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The Guardian - UK
The Guardian - UK
National
Alex Bellos

Did you solve it? Nick Berry, data dude

Nick Berry on Ilkley Moor
Nick Berry on Ilkley Moor Photograph: Nick Berry

Earlier today I set you these three problems from Nick Berry’s fabulous DataGenetics blog. Nick died last week aged 55, as I wrote about in the original post.

1. No-zero heroes

Write 1,000,000 as the product of two numbers; neither of which contains any zeroes.

(You may be interested to know that 10 x 10 x 10 x 10 x 10 x 10 = 1,000,000)

Solution 15625 x 64

Since 10 = 2 x 5, then a million is 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5 x 5 x 5. If we group the fives together and the twos together, we get the answer. If neither number is to have a zero (meaning they cannot have 10 as a divisor), then neither of the numbers can have both a 2 and a 5 as a factor.

2. Lucy’s secret number

You are at a party and overhear a conversation between Lucy and her friend. In the conversation, Lucy mentions she has a secret number that is less than 100.

She also confesses the following information: “The number is uniquely describable by the answers to the following four questions:”

Q1) Is the number divisible by two?
Q2) Is the number divisible by three?
Q3) Is the number divisible by five?
Q4) Is the number divisible by seven?

She then proceeds to whisper the answers to these questions to her friend. Unfortunately, because of the ambient noise at the party, you only hear the answer to one of the questions. Knowing this one answer allows you to determine the secret number. The answer you hear is ‘“yes.” What is Lucy’s secret number?

Solution 70

Since there are four questions, and each answer can be yes or no, there are sixteen possible combinations of answers. Lucy said that the answers to the questions uniquely determine her number. So we need to look for combinations that uniquely determine a number.

Let’s start with the combination of No, No, No, No. This combo allows the numbers 11, 13, (and several more), so we can eliminate it.

Next, let’s try No, No, No, Yes. This combo allows, 7, 49 (and a couple more), so we can eliminate it too.

Going through all the combos, there are only two that fix a single number:

No No Yes Yes determines 35

Yes No Yes Yes determines 70

The secret is therefore 35, or 70.

However, you will see that both of these solutions have the same answers to the questions Q2 “divisible by three”, Q3 “divisible by five” and Q4 “divisible by 7”. Therefore, if knowing the answer of only one question determines the number, the question has to be Q1 “divisible by 2”. We are told the answer is “Yes”, so Lucy’s secret number is 70.

3. Naughty maths elves

I write the whole numbers from 1-9999 (inclusive) on a huge chalkboard. Each number is written once.

During the night the board is visited by a series of naughty math elves. Each elf approaches the board, selects two numbers at random, erases them, and replaces them with a new number that is the absolute difference of the two numbers erased.

This vandalism continues all night until there is just one number remaining.

I return to the board the next morning and find the single number of the board. Is this remaining number odd or even?

Solution An even number

This problem can be solved, simply, with application of parity. A number is either odd or even. At the start of the night there are 9,999 numbers written on the board. Of these numbers, 5,000 are odd, and 4,999 are even.

When an elf selects a pair of numbers, there are three possible permutations. He can select two odd numbers, two even number, or one of each.

  • If he selects two odd numbers, the absolute difference between two odd numbers is always an even number. What has happened is that the quantity of odd numbers remaining has been reduced by two.

  • If he selects two even numbers, the absolute difference between a pair of even numbers is also even. The quantity of odd numbers remaining stays the same.

  • If he selects an odd and an even number, the absolute difference between this pair is odd. We lost one odd number, but gained a new one, so the quantity of odd numbers remaining stays the same.

From this you can see that either the quantity of odd numbers either stays the same, or reduces by two, with every act of defacement.

The quantity of odd numbers on the board started at 5,000, which is even, and has to remain even.

If we get down to a single number, it has to be an even number.

I hope you enjoyed the puzzles. I’ll be back in two weeks.

Thanks to Ian Mercer for help with today’s column.

Nick Berry’s Datagenetics blog is a treasure trove of material and is highly recommended.

I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.

I give school talks about maths and puzzles (online and in person). If your school is interested please get in touch.

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