At Last!! Encoded Totals Second Addition

The Long-awaited Sequel to Have Some Sums to Solve

Steven Kahan

• Publisher: Routledge
• Year: 2019
• Print ISBN:
  9780415785747
• eBook ISBN:
  9781351845823
• Topic: Mathematics
• Subtopic: Cryptography
• Index: Mathematics

How long have you been thirsting for tempting, tantalizing teasers, craving for challenging cryptographic conundrums? A sequel to ""Have Some Sums to Solve"", this work can satiate the desires of even the most prolific puzzle enthusiast.

SECTION 1

PUZZLES

even odds

directed approach: page 86

solution: page 115

even odds

Two words that have different sounds and meanings but have identical spellings are called heteronyms. If you ran across a word such as “sewer” out of context and were asked to pronounce it, you would have only

of doing so correctly.

Decode this alphametic without using the digit zero in your solution. Then consider doing something more.

THE OBJECT OF THIS QUIZ IS TO RECORD ALL OF THE HETERONYMS THAT APPEAR IN THIS PARAGRAPH. WHERE DOES THIS SEARCH LEAD? WELL, IF YOU READ VERY, VERY CAREFULLY, AND YOU USE ALL OF YOUR POWERS OF OBSERVATION, YOU SHOULD HAVE NO EXCUSE BUT TO FIND THE TOTAL NUMBER OF HETERONYMS PRESENT HERE. IF YOU WIND UP GETTING THEM ALL, TAKE A BOW; IF NOT, DON’T SHED A TEAR!

How many have you discovered?

go for the gold

directed approach: page 86

solution: page 117

go for the gold

Here’s a chance to be rewarded for cleverness. You are confronted with nine identical-looking bags, each of which contains ten gold nuggets. If the truth be told, however, only one of these bags contains real gold; the other eight are filled with worthless fool’s gold. A nugget of fool’s gold is known to weigh exactly 1 ounce, whereas a nugget of real gold weighs precisely

of an ounce. Available to you is an ordinary single-pan scale, with which you are permitted to perform one and only one weighing. With nine bags before you, you can reap quite a windfall if you can determine

The product of four consecutive positive integers can never be equal to a perfect square.

i.d.t. – ninety

directed approach: page 87

solution: page 119

i.d.t. – twenty

directed approach: page 87

solution: page 116

i.d.t. – ninety

i.d.t. – twenty

wholes with holes

directed approach: page 88

solution: page 114

wholes with holes

A perplexing geometric construction problem involves the positioning of four congruent quadrilaterals to create various configurations. Trace the shape shown below and cut out four copies of it.

Then arrange these four pieces into the following four patterns:

a. a square with a small square cut from its center;

b. a square with a large square cut from its center;

c. a rectangle with a rectangle cut from its center; and

d. a parallelogram with a parallelogram cut from its center.

If you accept this challenge, then

i.d.t. – sixtyone

directed approach: page 88

solution: page 120

i.d.t. – sixtyone

For no other value of n is it true that the sum of the first n perfect squares is itself a perfect square.

1,634 is the smallest four-digit number that equals the sum of the fourth powers of its digits.

54,748 is the smallest five-digit number that equals the sum of the fifth powers of its digits.

548,834 is the smallest six-digit number that equals the sum of the sixth powers of its digits.

1,741,725 is the smallest seven-digit number that equals the sum of the seventh powers of its digits.

24,678,050 is the smallest eight-digit number that equals the sum of the eighth powers of its digits.

division decisions

The number 27,720 has the distinction of being the smallest positive integer that is divisible by each of the first twelve counting numbers. (A considerably larger number with this property is 12! = 479,001,600.) Based upon the notion of congruence, tests are available that enable us to predict divisibility without actually performing the division. Let us introduce some notation.

We will write N = “a_na_n−1a_n−2 … a₂a₁a₀” to represent the integer under consideration, where a₀ is its units digit, a₁ its tens digit, and so forth. Then N is divisible by

2 if and only if a₀ is even;

3 if and only if a_n + a_n−1 + a_n−2 + … + a₂ + a₁ + a₀ is divisible by 3;

4 if and only if “a₁a₀” is divisible by 4;

5 if and only if a₀ = 0 or 5;

6 if and only if it is divisible by both 2 and 3;

8 if and only if “a₂a₁a₀” is divisible by 8;

9 if and only if a_n + a_n−1 + a_n−2 + … + a₂ + a₁ + a₀ is divisible by 9;

10 if and only if a₀ = 0;

11 if and only if a_n − a_n−1 + a_n−2 − … + (−1)ⁿa₀ is divisible by 11; and

12 if and only if it is divisible by both 3 and 4.

The divisibility test for 7 is not very well known. Specifically, N is divisible by 7 if and only if “a₂a₁a₀” – “a₅a₄a₃” + “a₈a₇a₆” – … is divisible by 7, where one or two leading zeros need to be introduced if the number of digits of N is not a multiple of 3. For instance, 3,511,539,579,751 must be divisible by 7, since 751 − 579 + 539 − 511 + 003 = 203 is divisible by 7.

division decisions

directed approach: page 89

solution: page 115

i.d.t. – fiftyone

directed approach: page 89

solution: page 113

The nicest part about this statement is that

Appropriately, use the last stated test to determine the solution in which DIVIDED is divisible by 7.

i.d.t. – fiftyone

optical allusion

directed approach: page 90

solution: page 116

optical allusion

Appearances can certainly be deceiving, as this reprint from a pre-1900 advertisement shows.

If your eyesight and insight are strong, you can literally turn this little frog into another of nature’s creatures. It’s not a prerequisite, but it certainly wouldn’t hurt if you happened to have

open and shut case

directed approach: page 90

solution: page 119

open and shut case

An innovative high school mathematics teacher posed the following problem to his honors class:

“There are one thousand lockers in our school, numbered consecutively from 1 to 1000. At the beginning of the school day, all of the locker doors are closed. Students enter the school in single file. The first student to enter will open each and every locker door. The second student to enter will then close all of the doors on the even-numbered lockers. The third student will next alter the status of the doors on all lockers ...

Table of contents

• Cover
• Half Title
• Title Page
• Copyright Page
• Dedication
• Table of Contents
• foreword
• preface
• narrative alphametics
• ideal doubly-true alphametics
• about the author