Blog Post #64: For Haley, Myriam, My Little Sister, Peter The Great AND a school pussy cat!

Dear Readers,

Happy Haley, a talented student at my school, recently surprised me with a lovely painting that she made as a New Year’s gift.

Haley enjoys solving puzzles, and so I’m featuring her in the following one together with Peter and Myriam, two adult friends of mine. Peter and Myriam have exactly the same birthday as each other (regarding the day, month and year), and in this puzzle let’s assume that their ages are currently whole numbers.

THE PETER-MYRIAM-HALEY BRAINTEASER

If P and M are the current ages of Peter and Myriam, and H is Haley’s favourite whole number, what is the last digit on the right of (P+M) to the power of (4xH) ? I had to include the number 4, otherwise the pussy cat might feel left out!!

SOLUTION (being posted on 27.2.2020)

As P and M are equal, positive, whole numbers, (P+M) must be a positive even number. There are now a few cases to consider:-

1. If H=0, then (P+M) to the power of (4xH) simply equals 1.
2. If H is not zero but (P+M) ends with last digit 0, then (P+M) to the power of (4xH) will also end with last digit 0.
3. If H is not zero and (P+M) ends with last digit 2, 4, 6 or 8, then (P+M) to the power of (4xH) always ends with last digit 6. That noteworthy fact often comes in handy when tackling number theory problems.

Pictured below: Peter The Great with a happy cat!

BONUS PUZZLE FOR MY LITTLE SISTER!

My youngest sister will turn 7×7 in 3×3 days from now. If we write dates in the six-digit form DDMMYY, then all dates this month will have the form DD0120. Only one of them is an exact multiple of 49, and that’s 190120 today! There are only two FIVE-DIGIT whole numbers ending in 120 which are exact multiples of 49. Can you find them without using a calculator?

SOLUTION (being posted on 27.2.2020)

190120-49000-49000=92120

and 92120-49000=43120

A CUTE CHECKMATE CHALLENGE!

Part 1: Imagine that White has one extra, invisible pawn somewhere on the board. That pawn must remain in the same place in Part 2 of the puzzle. Where exactly is the invisible pawn if (immediately after placing it on the board) White can deliver checkmate right away in one move?

Part 2: Suppose instead that White has the same invisible pawn in the same place as in part 1, but one of White’s other pawns vanishes, and so does one of Black’s pawns! Exactly which pawns vanish, if it is now Black to move and force checkmate in at most 5 moves?

SOLUTION (being posted on 27.2.2020)

With an extra white pawn on e6, 1 Na7# makes a pretty checkmate.

However, if we remove the pawns from d5 and b7, then 1…Qb7+ leads to a forced mate, with the longest line being 2 Kg1 Rg3+ 3 Kf2 Rg2+ 4 Ke1 Qe4+ 5 Kd1 Qb1#.

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Solutions to the puzzles will be published at the time of the next blog post, and in the meantime I wish you lots of happiness…and great fun with the puzzles!

With very best wishes as always,

Paul Motwani xxx