Puzzle 13
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.
Puzzle Copyright © Kevin Stone
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Hint
Start by labelling the number as ABCDE.
Answer
24976.
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, then B = 2, and E = 3, but 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.
Puzzle 14
What 5-digit number am I thinking of?
It has two prime digits.
Digit 3 is the highest digit.
Digit 2 is the lowest digit.
Digit 1 is higher than the sum of digits 4 and 5.
Digit 5 is half of digit 4.
Digit 1 is one smaller than digit 3.
Digit 5 is between digit 2 and digit 1
The digit 0 doesn't appear, and there are no duplicates. Don't forget that 1 isn't prime, but 2, 3, 5 and 7 are prime.
Puzzle Copyright © Kevin Stone
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Hint
What can digits 4 and 5 be?
Answer
71842.
Reasoning
The number can be represented as:
- - - - -
By (5), Digit 5 is half of Digit 4, so we have four possibilities:
- - - 2 1
- - - 4 2
- - - 6 3
- - - 8 4
By (4), Digit 1 is higher than the sum of Digit 4 and Digit 5, so we can eliminate two possibilities, leaving:
- - - 2 1
- - - 4 2
By (7), Digit 5 is between Digit 2 and Digit 1, so we can eliminate - - - 2 1, leaving:
- - - 4 2
By (4), Digit 1 must be higher than the sum of Digit 4 and Digit 5, so we have three possibilities:
7 - - 4 2
8 - - 4 2
9 - - 4 2
By (6), Digit 1 is 1 smaller than Digit 3 we can eliminate 9 - - 4 2, and fill in Digit 3, leaving:
7 - 8 4 2
8 - 9 4 2
By (3), Digit 2 is the lowest digit, and must be 1:
7 1 8 4 2
8 1 9 4 2
By (1), the only possibility is:
7 1 8 4 2
Puzzle 15
Last week I drove from Aardvark to Beeville.
On the first day, I travelled 1/3 of the starting distance.
On day two, I travelled 1/2 of the remaining distance.
On day three, I travelled 2/3 of the remaining distance.
At the end of day four, after travelling 3/4 of the remaining distance, I was still 5 miles away from Beeville.
How many miles had I travelled so far?
Puzzle Copyright © Kevin Stone
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Hint
Try working backwards.
Answer
175 miles.
The total distance was 180 miles, but as I still had 5 miles to go, the required answer is 175 miles.
Reasoning
Let's try working backwards.
Day 4: I travelled 3/4, which means that the other 1/4 was 5 miles – so 20 miles were left at the start of day four.
Day 3: I travelled 2/3, which means that the other 1/3 was 20 miles – so 60 miles were left at the start of day three.
Day 2: I travelled 1/2, which means that the other 1/2 was 60 miles – so 120 miles were left at the start of day two.
Day 1: I travelled 1/3, which means that the other 2/3 was 120 miles – so 180 miles were ahead at the start of day one.
The full distance was therefore 180 miles, but we were asked how many miles I had travelled so far, and as I still had 5 to go the answer is 175 miles.
Double-Checking
Start distance: 180 miles.
Day 1: I travelled 1/3 of the distance (60 miles), leaving 120 miles.
Day 2: I travelled 1/2 of the distance (60 miles), leaving 60 miles.
Day 3: I travelled 2/3 of the distance (40 miles), leaving 20 miles.
Day 4: I travelled 3/4 of the distance (15 miles), leaving 5 miles.
Puzzle 16
Imagine a chessboard of size 20 x 20 instead of the normal 8 x 8.
How many squares, of any size, are there in total?
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Hint
How many 1 x 1 squares are there, how many 2 x 2?
Answer
2,870 squares.
Reasoning
We have squares sizes from 1 x 1 up to 20 x 20, and these are the counts:
1 x 1 = 400
2 x 2 = 361
3 x 3 = 324
4 x 4 = 289
5 x 5 = 256
6 x 6 = 225
7 x 7 = 196
8 x 8 = 169
9 x 9 = 144
10 x 10 = 121
11 x 11 = 100
12 x 12 = 81
13 x 13 = 64
14 x 14 = 49
15 x 15 = 36
16 x 16 = 25
17 x 17 = 16
18 x 18 = 9
19 x 19 = 4
20 x 20 = 1
Total = 400 + 361 + 324 + 289 + 256 + 225 + 196 + 169 + 144 + 121 + 100 + 81 + 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 = 2870.
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