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The beginning of a well known book has been encrypted below. Each letter of the alphabet has been replaced by another using a simple substitution technique. For example, A may always be represented by G, etc. Can you name the book?
Mhk pyodm gzbak mhbm Y abr jkzz okqkqlko jbd b zbowk gzkbdbrm qkbcej jymh b gerc ep azkbo jbmko yr ym. Deqk dhbcx mokkd zkbrkc eiko ym, brc otdhkd brc jbmkozyzykd wokj bm mhk ckkg krc. Eiko mhk hkcwk er erk dyck jk zeenkc yrme b gzetwhkc pykzc, brc er mhk emhko jk zeenkc eiko b wbmk bm eto qbdmko'd hetdk, jhyah dmeec lx mhk oebcdyck; bm mhk meg ep mhk qkbcej jbd b woeik ep pyo mokkd, brc bm mhk lemmeq b otrryrw loeen eikohtrw lx b dmkkg lbrn.
Jhyzk Y jbd xetrw Y zyikc tger qx qemhko'd qyzn, bd Y aetzc rem kbm wobdd. Yr mhk cbxmyqk Y obr lx hko dyck, brc bm rywhm Y zbx cejr azedk lx hko. Jhkr ym jbd hem jk tdkc me dmbrc lx mhk gerc yr mhk dhbck ep mhk mokkd, brc jhkr ym jbd aezc jk hbc b ryak jboq dhkc rkbo mhk woeik.
The piece of text is from the original book and should read:
The first place that I can well remember was a large pleasant meadow with a pond of clear water in it. Some shady trees leaned over it, and rushes and waterlilies grew at the deep end. Over the hedge on one side we looked into a ploughed field, and on the other we looked over a gate at our master's house, which stood by the roadside; at the top of the meadow was a grove of fir trees, and at the bottom a running brook overhung by a steep bank.
While I was young I lived upon my mother's milk, as I could not eat grass. In the daytime I ran by her side, and at night I lay down close by her. When it was hot we used to stand by the pond in the shade of the trees, and when it was cold we had a nice warm shed near the grove.
a has been represented by b.
b has been represented by l.
c has been represented by a.
d has been represented by c.
e has been represented by k.
f has been represented by p.
g has been represented by w.
h has been represented by h.
i has been represented by y.
j has been represented by s.
k has been represented by n.
l has been represented by z.
m has been represented by q.
n has been represented by r.
o has been represented by e.
p has been represented by g.
q has been represented by u.
r has been represented by o.
s has been represented by d.
t has been represented by m.
u has been represented by t.
v has been represented by i.
w has been represented by j.
x has been represented by f.
y has been represented by x.
z has been represented by v.
Puzzle 46
Cen yee rell wrer er wreng werr rrer peregrepr? Er ferrr glence yee mey rrenk rrer jerr rre vewelr reve preblemr, ber ne! Rrree cenrenenrr elre reve preblemr, rrey reve been repleced werr rre lerrer R, ber wrecr ener ere rrey?
Hint: It's not just the vowels that have a problem.
Answer: H, S and T.
It should read:
Can you tell what is wrong with this paragraph? At first glance you may think that just the vowels have problems, but no! Three consonants also have problems, they have been replaced with the letter R, but which ones are they?
Each number in the sequence is a representation of the number 32 in different bases, starting with base 10.
Base 10: 32 = 3 x 10^1 + 2 x 10^0 = 3 x 10 + 2 = 32
Base 9 : 35 = 3 x 9^1 + 5 x 9^0 = 3 x 9 + 5 = 32
Base 8 : 40 = 4 x 8^1 + 0 x 8^0 = 4 x 8 + 0 = 32
Base 7 : 44 = 4 x 7^1 + 4 x 7^0 = 4 x 7 + 4 = 32
Base 6 : 52 = 5 x 6^1 + 2 x 6^0 = 5 x 6 + 2 = 32
Base 5 : 112 = 1 x 5^2 + 1 x 5^1 + 2 x 5^0 = 1 x 25 + 1 x 5 + 2 = 32
Base 4 : 200 = 2 x 4^2 + 0 x 4^1 + 0 x 4^0 = 2 x 16 + 0 + 0 = 32 (as required)
Puzzle 48
During a recent Formula 1 race meeting, Sterling managed to complete a lap with an average speed of 150 mph. He managed to complete the first two fifths of the lap length at a speed of 123 mph and the second two fifths of the lap length at a speed of 164 mph. At what speed was the final one fifth of the lap length covered  the answer is not 176 mph?
we have had a lot of queries on this puzzle and many methods have been suggested but only the one below gives the correct answer.
If we take the number of miles in a lap to be x miles and using time = distance / speed.
The first 2/5 are covered at 123 mph. Therefore, the time taken was [(2/5) * x] / 123 hours.
The next 2/5 are covered at 164 mph. Therefore, the time taken was [(2/5) * x] / 164 hours.
The last 1/5 is covered at y mph. Therefore, the time taken was [(1/5) * x] / y hours.
The entire trip is covered at an average of 150 mph. we can conclude that:
2x 2x x x
 +  +  = 
5*123 5*164 5*y 150
Working this out we get:
1 2 2 1
     = 
30 123 164 y
The means that y = 1 / (1/30  2/123  2/164)
Therefore y = 205.
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