Dear All,
Here in Blog Post #154, with the coming (in just a few days’ time) of Easter–which celebrates Jesus Christ’s triumph over death, and the gift of eternal life to faithful followers–a perfect Bible verse to recognise properly with gratitude and joy is: “Then, when our dying bodies have been transformed into bodies that will never die, this Scripture will be fulfilled: Death is swallowed up by triumphant Life.”–1 Corinthians 15:54 ♥
The above truly is infinitely more important than anything at all relating to Chess or Mathematics, for example, but still I’m now going to happily share a feast of nice, fresh puzzles 😊💖
FUN BRAINTEASER 🎁
A girl makes a lovely cuboid-shaped gift box for her father for Easter. The total outer surface area of the closed box is 386 cm2. Its dimensions (length, width and height) in centimetres are each different positive whole numbers, which are also the ages (in years) of the girl and her two younger siblings.
Your fun brainteaser is to figure out all of their ages. The puzzle can be solved using neat algebra 👍✔
A 2023 PUZZLE 😎
If S is the smallest positive whole number for which the sum of the digits equals 2023, then what is the sum of the digits of S+1 ?
A PIECE OF CAKE BRAINTEASER! 🎂
On his birthday next year, in 2024, Nathan will be N years old. He looks forward to then having a cake with N candles on it. Imagine making triangles by connecting every possible distinct trio of three candles on the cake. Then there would be precisely 364 different triangles, to match the days in 2024 ! 😃
The brainteaser is to figure out the value of N.
JOKE 😁
What did the Chinese philosopher Li Ping say about the other two days…?
“Did you miss today’s joke?”
but I thought she said, “Did you miss two days…joke?!!” 😂
AS EASY AS ABC…! 😉
Which three-digit positive whole number ABC is exactly equal to AA + BB + CC? Can you prove that the solution is unique?
A THREE-PEOPLE PUZZLE FROM 11 YEARS AGO 💖😍💖
I am six years older than Jenny, and she is thirty years older than Michael. How old will Michael be when (in future, God-willing) the total of our three ages will be as close as possible to 154?
NUMBER WORD PUZZLE 💖
The number-word ONE HUNDRED AND FIFTY-FOUR contains all the five official English alphabet vowels, A, E, I, O, U. What is the smallest positive whole number for which its number-word contains all the vowels A, E, I, O, U exactly once each? (Hint: The correct number is greater than 154.)
FAST EASTER WORD PUZZLE 😊
Rearrange the letters of EASTER to make a different, proper six-letter English word that does not begin with E.
AN ELEGANT CHESS PUZZLE ♟
It’s my intention to publish solutions to all the puzzles around the time that blog post #155 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 😊.
I would like to round off this article now by most sincerely wishing you a very blessed month of April, with lots of happiness in everything that you do ❤.
With kindest wishes as always,
Paul M😊twani ❤
P.S. = Puzzle Solutions (being posted on 19.4.2023)
In the cuboid brainteaser, 2(LW+LH+WH)=386→LW+LH+WH=193. So, 193+H2 = LW+LH+WH+H2 = L(W+H)+H(W+H) = (L+H)(W+H), a neat and noteworthy algebraic trick 😍✔
Now trying H=1: 194 = (L+1)(W+1)→2 x 97 = (L+1)(W+1), but then L = 1 and W = 96 (or vice-versa), and that’s not OK because H & L should be different (like the siblings’ ages), and also none of them should be 96 in this puzzle!!
Trying H=2: 197 is a prime number; it can’t equal (L+2)(W+2), since L & W have to be positive. (We can’t allow L+2=1 & W+2=197, for example, because we’d get L=-1.)
Trying H=3: 202 = 2 x 101, which can’t equal (L+3)(W+3) in any suitable way.
Trying H=4: 209 = 11 x 19 = (L+4)(W+4) with L=7 & W=15, for example.
So, a solution for the siblings’ ages is 4, 7 and 15, with the ‘main’ girl in the puzzle being 15 (the oldest one).
If we were to investigate further, we wouldn’t find any truly different solutions, though of course H=7 or H=15 are possible, and would lead to L & W being 4 & 15 or 4 & 7. We still end up with the same set of dimensions (and siblings’ ages): 4, 7 & definitely 15 for the girl 👍
The ‘2023’ puzzle can be solved nicely like this: since 2016 is a multiple of 9 (224 x 9=2016) and 2023 = 7 + 2016, then the number S = 7999…999 (starting with a 7 followed by 224 nines) will be the smallest number–using the least-possible number of digits–whose digits sum equals 2023. Therefore, S+1 = 8000…000 (starting with an 8 followed by 224 zeroes), and so its digits sum is simply 8. 😎
When there are N candles on Nathan’s cake, the number of different trios of candles that we could connect to get triangles is N x (N-1) x (N-2) ÷ 6. (We divide by 6, because we don’t want to miscount any same trio 6 times just because that same trio could be listed in 6 different orders.) We were basically given that N x (N-1) x (N-2) ÷ 6 = 364, and so N x (N-1) x (N-2) = 2184. We observe that N-1 should be pretty close to the cube root of 2184, which is 12.974…, very close to 13. That suggests N = 14, and indeed 14 x 13 x 12 = 2184. So, it’s confirmed: N = 14. 🎂
In the ‘ABC’ puzzle, ABC = 198, the unique solution 😊
Proof: Given ABC = AA + BB + CC in this particular puzzle, then consideration of the true ‘place values’ of the digits leads to→100A + 10B + C = 10A + A + 10B + B + 10C + C, which rearranges to→89A = B + 10C. Since 10C + B can’t be big enough to be a three-digit number (because C & B are numbers not bigger than 9), it follows that 89A has to just be a two-digit number, and therefore A = 1. Then 10C + B = 89, which forces C = 8 & B = 9. It’s confirmed: ABC = 198. 💖 That lovely, special number always reminds me of 1998, the year when my son, Michael, was born 😍
In my family puzzle, when Michael is 29 (God-willing), Jenny 59 & Paul 65, the total of the three ages will be 153, which is the closest it can get to 154 before shooting past there! 💕
TWO HUNDRED AND SIX contains vowels A, E, I, O & U exactly once each. 😊
EASTER→TEASER 😁
In the Chess puzzle…
White forces checkmate quickly, with the ‘main’ line being 1 Qg8+ Kd6 2 Qd8+ Ke6 3 Qc8+ Kd6 4 Bc5#, provided that Black’s invisible b-pawn is not on b6 !! 😃🤣
Paul Motwani 