Download Mindstretching Puzzles by Terry H. Stickels PDF
By Terry H. Stickels
Includes ninety puzzles chosen to extend your problem-solving powers. They comprise cryptarithms, asphmetics, simultaneous equations and layered considering - extra complex puzzles than commonplace, difficult you to exchange letters with digits, holiday codes and lots more and plenty extra. solutions are integrated.
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Additional info for Mindstretching Puzzles
Example text
His logic brainteasers offer a clear, straightforward presentation of the puzzle, yet fully test the deductive reasoning process of even the best puzzle enthusiasts. His book The Great Book ofMind Teasers & Mind Puzzlers will keep you busy for days. 36 In one of his creations, which could be called the " letter cross," letters represent numbers, and you must make several deductions to come up with the value of each letter. Here is a version of a letter cross puzzle. Although not particularly difficult, it still requires several steps for its solution.
Its remainder is 7 when divided by 10. 14. Normally, newspapers are printed on one large sheet. In a finished section, the first and second pages are printed on half of the sheet, and the second to last and last page are on the front and back of the other half. Therefore, the final page of a section of newspaper is usually a multi ple of 4. In this case, pages 1 and 2 are attached to 39 and 40 (since the section contains 40 pages). The rest of the pages are attached like this. 1-2 3-4 5-6 7-8 9-10 11-12 13 15-16 17-18 19-20 @ 39-40 37-38 35-36 33-34 3 1-32 29-30 (27-28) 25-26 23-24 21-22 The three missing pages are 14, 27, and 28.
The equation then becomes B+ P+ T+ A=24 or B+ P + T =22, since A =2. Then we have: + B+ P + 2=T - B- P+ 22=T 24 =2T So, T=12. Now, 12+ A= 14=E and Q- 12=17. So, Q =19, and, therefore, S =5. 5. View C is not correct. 57 6. Besides the one shown in Puzzle 5 and the two in this puzzle, eight other ways are possible. 58 7. It is always helpful to set up a legend of what is given and to work from there. 50Z = 100 Now, we need at least one of the values to drop out in order to consider the other two.