B-10B-A=45 so 9(A-B)=45. This gives A-B=5. Of the pairs 5,0; 6,1; 7,2; 8,3; 9,4 only 7,2 are both prime numbers
Jack and Jill
Without the smaller jug, the larger one contains 8 units in the position shown. The introduction of the smaller jug as shown displaces 3/4 of its own volume (3 units) leaving 5 units.
It took 4 hours 34 minutes for one cannibal to remain. It took 5 cannibals, 2/5 hours to eat the first victim, four cannibals 2/4 hours to devour the second, three cannibals 2/3 hours for the third, two cannibals 2/2 hours for the fourth and one cannibal 2/1 hours for the fifth. One could argue that stomach contents increase a cannibal's edible volume when he is the victim. However, if the further interpretation is granted that these contents also turn a cannibal as consumer into more than one cannibal, the original solution is restored!
The red boxes totalled $40000, the blue ones $35000 and the green boxes $25000. Suppose the red statement is true then a red and green each have $15000 so that the cases are as follows:
BLUE (false) (a) 25 + 25; (b) (10 + 25)/(10 + 10)
GREEN (false) (a) 15 + 10 ; (b) (15 + 10)/(15 + 25)
RED (true) (a) 15 + 10; (b) 15 + 25
Here the possibility arises for the true statement (red) to have the highest total of the three and the first part of case (b) applies for blue and green. Checking the cases blue true or green true we see the same possibility does not arise.
In descending order, the ranking is B,D,C,A,E; the first two lied. With the particular questions asked, a completely determined order only arises from having exactly one "yes" or one "no" in five truthful answers. So the two wrong answers must be a "no", "no" or a "yes", "no". Only inverting each of the last pair keeps exactly one "yes". So inverting the reported "yes" means either C or D lies. If it's C, only A as the other liar gives exactly two inversions. If it's D then B is the other. Fot these two cases, NNNNY and NYNNN are the deduced correct answers. Only E is trustworthy being in both cases, so his correctly reported "no" gives NYNNN. The solution follows.
The distribution is as follows: A = 2, B = 0, C = 2. The possible combinations can only be 1, 1, 2 or 0, 2, 2. For the first case, to avoid adjacent boxes having the same number, the order must be 1, 2, 1, but then all three numbers are falsely stated (invalid). So it must be the second case. Again to avoid the adjacency condition we must have 2, 0, 2 which allows only C to be true (valid).