Mathematical Puzzles 

Workings

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.

Workings

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

Workings

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.

Workings

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.