Puzzles
12 December
Mary uses the digits 1, 2, 3, 4, 5, 6 and 7 to make two three-digit numbers and a one-digit number (using each digit exactly once). The sum of her three numbers is 1000.
What is the smallest that the larger of her two three-digit numbers could be?
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The sum of the units digits of the three numbers must be 10 (as the units digit of the result must be 0 and the number are not big enough to allow the sum to be 20). Similarly, the sum of the
two tens digits must be 9, and the sum of the two hundreds digits is 9 (so that 10s are made when the 1 is carried).
The> hundreds digits could be 2 and 7, 3 and 6, or 4 and 5. As we're looking to make the larger number as small as possible, we should pick 4 and 5.
The tens digits could be 2 and 7 or 3 and 6. We pick 2 and 7 and as 2 is the smallest possible number here.
Finally, the units digits are 1, 3 and 6. We use 1 for our number as it's smallest, leading to 521.
4 December
Some numbers can be written as the sum of four consecutive numbers, for example: 142 = 34 + 35 + 36 + 37.
What is the mean of all the three-digit numbers that can be written as the sum of four consecutive numbers?
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The sum of four consecutive numbers starting at \(n\) is \(n + (n + 1) + (n + 2) + (n + 3) = 4n+6=4(n+1)+2\). This tells us that the numbers
that are the sum of four consecutive numbers are the numbers that are two more than multiples of 4. The smallest three digit number of this form is 102 and the largest is 998: the mean of these
two is 550. The second smallest and second largest are 106 and 994: the mean of these is also 550. All the numbers can be paired up in this way except for 550 itself, and so the mean of all the
numbers is 550.
Extension
What is the mean of all 3-digit numbers that can be written as the sum of \(k\) consective numbers, for different values of \(k\)? When is the answer an integer and when is it not?
3 December
Holly picks the number 513, reverses it to get 315, then adds the two together to make 828.
Ivy picks a three-digit number, reverses it, then adds the two together to make 968. What is the smallest number that Ivy could have started with?
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If Ivy's number is ABC, then A+C must be 8 or 18 (as this makes this final digit of the sum correct). A+C cannot be 18 of the sum would be bigger than 1000, so A+C must be 8.
The middle digit tells us that 2×B must be 6 or 16. As the first digit of the sum is 9 and not 8, there must be a 1 carried over into the hundreds column, so 2×B = 16 and B = 8.
Overall, we know that Ivy's number is A8C, with A+C = 8. The smallest possible number Ivy could have started with is therefore 187.
12 December
Holly picks a three-digit number. She then makes a two-digit number by removing one of the digits.
The sum of her two numbers is 309. What was Holly's original three-digit number?
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Holly's sum is odd, so she must have removed the units digit and so her calculation was:
$$
\begin{array}{cccc}
a&b&c\\
&a&b&+\\
\hline
3&0&9
\end{array}
$$
\(a\) and \(b\) cannot both be 0, so the must sum in the tens column must've caused a carry into the hundreds column. This means that \(a\) must be 2, and the calculation is:
$$
\begin{array}{cccc}
2&b&c\\
&2&b&+\\
\hline
3&0&9
\end{array}
$$
Two single digits cannot add to 19, so there can't be a carry from the units column into the tens column. This means that \(b\) is 8:
$$
\begin{array}{cccc}
2&8&c\\
&2&8&+\\
\hline
3&0&9
\end{array}
$$
We can see now that \(c\) was 1, so Holly's three digt number was 281.
5 December
The sum of 11 consecutive integers is 2024. What is the smallest of the 11 integers?
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Call the smallest number \(n\). The sum of the 11 integers is:
$$n + (n+1) + (n+2) + \dots + (n+10)$$
This simplifies to:
$$11n +55$$
If \(11n+55=2024\), then \(n\) is 179.
3 December
There are 5 ways to write 5 as the sum of positive odd numbers:
- 1 + 1 + 1 + 1 + 1
- 1 + 1 + 3
- 3 + 1 + 1
- 1 + 3 + 1
- 5
How many ways are there to write 14 as the sum of positive odd numbers?
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This can be solved by working it out for some examples then looking for the pattern:
| Total | Ways to make | Number of ways |
| 1 | 1 | 1 |
| 2 | 1+1 | 1 |
| 3 | 1+1+1, 3 | 2 |
| 4 | 1+1+1+1, 3+1, 1+3 | 3 |
| 5 | 1+1+1+1+1, 1+1+3, 1+3+1, 3+1+1 | 5 |
| 6 | 1+1+1+1+1+1, 1+1+1+3, 1+1+3+1, 1+3+1+1, 3+1+1+1, 3+3, 5+1, 1+5 | 8 |
The looks like the Fibonacci numbers: every term is the sum of the previous two terms.
Continuing the pattern gives 377 ways to make 14.
To justify why the answer is the Fibonacci numbers, notice that you split the sums for a number n into two sets:
those that end with "+1" and those that end with something else.
Those that end with "+1" are a way of making n-1, plus the one on the end.
Those that don't end with "+1" are a way of making n-2, with two added to the final number.
So the number of ways of making n is
the number of ways of making n-1 plus the number of ways of making n-2.
1 December
Eve writes down five different positive integers. The sum of her integers is 16. What is
the product of her integers?
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The only five different positive integers with a sum of 16 are 1, 2, 3, 4, and 6.
The product of these is 144.
16 December
Some numbers can be written as the sum of two or more consecutive positive integers, for example:
$$7=3+4$$
$$18=5+6+7$$
Some numbers (for example 4) cannot be written as the sum of two or more consecutive positive integers.
What is the smallest three-digit number that cannot be written as the sum of two or more consecutive positive integers?
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Every number except the powers of 2 can be written as the sum of two or more consecutive positive integers. The smallest
three-digit power of 2 is 128.
Extension
Why can every number that isn't a power of 2 be written as the sum of two or more consecutive positive integers?
