Puzzles

Holly's sum is odd, so she must have removed the units digit and so her calculation was:

\(a\) and \(b\) cannot both be 0, so the must sum in the tens column must've caused a carry into the hundreds column. This means that \(a\) must be 2, and the calculation is:

Two single digits cannot add to 19, so there can't be a carry from the units column into the tens column. This means that \(b\) is 8:

We can see now that \(c\) was 1, so Holly's three digt number was 281.

Call the smallest number \(n\). The sum of the 11 integers is:

This simplifies to:

If \(11n+55=2024\), then \(n\) is 179.

This can be solved by working it out for some examples then looking for the pattern:

The looks like the Fibonacci numbers: every term is the sum of the previous two terms.

Continuing the pattern gives 377 ways to make 14.

To justify why the answer is the Fibonacci numbers, notice that you split the sums for a number n into two sets: those that end with "+1" and those that end with something else.

Those that end with "+1" are a way of making n-1, plus the one on the end.

Those that don't end with "+1" are a way of making n-2, with two added to the final number.

So the number of ways of making n is the number of ways of making n-1 plus the number of ways of making n-2.

The only five different positive integers with a sum of 16 are 1, 2, 3, 4, and 6. The product of these is 144.

Every number except the powers of 2 can be written as the sum of two or more consecutive positive integers. The smallest three-digit power of 2 is 128.

Why can every number that isn't a power of 2 be written as the sum of two or more consecutive positive integers?

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