Christmas (no less than any other time) offers you, me, and the whole world a perfect opportunity to recognise clearly, and celebrate with joy and thanks, God’s perfect love for us in having sent ♥ Jesus as the one true Saviour ♥ that we all need if we are to go to Heaven after our short lives now. In comparison to that all-important truth, I realise that anything that I might know about Mathematics, Chess, or other subjects, is practically insignificant.
Still, God purposely gave us all personal gifts to be used well, and so I will now offer some beautiful puzzles which I hope that many readers will enjoy solving.
The first puzzle (below) is one that I thought of just yesterday, and of all the puzzles that I have ever offered, it’s definitely one of my absolute favourites.
♥ Christmas Love Celebration Brainteaser ♥
Your Christmas Love brainteaser challenge is this: If the total sum of all the fifteen number values (for C, H, R, I, S, T, M, A, L, O, V, E, and also 52, 31 and the red heart) equals 141, then what must be the exact number value of the red heart?
NOTE WELL: Each item anywhere above the bottom row of the pyramid always equals the sum of the two items just below it. For example, I = M + A, because M & A are just under the letter I there. Similarly, R = S + T, and so on.
The number 141 has specially appeared in my life on literally hundreds of occasions within my sixty years, so far, and I couldn’t help noticing that the total sum of the eleven numbers in the display column of the next photo (here in blog post #141) is again 141 😊
♥ A Second Lovely Puzzle ♥
From the display board, I’m thinking of two particular numbers. If I increase the numbers by 10 each, their product increases by 300. What are my two starting numbers?
The word ‘CARDS’ may be closely connected to ‘CARES’, and ‘REASON’ is certainly connected to ‘SEASON’ through the perfect truth given in the next photo ♥
♥ A Wonderful Chess Study ♥
It’s now 22 days until Christmas 2022. Sometime before then, I hope and intend (God-willing, as always) to publish solutions to the puzzles within this post, but please do feel free to send me your best solutions to any/all of the puzzles, if you like 😊
Here’s a beautiful detail regarding the birthday tomorrow (4 December, 21 days before Christmas) of a lady I know, born in 1965… I shared with her that 21 x 1965 = 41265. We noted that, even I were just to write 21 x 19XY = D12XY, it can be proven that the day number D has to be 4 and the year has to be 1965; that’s the unique solution ♥
To everyone whose birthday is yet to come this month, I wish you a wonderful, happy time. Most of all, I wish everyone a very blessed Christmas soon.
With love and kindest wishes as always,
Paul M😊twani ♥
P.S. = Puzzle Solutions (being published now on 12.12.2022)
The Addition Pyramid Brainteaser can actually be solved rather quickly by considering the sums of the entries, one row at a time…
Top Row: 52
2nd Row: 52
(because C + H must exactly match the number 52 positioned right above the letters C & H)
3rd Row: 52 – ♥
(because R + ♥ + I = (R + ♥) + (♥ + I) – ♥ = C + H – ♥ = 52 – ♥)
4th Row: 52 – ♥ – ♥
(because S + T + M + A = (S + T) + (M + A) = R + I = (R + ♥ + I) – ♥ = (52 – ♥) – ♥)
Bottom Row: 52 – ♥ – ♥ – ♥ + 31
(because L + O + 31 + V + E = (L + O) + (V + E) + 31 = S + A + 31 = (S + T + M + A) – (T + M) + 31 = (52 – ♥ – ♥) – ♥ + 31)
So, the grand total sum of all the rows is
52 + 52 + 52 – ♥ + 52 – ♥ – ♥ + 52 – ♥ – ♥ – ♥ + 31 = 141 (given information)
→(52 x 5) -6 x ♥ + 31 = 141
→260 – 6 x ♥ + 31 = 141
→291 – 6 x ♥ = 141
→6 x ♥ = 150
→♥ = 150 ÷ 6 = 25.
♥ = 25, perfect for Christmas 😊
In the second puzzle, suppose that X & Y represent the two numbers that I was thinking of. Then, from the information given in the puzzle,
→10*X+10*Y+100=300 (after cancelling X*Y terms)
→X+Y=20 (after ÷ by 10)
The only numbers on the display board which could satisfy the requirement X+Y=20 were 9 & 11.
Warm congratulations to Jens Van Steerteghem, who solved that puzzle and the Addition Pyramid ♥ Brainteaser! 😊♥😊
In the Chess study, White wins with 1 c4! (threatening 2 Qd5#) 1…Qe4 2 d4+!, after which either 2…cxd4 or 2…Qxd4 is met by 3 Qe7#.
Alternatively, 1 c4! Qh1 2 d4+! cxd4 3 Qe7#, or 2…Ke4 and Black’s queen will be lost after the skewering check 3 Qd5+.