*many*math and logic puzzles which he had collected or made up and delighted in sharing.

From the time I was in eighth grade through my college years, Papa, as all the grandchildren referred to him, wrote weekly letters that he then photocopied at a local library and sent individually to his two daughters and all the members of his extended family. I saved most of his letters in cardboard shoeboxes, but inevitably many were lost. Fortunately, my mother and Aunt have since recovered, organized, and retyped all the handwritten letters, and with the help of my jack-of-all-trades cousin, they will soon self-publish these letters as a 350-page book for family members here and abroad. The letters capture the essence of a remarkable man, and the book will surely be passed down to my children, and someday from them to theirs.

Papa came from a respected Orthodox family in London. His father, born in Mogilev in Eastern Belarus,was a Torah scholar who served as the

*Rosh Hashochetim*(head trainer, inspector, and

*Kabbala*[license] provider for kosher meat production) for all of England and Ireland. Rav Kook, then chief rabbi in Palestine and one of the great Torah scholars of his time, wrote a glowing endorsement of my great-grandfather which might have helped him secure his position.

Papa left England as a young man, worked at a Hebrew Orphanage, taught math at Temple University, and later became an actuary. If you own a variable annuity or universal life insurance policy, then you have my grandfather to thank (or blame), because he pioneered the development of both. In his obituary, the Council of Actuaries credited him with developing the concept of a flexible life insurance plan that was eventually marketed as universal life. Additionally, he published the first actuarial paper on the variable annuity. Above all, Papa was a philosopher, a man who loved to share his mastery of logic and reasoning through clever puzzles and thought-provoking essays.

In another column, I will share snippets of Papa’s thoughts about a range of ethical and philosophical issues. For now, here are six of his countless math and logic puzzles which he shared with his extended family for them to ponder in their spare time. Papa invented some of these puzzles, and found others in puzzle books. Incidentally,

*most*of them can be solved with basic arithmetic skills, which

*doesn't*mean that all of them are easy (at least not for me).

If you enjoy puzzles I invite you to try to solve at least one of them! If you want to post an answer or explanation, please refer to the puzzle by number.

______________________

Puzzle 1 – HOW LONG IS THE SHELF

A shelf is exactly filled with books of equal thickness.

If the books were 1” thinner, the shelf would accommodate 6 more books.

If the books were 1” thicker, then there would be no room for 3 books.

How many inches long is the shelf?

______________________

Puzzle 2 – HOW OLD ARE BERT'S CHILDREN?

Bill is the Insurance Agent. Bert is the prospect for Insurance.

Bill: So you are just 40 – with three kids. How old are they?

Bert: Figure it out for yourself. The three ages add up to the street

number of this house. If you multiply their ages together, the

result will be my age.

Bill: I still can’t tell their ages.

Bert: Forget it then. The two elder kids will be walking back from

school now; so you will meet them.

Bill: That’s all I need to know.

Bill then gave the three ages without delay.

What are the three ages?

______________________

Puzzle 3: HANGED OR DROWNED?

A man has committed a crime punishable by death.

He is to make a statement.

If the statement is true, he is to be drowned.

If the statement is false, he is to be hanged.

What statement did he make to confound his executioners?

______________________

Puzzle 4: STRAIGHTFORWARD MATH?

What number leaves a remainder of 1 when divided by 2, 3, 4, 5, 6, 7, 8 and 9, respectively, but leaves no remainder when divided by 11?

______________________

Puzzle 5: MEASURING THE BOOKWORM

On January 1, I started to read a book and by reading the same number of pages each day of the month, I managed to finish it on the 31st of January (include the 31st).

If I had started by reading a quarter of that number of pages on January 1, and on each succeeding day, one page more than on the preceding day, I should also have finished it on January 31 (include the 31st).

How many pages did the book contain?

______________________

Puzzle 6: WHO DONE IT?

Four men were eating dinner in a restaurant when one of them suddenly struggled to his feet, cried out: “I’VE BEEN POISONED” and fell dead.

His companions were arrested on the spot and under questioning made the following statements, exactly one of which is false in each case.

WATTS: I didn’t do it

I was sitting next to O’NEIL.

We had our usual WAITER today.

ROGERS: I was sitting across the table from SMITH.

We had a new WAITER today.

The WAITER didn’t do it.

O’NEIL: ROGERS didn’t do it.

It was the WAITER who poisoned SMITH.

WATTS lied when he said we had our usual WAITER today.

Assuming that only SMITH’s companions and the WAITER are implicated, WHO WAS THE MURDERER?

## 12 comments:

Brilliant stuff, David. Sounds like your grandfather was a wonderful man. I need to spend some time on these!!

I would reword the problem a little:

A shelf is exactly filled with books of equal thickness.

If the books were 1" thinner, the shelf would accommodate 6 more books (BUT NOT SEVEN).

If the books were 1" thicker, then there would be no room for 3 books (BUT IT COULD FIT TWO).

How many books are on the shelf?

Answer: 4 books are on the shelf. The books are between 11/7 and 10/6 inches long. The

shelf is between 44/7 and 40/6 inches long.

Let N be the number of books, T be their thickness, and L=N*T be the length of the shelf.

The 1" thinner condition tells us that:

(N+6)*(T-1) <= L=N*T < (N+7)*(T-1)

This can be rewritten as:

[*] 1 + N/7 < T <= 1 + N/6

The 1" thicker condition tells us that:

2*(T+1) <= L=N*T < 3*(T+1)

The can be rewritten as:

[**] 2/(N-2) <= T < 3/(N-3)

There can't be 3 or fewer books on the shelf, because then the upper bound in [*] tell us that T is smaller (or equal to) 1 1/2 and the

lower bound from [**] says that T is bigger (or equal) to 2.

Similarly, there can't be 5 or more books on the shelf, because then the lower bound in [*] tells us that T is larger than 12/7 and the upper bound in [**] tells us that T is smaller than 3/2.

So there must be 4 books on the shelf. Condition [*] tells us that 11/7 < T <= 10/6. Condition [*] tells us that 1/2 <= T < 3.

-----

Here is a variant of the above problem suitable for interest elementary schoolers.

A shelf is exactly filled with books of equal thickness. The thickness is some whole number (no fractions) of inches.

If the books were 1" thinner, the shelf would accommodate 2 more books (exactly, with no space left over).

If the books were 1" thicker, the shelf would accomodate one fewer books (exactly, with no space left over).

How many books are on the shelf? How thick are the books? How long is the shelf?

2: 8, 5, and 1 it seems.

3: "I will be hanged."

Will try to look at the rest when I have more time.

Thanks - to you and your grandfather. If he was around now he probably would be blogging.

Steve

I should have said in my previous post that your grandpa seems fascinating and had a good selection of problems. (Also I should have given a spoiler alert that my response was a solution to Puzzle 1.)

So now I give a spoiler alert that the following is my solution for Puzzle 4.

The answer is of form 1+ x*(5*7*8*9) for x chosen so that answer is a mutiple of 11. For this to be the case, the remainder of x*(5*7*8*9) divided by 11 must be -1. Since the remainder of 5*7*8*9 divided by 11 is 1, we can take x=-1. For x=-1 we get the answer -2519. We can add any multiple of 11 to x. For x=10, the we get the answer 25201.

b is number of books

w is width of a book

the width of the shelf is thus b * w

the width of the shelf is ALSO (b+6) * (w - 1)

the width of the shelf is ALSO (b-3) * (w + 1)

so

b* w = (b+6) * (w - 1)

w/(w-1) = (b+6)/b = 1 + 6/b

w/(w-1) - 1 = 6/b

and

b * w = (b-3) * (w + 1)

w/(w+1) = (b-3)/b = 1 - 3/b

1 - w/(w+1) = 3/b

2 - 2w/(w+1) = 6/b

combining the previous 2 results, with 2 expressions in w both equal to 6/b

w/(w-1) - 1 = 2 - 2w/(w+1)

multiplying each side by (w-1)*(w+1)

w(w+1) - (w-1)(w+1) = 2(w-1)(w+1) - 2w(w-1)

w*w + w - (w*w - 1) = 2w*w - 2 - (2w*w - 2w)

w + 1 = 2w - 2

w = 3

substituting 3 for w in the first equation

b*3 = (b+6)*(3-1)

b*3 = b*2 + 12

b = 12

so the shelf width = 3 * 12 = 36

Puzzle 3 - Stymie the judges with this: "This statement is false"

Puzzle 2: I also get 8, 5 and 1... tho I suppose it could be 4, 5 and 2. Depends on whetehr you want your 4 year old walking home from kindergarten.

Puzzle 1:

x=thickness of book, in inches

y=number of books

first case:

xy=(x-1)*(y+6)

xy=xy+6x-y-6

6x=y+6

second case:

xy=(x+1)*(y-3)

xy=xy-3x+y-3

3x=y-3

6x=2y-6

combining:

6x=y+6=2y-6

y=12

6x=12+6

x=3

bookshelf=xy=36 inches

Puzzle 2:

combinations yielding 40, when multiplied:

20*2*1 (20 is too old for school, except in parts of Kentucky)

10*2*2 (2 is too young for school, unless your dad is Stephen Hawking)

10*4*1 (4 is too young for school)

5*4*2 (2 is too young for school)

So, it must be the only other unique combination, with two ages being school ages:

8*5*1

Puzzle 3:

No answer... logic is for those who can't think straight.

Puzzle 4:

(2*3*4*5*6*7*8*9)-1=362879

Puzzle 5: (I admit that after all these years I had to look up the formula for a finite series)

x=pages per day

sum of a + kd, for k=0 to n-1, where a=1/4x,d=1,n=31

(n/2)*(2a+(n-1)d)=(31/2)*(2*(1/4x)+(31-1)*1)=31/4x+1860

so,

31x=31/4x+465

124x-31x=1860

93x=1860

x=20

pages in book=20*31=620

Puzzle 6:

See answer for puzzle 3.

Hang my head in shame... I normally pride myself in mastery of patterns and sequences, but I rushed on number 4 (whilst working at night), and didn't test my work... this sort of oversight is what leads to satellites plunging into deep space, if your line of work is avionics for NASA. So, my incorrect response came to me as I was driving into Cambridge this morning. The correct answer is:

(2*3*4*5*6*7*8*9*10)+1=3628801

The remainder piece is obvious, because the number is one greater than a number which is divisible by all factors. The divisible by 11 piece, with no remainder, is only slightly more obtuse.

Puzzle #6

This is actually very easy just some simple logic which they don't teach at school. I leave out some details but you should be able to follow. Let me know before posting.

Consider

Statement:

O1 - O'Neil statement 1

O2

O3

R1

R2

R3

W1

W2

W3

R3 and O3 say the same thing

If both are correct then W3 is false, which means W1 and W2 are true.

If both are false (R3 and O3) then either R2 or O2 is false which cannot be by the rules.

So if W3 is false then Watts did not do it and he was sitting next to O'Neil

This makes R1 false, which means the waiter did not do it by R3

This makes O2 false.

So waiter did not do it, Watts did not do it W1, Rogers did not do it O1.

This leaves only O'Neil.

SPOILER ALERT!

ANSWERS TO FOLLOW!

The responses in the comments section confirmed what I already knew; I have some very math-smart readers! Instead of providing my answers to each question, I will refer to answers and explanations in this comment section (since they are way better than anything I could come up with).

There are 11 comments (including this one); I will refer each comment by its number order.

(Incidentally, you are reading comment #11)

Question 1, “How long is the Shelf?” - ssancetta (#5) and anonymous (#8) gave clear explanations for the problem as most people would read it.

Mathematician Scott Axelrod (#2) noticed an ambiguity in the wording of the problem (like a true mathematician), reworded the problem, and then solved the reworded problem, providing a different answer than #5 and #8. I am sure Scott is right because this kind of problem was already too easy for him in elementary school.

Question 2, “How old are Bert’s children?” - Of all the correct answers, anonymous (#8) gave the most colorful explanation.

Question 3, “Hanged or drowned?” - Two different solutions. Steve (#3) and ssancetta (#6).

Question 4, “Straightforward math?” - I couldn’t figure this out, until my math-smart Russian neighbor explained it to me (yes, sometimes stereotypes are accurate!). Anyway, Scott Axelrod (#4) explains the answer.

Question 5, “Measuring the bookworm.” - Anonymous (#9).

Question 6, “Who done it?” - Anonymous (#10), who plays a mean game of chess, got the same answer as my jack-of-all-trades cousin and I. Interestingly, this is the only puzzle for which we couldn’t find Papa’s answer.

David -

Thanks for the puzzles.

I wanted to add one note to the puzzle of the children's ages. School age isn't really the key factor here. The choices for ages were (see #8):

1,2,20

1,4,10

1,5,8

2,2,10

2,4,5

plus (theoretically) 1,1,40.

The clue is that Bill couldn't tell the ages just by knowing the product and sum. (Bill presumably knows the house number even though we don't; after all, he's at Bert's house.) Only 1,5,8 and 2,2,10 give the same sum, so the house number must be 14 if Bill couldn't guess the ages. However, when Bert mentions "the two elder kids" that eliminates the 2,2,10 combination, leaving 1,5,8.

Post a Comment