Puzzle 1
In the following sum, the digits 0 to 9 have all been used, and the top row's digits add to 9.
Knowing that O = Odd, and E = Even (zero is even), can you determine each digit?
Puzzle Copyright © Kevin Stone
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Hint
The answer is 4 digits long, so what must G equal?
Answer
423 + 675 = 1098.
Reasoning
Remembering that:
even + even = even
odd + odd = even
even + odd = odd
To discuss individual letters, it's easiest to represent the sum as:
A B C
D E F +
————————
G H I J
A + D has to be over 9, which means that G = 1.
B + E = I, is even + odd = odd, which means that we can't have a carry from C + F (otherwise it would have been even + odd + 1, which is even).
The 1 has already gone, so the smallest possible value for either C or F is 3, which means that the other can't be 7 or 9 (otherwise we'd have a carry).
Therefore, C and F are 3 and 5, but we don't know which is which. But we do now know that J = 8.
A + D = H, is even + even = even, which means that we can't have a carry from B + E. Therefore, E can't be 9, as this would force a carry. So E = 7.
I is the only remaining odd number, so I = 9.
Which means that B = 2.
Neither A nor D can be 0 (otherwise we would have two of the same digit). So, H = 0.
Therefore, A and D are 4 and 6 (but we don't yet know which is which).
Since the top row's digits have to add to 9, A can't be 6, so A = 4, making C = 3.
This makes the sum 423 + 675 = 1098.
Puzzle 2
My current age is the age of my favourite book (which is 14 years old) plus one third of my age.
How old will I be when my favourite book is twice its current age?
Puzzle Copyright © Kevin Stone
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Hint
A little algebra might help.
Answer
I will be 35 in 14 years.
Reasoning
I am currently 21.
If X is my current age, then:
X = 14 + X
———
3
so
2X = 14
———
3
and
2X = 42
So
X = 21
I am currently 21 and my favourite book is currently 14 years old, when it is 28 years old (double its age) I will be 35.
Puzzle 3
Starting in the bottom left corner and moving either up or right, adding up the numbers along the way, what is the largest sum that can be made?
| 2 |
3 |
5 |
4 |
1 |
| 3 |
2 |
4 |
3 |
4 |
| 5 |
1 |
3 |
5 |
2 |
| 3 |
3 |
2 |
3 |
1 |
| 1 |
4 |
2 |
4 |
4 |
Note: this puzzle is not interactive, and the squares cannot be clicked.
Puzzle Copyright © Kevin Stone
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Answer
The largest sum that can be made is 28.
Puzzle 4
Yesterday I went for a short bicycle ride around the local lakes. As the weather was very hot, I rode in shorter stages than I'd normally ride.
In Stage One, I rode half of the overall distance.
Stage Two saw half of the remaining distance plus 35 metres covered.
Stage Three covered three-quarters of the remaining distance.
Stage Four completed half of the remaining distance, plus 75 metres.
Stage Five completed the journey with a final burst of 150 metres.
How far did I cycle in total?
Puzzle Copyright © Kevin Stone
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Hint
Try working backwards.
Answer
7,340 metres.
Reasoning
Stage 5's clue tells us that 150m was left. So the distance left at the start of Stage 4 must have been:
Dist4 = Dist4 + 75 + 150
—————
2
Which simplifies to give Dist4 = 450m.
Stage 3's clue tells us that 450m was one quarter of the remaining distance, so Dist3 = 1,800m.
Stage 2's clue tells us that the distance at the start of Stage 2 was:
Dist2 = Dist2 + 35 + 1800
—————
2
Which simplifies to give Dist2 = 3,670m.
Stage 1's clue tells us that 3,670m was half the overall distance, which means the entire ride was 7,340m.
Double-Checking
Starting with 7,340m:
Stage 1: cycled 3,670m, leaving 3,670m
Stage 2: cycled 1,870m, leaving 1,800m
Stage 3: cycled 1,350m, leaving 450m
Stage 4: cycled 300m, leaving 150m
Stage 5: cycled 150m, leaving 0m
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