Dear All,

“He will be a joy and a delight to you, and many will rejoice because of his birth.”–Bible verse, Luke 1:14 π

Given that we are children of God, created in His image, it’s lovely to get to know one another, and gently help each other to become ever more what our Creator made us to be. When we use well and share our God-given gifts for the benefit of others, there is no limit to the good that can be accomplished.

Staying ‘tuned in’ to the Holy Spirit and coming to know God’s word and His will through Bible study and prayer also helps us to understand more about ourselves and each other. When we then interact with gentle, loving patience and kindness, problems are often solved very happily with miraculous ease! ππ

**Nice Word Puzzle** π

*Use all the letters of ARC SMILE to make a proper 8-letter English word.*

There are (at least) three possible solutions!

I recently promised a friend that this article being offered freely here would include some good, fun number/number-theory puzzles. I wish you lots of enjoyment with them now. πβ

**Power of Friends Brainteaser** π

The ages of five special friends are consecutive whole numbers, represented here by A, B, C, D and E. Andy is the youngest, followed by Beth, Chris, Dee and Eric, the oldest. The five friends are staying together at a large holiday house until 31 May. The house number (on the front door) is represented here by H Λ 1.

A remarkable detail is the fact that **31 x 5 x (H ^{A} + H^{B} + H^{C} + H^{D}) = H^{E} β H^{A}.**

** Your super-fun brainteaser is to figure out, with proof, the exact value of the house number H.** π

**New 2023 Brainteaser** π

** The 2023 brainteaser is to figure out, with proof, all possible positive whole number values for N such that N – 1 is a factor of N^{2} + 2023**.

**Double All-Play-All Chess Brainteaser (with no Chess moves!! πβπ€£)**

Suppose that P players are having a double all-play-all Chess tournament, and P Λ 2. Each player will play against each other player twice: once with White, and once with Black. In each game, if there’s a winner, he/she gets 1 point and the loser 0 points; if the game ends in a draw, then the two players involved receive 0.5 points each. At the end of the tournament, each player’s final score will be the total sum of the points that he/she received in his/her games.

Suppose that the gold medal winner in 1st place actually achieves the very highest possible score π, and the silver medal winner in 2nd place achieves what is then the highest-possible runner-up score (lower than the winner’s score, though).

*It turns out that the number obtained by dividing the winner’s score by the runner-up’s score does not involve more than one distinct, different digit when you write it out.*

**The brainteaser is to figure out, with proof, three possible values for P, the number of players.** ππ

**A Double High-Five Brainteaser!!** ππ

Consider all non-negative whole numbers n with not more than D digits.

*The brainteaser is to figure out (as an expression, with proof) how many of them are such that n^{10} + 1 is exactly divisible by 10.*

**Advanced Algebra Triple Brainteaser**!!! πππ

Suppose that ‘a’ and ‘b’ are distinct non-zero numbers such that

a^{2} + 2ab – 3b^{2} = p & 2a^{2} – 3ab + b^{2} = q.

*The triple challenge is to find a neat, simplified expression for ‘a’ in terms of p, q & b AND to discover three values for p Γ· q which are NOT possible in this brainteaser*

*AND to find a proper, mathematical, English word which contains a, b, p and q (not necessarily in that order) at least once each among its letters!!!*

**Chess Puzzle (with moves this time!! ππ)**

It’s my intention to publish solutions to all the puzzles around the time that blog post #157 comes out, God-willing as always.

In the meantime, please do feel free to send me your best solutions to any or all of the puzzles, if you like π.

I would like to round off this article by most sincerely wishing you a very blessed month of May and a wonderful weekend now, with lots of happiness in everything that you do β€.

**Beautiful Bonus! **πππ

Extra-special wishes to lots of people I know who are celebrating their birthdays today! You can write down *any* proper three-digit whole number that you like (e.g. 523). Repeat it to get a six-digit number (e.g. 523523). Divide by 2 in honour of our friendship. Now divide by 7, the number of letters in FRIENDS. Also divide by 13, the total number of letters in HAPPY BIRTHDAY. To finish, divide by the three-digit number that you started with…

The final result of…π 5.5…is to wish you a beautiful birthday now on 5 May πππ

With kindest wishes as always,

Paul Mπtwani β€

“For shoes, put on the peace that comes from the Good News, so that you will be fully prepared.”–Bible verse, Ephesians 6:15 π

**P.S. = Puzzle Solutions** (being posted on 12.5.2023)

ARC SMILE rearranges to make **MIRACLES** (and you could claim that it makes CLAIMERS or RECLAIMS too!! π).

In the big H brainteaser, we were given that

31 x 5 x (H^{A} + H^{B} + H^{C} + H^{D}) = H^{E} β H^{A}

where A, B, C, D & E are * consecutive whole numbers in increasing order*, and so

155 x H^{A} x (1+H+H^{2}+H^{3}) = H^{A} x (H^{4} – 1)

and ‘cancelling’ the common factor H^{A} on both sides of the equation gives

155 (1+H+H^{2}+H^{3}) = (H^{4} – 1)

and then factorising the expression on the right-hand side leads to

155 (1+H+H^{2}+H^{3}) = (H-1) (1+H+H^{2}+H^{3})

which simplifies to

155 = H-1

so ** H = 156**, which is nice here in blog post #156 π

In the ‘2023 Brainteaser’, we were given that

N – 1 is a factor of N^{2} + 2023, and since N^{2} + 2023 = (N-1)(N+1)+2024,

N-1 must also be a factor of 2024β1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012 or 2024 for N-1,

and so

** N can be any of the sixteen numbers 2, 3, 5, 9, 12, 23, 24, 45, 47, 89, 93, 185, 254, 507, 1013 or 2025**.

In the Double All-Play-All Chess Tournament with P players,

the maximum-possible score (for someone who wins every single game) is 2(P-1), and in that case the maximum-possible runner-up score (for someone who wins every single game apart from the two games lost against the super champion!) is 2(P-2).

The winner’s score divided by the runner-up’s score simplifies to (P-1)/(P-2).

There are three possible values of P for which (P-1)/(P-2) produces positive results that in each case don’t require more than one distinct, different digit when written out. They are ** P = 3 or 11 or 12**, and

**.**

*the corresponding (P-1)/(P-2) values are 2, 1.111… repeated and 1.1 exactly*In the Double High-Five Brainteaser, n^{10} + 1 is exactly divisible by 10

βn^{10} + 1** **ends with a digit 0βn^{10} ends with a 9βn^{5} ends with a 3 or a 7βn itself ends with a 3 or a 7; that is precisely two out of each ten numbers ending with 0 to 9, because only the ones ending in 3 or 7 actually work here. Now, the total number of non-zero whole numbers with not more than D digits is 10^{D}. So, the total number of suitable numbers for n is 10^{D} x 2 Γ· 10,

which simplifies to **2 x 10 ^{D-1}** πβ

In the advanced Advanced Algebra Triple Brainteaser, where

‘a’ and ‘b’ are **distinct non-zero numbers** such that

a^{2} + 2ab – 3b^{2} = p & 2a^{2} – 3ab + b^{2} = q,

factorisation leads to

(a-b)(a+3b) = p & (a-b)(2a-b) = q

so p/q = (a+3b)/(2a-b). Since a is *not *equal to zero, p/q is *not* equal to -3; since b is *not *equal to zero, p/q is *not* equal to 1/2; since a is *not *equal to b, p/q is *not* equal to 4. In summary, **p/q cannot equal -3, 1/2 or 4**.

Also, the equation p/q = (a+3b)/(2a-b) can be rearranged to make ‘a’ the subject. We get **a = (p+3q)b/(2p-q)**. That equation helps to confirm again that, since a is not zero, p+3q is not zero, so **p/q is not -3**. Similarly, 2p-q can’t be zero, so **p/q can’t equal 1/2**. Also, a is not allowed to equal b in this puzzle, so (p+3q)/(2p-q) can’t equal 1βp+3q **β ** 2p-qβ4q **β ** pβ**p/q β 4**.

Mathematical words containing the letters a, b, p and q include ** equiprobable** and

**.**

*equiprobability*In the Chess puzzle,

White wins beautifully with **1 Qxh6+!! Kxh6 2 Rh3+ Kg5** (2…Kg7 3 Rh7#) **3 f4+! Kxf4** (or 3…Kf5 4 g4+ Kxf4 5 Rf1+ Kg5 6 Rh7 and 7 h4#) **4 Rf1+ Kg5 5 Rh7! followed by 6 h4#!** π