# Enigmatic Code

Programming Enigma Puzzles

## Puzzle #35, #36, #37, #38, #39: A bunch of brain teasers

Q1: By the twelfth day of Christmas, my true love has given me 12 partridges in a pear tree. But which gifts have I received the most of?

Q2: I want to give all the gifts back. Starting on 26 December 2019, I am going to give one of them to my true love every day. On which date will I give them my final gift?

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#### Puzzle #36: All squares (1)

I met Natalie the other day. She wasn’t prepared to tell me her age, but she did tell me that in the year N², she will turn N years old.

In what year was she born?

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#### Puzzle #37: All squares (2)

Can you work out (68² – 32²)/(59² – 41²) without using a calculator?

And can you do it without having to square any of the numbers?

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#### Puzzle #38: Meaningful matches (1)

The figure below has four equilateral triangles. Move two matchsticks to get only three equilateral triangles.

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#### Puzzle #39: Meaningful matches (2)

The figure below is composed of 29 matchsticks. Move two matchsticks to get a correct multiplication result.

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[puzzle#35] [puzzle#36] [puzzle#37] [puzzle#38] [puzzle#39]

### One response to “Puzzle #35, #36, #37, #38, #39: A bunch of brain teasers”

1. Jim Randell 21 December 2019 at 9:57 pm

Most of these puzzles are fairly straightforward:

#35.1: The gifts accumulate at 12×1 + 11×2 + 10×3 + … + 2×11 + 1×12. Which is 12 + 22 + 30 + … + 22 + 12 (= 364). In the middle these reach a maximum of 6×7 = 42 and 7×6 = 42, so you get an equal number of type 6 (geese) and type 7 gifts (swans).

#35.2: As noted above there are 364 gifts received in total. Starting on 26th December 2019 we can return 1 gift a day until 23rd December 2020 (as 2020 is a leap year).

#36: 44² = 1936, 45² = 2025. So Natalie was born in 45² – 45 = 1980.

#37: We can express (a² – b²) as (a + b)(a – b) giving:

(68² – 32²) / (59² – 41²)
= ((68 + 32)(68 – 32)) / ((59 + 41)(59 – 41)
= (100×36) / (100×18)
= 36 / 18
= 2

#38: There are many ways a few matches can be removed to leave only 3 equilateral triangles. Removing 1, 2, or all 3 of the matches that make up the middles of the sides of the large equilateral triangle will leave only the three smaller equilateral triangles.

#39: Keeping the multiplication sum in the form: AB × C = DE the only solution is:

23 × 4 = 92

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