Each term is a factorial, 6! = 1 x 2 x 3 x 4 x 5 x 6 = 720.
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Puzzle 27
Can you find a five-digit number that has no zeros nor ones in it and no digit is repeated, where:
The fourth digit is a quarter of the total of all of the digits.The second digit is twice the first digit.The third digit is the largest.The 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.
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Puzzle 28
A grocer has to place 28,121 carrots into bags, each bag containing the same number of carrots, and using as few bags as possible.
If you think that one big bag will suffice, then feel free to calculate its size!
If we assume that more than one bag is used, how many are required?
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