# Enigmatic Code

Programming Enigma Puzzles

## Enigma 540: Icarus allsorts

“It takes all sorts to make a hang-gliders’ convention go with a whirl,” confided Icarus Thorns to the Rev E. B. Inept as they dallied over their tea on a mountain top.

“You can say that again,” answered his companion, dipping into the bag of proffered sweets and adding absent-mindedly another two lumps to his already sweetened tea. “You can say that again,” echoed the mountains faintly.

“This puzzle, unlike our jump, needs little preamble. In the (correct) equation:

ABC – D – E – F – G – H – I – J = 100

each of the letters stands for a different digit. All you have to do is deduce the value of:

(D × E × F × G × H × I × J) ÷ ABC .”

So saying, the intrepid Inept threw himself down off the outcrop, his wings glinting in the sun.

Can you deduce the answer before he reaches the bottom?

[enigma540]

## Enigma 536: A strange encounter

At a Halloween party I met a strangely beautiful red-haired woman who was gazing with longing into a bonfire. To my inquiries she replied: “(My telephone number) 2 = AABCCD and the exchange is 04617. I was born in the year given by: (AE) F = EDGE”

I realised that it was one of those ‘letters-for-digits-and-digits-for-letters’ puzzles, but I must have looked confused for she added: “The other letters are HIS.” So saying she vanished into thin air.

Given that in the above relations the same letter stands for the same digit and the same digit stands for the same letter; that different digits stand for different letters and vice versa; and that 04617 stands for the place where she now resides:

Can you solve the mystery and write down the digits 0 to 9 in the alphabetical order of the letters that represent them?

[enigma536]

## Enigma 531: Petits fours

From New Scientist #1683, 23rd September 1989 [link]

“Four-armed is four-warmed,” declared Professor Törqui as he placed the petits fours in the oven in his lab at the Department of Immaterial Science and Unclear Physics. “There are 4444 of them: a string of 4s. By which I mean, naturally enough, a number in base 10 all of whose digits are 4. Do you like my plus fours? [*] Speaking of 10s and plus fours, you can hardly be unaware of the fact that all positive integral powers of 10 (except 10¹, poor thing) are expressible as sums of strings of 4s.”

“The most economical way of expressing 10² as a sum of strings of 4s (that is, the one using fewest strings and hence fewest 4s) uses seven 4s:”

10² = 44 + 44 + 4 + 4 + 4.

“The most economical means of expressing 10³ as a sum of strings of 4s requires sixteen 4s:”

10³ = 444 + 444 + 44 + 44 + 4 + 4 + 4 + 4 + 4 + 4.

“Now, it’s four o’clock, and just time for this puzzle: Give me somewhere to put my cakestand and I will make a number of petits fours which is an integral positive power of 10 such that the number of 4s required to write it as a sum of strings of 4s in the most economical way is itself a string of 4s.”

What is the smallest number of petits fours Törqui’s boast would commit him to baking? (Express your answer as a power of 10.)

[*] £44.44 from Whatsit Forum.

[enigma531]

## Enigma 522: As easy as ABC

From New Scientist #1674, 22nd July 1989 [link]

“Perhaps this puzzle will please”, piped up the puzzling person from Pembroke. “It contains three digits only.”

“At first sight the sum:

looks impossible. Until I tell you, that is, that the squiggly brackets signify that a 3-digit number is to be made by writing down the enclosed digits in some definite permutation.”

“You mean”, I asked, “that if A, B and C happened to be 1, 4 and 7 then {ABC} could, in a given line of the sum, be either 147, 174, 417, 471, 714 or 741?”

“Yes. But that won’t do, of course, since no 3-digit number comprising a 1, 4 and a 7, when added to a 3-digit number comprising a 1, 4 and a 7 will yield a 3-digit number comprising a 1, 4 and a 7. You must find three digits A, B and C such that a 3-digit number comprising A, B and C when added to a 3-digit number comprising A, B and C will give a 3-digit number comprising A, B and C.”

Work out the answer to the sum in question.

[enigma522]

## Enigma 518: Day of reckoning

From New Scientist #1670, 24th June 1989 [link]

Nugel, Chancellor of the Exchequer of Planet-X, was to bring before the Prime Minister the results of his latest calculations of the planetary trade deficit. The calculation had proved awkward (but not as awkward, he thought ruefully, as trying to account for them was going to be) and he had asked the Minister for More Mathematics (who had attended the same school many moons ago) to process the figures in secret on his Magathon Computer. The result arrived while the Chancellor was already waiting to be admitted to The Presence and he read the missive at once:

“A boring task, my dear Nugel, but an aesthetic result: your trade deficit in Zloots is represented by the smallest integer which is simultaneously half a perfect square, a third of a perfect cube, a fifth of a perfect fifth power and a seventh of a perfect seventh power.”

Seventh heaven was where Nugel wasn’t when close inspection of the letter failed to reveal the number. He felt let down by his ex-prefect. What on Planet-X was the coot on about? With a sigh and resourceful to the last, he took out his battered Clackulator and set to work. In the background he could hear the voice of the only minister (Prime or Composite) to have survived 19 terms in office.

Can you find the trade deficit in Zloots (give your answer as a product of powers of prime numbers) before Nugel becomes the ex-Chancellor of the Exchequer?

[enigma518]