Solutions to Puzzles to Puzzle you
1. Circle the Earth:
We know that the circumference of a circle is given by 2 pi R. So the length of Lekha's string is just the circumference, L = 2 pi R, where R is the radius of the Earth. Now, Pari increased the length by 10 m, so the total length of the string is L+10. When spread around the Earth it makes a circle of radius r, whose circumference is L+10. So L+10=2 pi r1. But L itself is 2 pi R, so substituting for L, we have L+10=(2 pi R)+10 = 2 pi r. So 10 = 2 pi (r-R), or r - R = 5/pi = 1.6 m roughly. So the extra radius or height of the new circle with the longer string is 1.6 m, just enough for you to duck under, if you are of medium height. Isn't it puzzling that the answer did NOT depend on the radius of the earth?
2. What's the probability?
Typically, if the stick is nice and thin and easy to break, you should be able to break it roughly in half, so the average length of each piece would be 50 cm. If you ask this question strictly mathematically, then there is an equal chance to break off any length at all if you are completely unbiased about it. Since the stick is one metre long, the longest length of the shorter half is 0.5 m. So, any answer between 0 and 0.5 m is possible, so on the average the shorter piece would be 0.25 cm long.
3. Non-linear candles
The key point to note is the following: although a fuse may take 5 minutes to burn half-way and then 55 minutes for the other half, it takes exactly 60 minutes or one hour for them to burn all the way. So if you burn a fuse from both ends, it will take exactly 30 minutes to burn through (although the two flames may not meet in the middle because of this unequal burning).
Once you see this point, the solution is clear: Light one fuse at both ends. At the same time, light another fuse at one end. After exactly 30 minutes, the first fuse is completely burned out and the second fuse has burned for 30 minutes as well (Note that the second fuse may not be burnt half-way through but that doesn't matter). As soon as the first fuse finishes burning, light the other end of the second fuse. Now both ends of the second fuse are burning, and it had already burned for 30 minutes. So with both ends burning, the second fuse will burn out in 15 minutes. The total time after the first and second fuses burn out is 30+15=45 minutes, as required.
4. Alphabet multiplication
ABCD × E = DCBA
. There is no option but to try all possibilities. Since anything multiplied by 1 is the number itself, E cannot be 1. So we should try E=2,3,...9. Also, the numbers are between 1 and 9, and they are all different, so the maximum value of any one multiplication is 9x8=72, that is, a 2-digit number. Also, the maximum value in the 10s place is 7.
. Let us separate the multiplication: DxE = M, where M has at most 2 digits of which the rightmost one is A since the answer has A in the rightmost place. Let us write M=10X + A, where X is at most 7. So DxE=10X+A. Now this X in the 10s place gets carried over and adds to the next multiplication which is CxE. So CE+X=10Y+B, where Y is the value of the answer in the 10s place. Similarly, BxE+Y=10Z+C.
. Finally, AxE+Z=D. There is no carry forward since the answer is also 4 digits, so AxE (with a possible carry of Z) is less than 10. Let us start with the last equation. E cannot be 1. If E=2, maximum value of A is 4 otherwise there will be a carry. For A=1,3,4 (A=2 is not allowed since all digits are different), we have AxE=2,6,8, so the possible solutions for D are D=2,6,8, with Z=0, with D=2 not allowed. So we have the choices (A,D)=(3,6),(4,8). So DxE=(6x2,8x2)=(12,16) but this does not satisfy DxE=10X+A. Hence, E=2 is not allowed. (If you take paper and pencil and try it out, it is quite simple). Going through all possible values of E which are consistent with both AxE=D and DxE=10X+A, we find only two possible solutions: (A,E,X)=(2,4,3) or (1,9,8). But for the second case with A=1, DxE=10X+A, so D=9, which is equal to E and is not allowed. So there is only one possibility, (A,E,X)=(2,4,3). From this, we use AxE+Z=D to get 8+Z=D. Since there is no carry, Z=0,1. If Z=0, D=8. If Z=1, D=9 which is not allowed. So we have Z=0,D=8.
. Now look at the other two equations. CxE+X= 10Y+B and BxE+Y=10Z+C with Z=0. Substituting the known values, we have 4C+3=10Y+B; 4B+Y=C. Substituting C=4B+Y in the first equation, we have (16B+4Y)+3=10Y+B or 15B+3=6Y. We can see that B is less than 6 otherwise it will become a 3-digit number. Already (A,D,E)=(2,8,4). So B can be one of 1,3,5. The corresponding solutions for Y are (18/6,45/6, 78/6)=(3,7.5,13). Since Y is an integer less than 10, the only solution is Y=3, so B=1. Substituting in 4B+Y=C, we have C=7.
. Hence our number is ABCD=2178, and E=4. Checking, 2178x4=8712=DCBA!