Puzzles

If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.

Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?

If Eve had written down an even number of numbers, then the mean of her numbers would be halfway between the two middle integers, so would be an integer and a half.

If Eve had written down two numbers, then the mean of those two numbers would be 422. This is not an integer and a half.

If Eve had written down three numbers, then the mean of those three numbers would be 281.3333. This is not an integer.

If Eve had written down four numbers, then the mean of those two numbers would be 211. This is not an integer and a half.

If Eve had written down five numbers, then the mean of those three numbers would be 168.8. This is not an integer.

If Eve had written down six numbers, then the mean of those two numbers would be 140.6666 This is an not integer and a half.

If Eve had written down seven numbers, then the mean of those two numbers would be 120.5714. This is an not integer.

If Eve had written down eight numbers, then the mean of those two numbers would be 105.5. This is an integer and a half, so this is possible. The eight consecutive integers with a mean of 105.5 are 105 and 106, 104 and 107, 103 and 108, and 102 and 109.

To be certain that this is the only answer, we should rule out Eve having written more than eight numbers.

The sum of the numbers from 1 to 41 is 861. This is greater than 844, so Eve must have written fewer than 41 numbers.

If Eve had written down an odd number of numbers, then the mean of her numbers would be an integer. This means that the number of numbers would be a factor of the sum of all the numbers. 844's prime factorisation is 2×2×211, and so its only odd factor is 211. This is greater than 41, so Eve didn't write down this many numbers.

Eve therefore must have written down an even number of numbers.

If Eve wrote down an even number of numbers, then the mean of her numbers would be an integer plus a half. This implies that half the number of numbers would be a factor of the sum, and the number of numbers cannot be a factor of the sum. This is only possible if there are 8 numbers.

836=2×2×11×19, and so for the product to be correct the integers could be:

The only one of these whose sum is 51 are 2, 11, and 38. Therefore the product of the largest and second-largest integers is 418

The sum of 13 consecutive integers is 13 times the middle number, so today's number is a multiple of 13. The sum of 12 consecutive integers is 6 times the sum of the two middle numbers, so today's number is also a multiple of 6. Therefore today's number is a multiple of 78.

78 and 156 do not work (the numbers would not all be strictly positive), so today's number is 234.

The digits of the number must be 1, 2 and 3, so the largest number is 321.

The only way to make 208 by adding consecutive numbers is 10+11+12+13+14+15+16+17+18+19+20+21+22. Therefore next year Noel will give his grandchildren a total of £221.

You can find my write up of a solution to this here. Pedro Freitas (@pj_freitas) sent me the following nice alternative solution:

Any number can be written as \(10a+27b\) with integer \(a\) and \(b\), since \(1=3\times27-8\times10\). So the problem may be thought of as asking when one of \(a\) and \(b\) must be negative.

Given one way of writing a number, you can get the others by shifting by 14*29. For example,

So the question now becomes: When does this adjustment fail to eliminate negative numbers?

This is when you are at what Pedro calls "limit coefficients":

So the answer is 233.

Let \(n\) and \(m\) be integers. What is the largest number that cannot be written in the form \(na+mb\), where \(a\) and \(b\) are nonnegative integers?

Yes: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.

It is clearly possible for the numbers 1 to 15 (remove the 16 from the end of the sequence above) and 1 to 17 (add 17 to the start of the sequence above).

The OEIS sequence A090461 gives other numbers for which this is possible. It starts 15, 16, 17, 23, then includes every number from 25 onwards. It is conjectured, but not proven, that it is possible for every number above 25.