But a more mathematical method might help to answer the Bonus Question, as this might take a while if we keep adding!
So, let's create a method by imagining that we are adding the numbers from 1 to 30.
1 + 2 + 3 + … + 28 + 29 + 30
If we now take the numbers in pairs, taking one from each end, we have:
(1 + 30) + (2 + 29) + (3 + 28) + … + (15 + 16)
Each pair adds to 31, and we have 15 pairs. So the total sum is 31 x 15 = 465.
The total sum from 1 to any number (N) can be found using this technique, and we will have:
Each pair adds to (1 + N), and there are N ÷ 2 pairs. So the total is:
(1 + N) x N ——— 2
In this puzzle, we know that this equals 276.
So:
(1 + N) x N = 276 ——— 2
We can expand the brackets, and multiply both sides by 2, to give:
N + N2 = 552
Rearranging we get:
N2 + N − 552 = 0
And 552 = 2 x 2 x 2 x 3 x 23, so this can be factorised as:
(N + 24) x (N − 23) = 0
Because we need to find a positive number of days, the only possible answer is:
(N − 23) = 0
So N = 23 days. Bonus Question
To answer the bonus question, we have:
(1 + N) x N = 56616 ——— 2
Rearranging we get:
N2 + N − 113232 = 0
And 113232 = 24 x 3 x 7 x 337, so this can be factorised as:
(N − 336) x (N + 337) = 0
Because we need to find a positive number of days, the only possible answer is:
(N − 336) = 0
So N = 336 days (I did say that I liked collecting leaves!).
??
Puzzle 198
From each of these words, remove one letter, and then rearrange the remaining letters to find a new word. The resulting ten new words are related to each other.
Can you find a five-digit number that has no zeros, no ones, no digit is repeated, and:
the fourth digit is a quarter of the total of all of the digitsthe second digit is twice the first digitthe third digit is the largestthe last digit is the sum of the first two digits
Reasoning
We can start by labelling the digits as ABCDE.
We know that:
(i) B = 2 x A
and:
E = A + B
And using (i) we get:
E = A + (2 x A) (ii) E = 3 x A
If A = 1, this isn't allowed (as there are no 1's in the puzzle).
If A = 2, then B = 4, and E = 6.
If A = 3, then B = 6, and E = 9, but this isn't allowed (as C has to be the largest digit).
So, A = 2, B = 4, E = 6, and we now have to find C and D.
We also know that:
D = (A + B + C + D + E) ÷ 4
And using (i) and (ii) we get:
D = [A + (2 x A) + C + D + (3 x A)] ÷ 4
so:
3 x D = (6 x A) + C
so:
(iii) D = [(6 x A) + C] ÷ 3
C can only be 7, 8 or 9 (as it's the largest digit, and we've already found 6) and (iii) tells us that it must be a multiple of 3, which means that C = 9. Leaving D = 7.
So the final number is: 24976.
Double-Checking
The answer is 24976.
The fourth digit is a quarter of the total of all of the digits.
A + B + C + D + E = 2 + 4 + 9 + 7 + 6 = 28, and 28 ÷ 4 = 7.
The second digit is twice the first digit.
4 = 2 x 2.
The third digit is the largest.
9 is the largest digit.
The last digit is the sum of the first two digits.
6 = 2 + 4.
