Puzzles
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.
10 December
A number is a palindrome if it's the same when its digits are written in reverse order.
What is the sum of all the numbers between 10 and 100 that are palindromes?
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The palindromes between 10 and 100 will be the numbers 11, 22, 33, ..., 99. The sum of these is 495.
9 December
Put the digits 1 to 9 (using each digit exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 4+3×2 is 14, not 10.
Today's number is the product of the numbers in the red boxes.
| – | | + | | = 8 |
| – | | ÷ | | ÷ | |
| + | | × | | = 9 |
| × | | ÷ | | ÷ | |
| ÷ | | × | | = 9 |
=
12 | | =
1 | | =
3 | |
6 December
The number n has 55 digits. All of its digits are 9.
What is the sum of the digits of n3?
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We can look for a pattern as we increate the number of 9s that make up n:
| n | n3 |
| 9 | 729 |
| 99 | 970299 |
| 999 | 997002999 |
| 999 | 999700029999 |
| 9999 | 999970000299999 |
If n has k digits, then n3 is
k-1 9s,
followed by a 7,
followed by k-1 0s,
followed by a 2,
followed by k 9s,
The sum of all these digits will by 18k.
Hence, the answer is 18×55 = 990.
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.
24 December
When written in binary, the number 235 is 11101011. This binary representation starts and
ends with 1 and does not contain two 0s in a row.
What is the smallest three-digit number whose binary representation starts and
ends with 1 and does not contain two 0s in a row?
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- 100 in binary is 1100100;
- 101 in binary is 1100101;
- 102 in binary is 1100110;
- 103 in binary is 1100111;
- 104 in binary is 1101000;
- 105 in binary is 1101001;
- 106 in binary is 1101010;
- 107 in binary is 1101011;
