Puzzles
5 December
I make a book by taking 111 sheets of paper, folding them all in half, then stapling them all together through the fold.
I then number the pages from 1 to 444.
Today's number is the sum of the two page numbers on the centre spread of my book.
4 December
Today's number is the number of 0s that 611! (611×610×...×2×1) ends in.
3 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.
| + | | + | | = 11 |
| - | | + | | × | |
| + | | - | | = 11 |
| - | | - | | - | |
| + | | + | | = 11 |
=
-11 | | =
11 | | =
11 | |
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| 4 | + | 5 | + | 2 | = 11 |
| - | | + | | × | |
| 8 | + | 9 | - | 6 | = 11 |
| - | | - | | - | |
| 7 | + | 3 | + | 1 | = 11 |
=
-11 | | =
11 | | =
11 | |
The product of the numbers in the red boxes is 108.
1 December
There are 5 ways to write 4 as the sum of 1s and 2s:
- 1+1+1+1
- 2+1+1
- 1+2+1
- 1+1+2
- 2+2
Today's number is the number of ways you can write 12 as the sum of 1s and 2s.
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233
Let \(f(n)\) be the number of ways to write \(n\) as the sum of 1s and 2s.
When you write \(n\) as the sum of 1s and 2s, the sum must end in either 1 or 2: if it ends in 1, then there are \(f(n-1)\) ways to write the
rest before this 1 as the sum of 1s and 2s; if it ends in 2, there are \(f(n-2)\) ways to write the rest.
This means that \(f(n)=f(n-1)+f(n-2)\), or in other words the Fibonacci numbers.
Square and cube endings
Source: UKMT 2011 Senior Kangaroo
How many positive two-digit numbers are there whose square and cube both end in the same digit?
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Only the units digit of the number will affect the last digit of the square and cube. This table shows how the last digits of the square and cube depend on the last digit of the number:
| Last digit of... |
| number | square | cube |
| 0 | 0 | 0 |
| 1 | 1 | 1 |
| 2 | 4 | 8 |
| 3 | 9 | 7 |
| 4 | 6 | 4 |
| 5 | 5 | 5 |
| 6 | 6 | 6 |
| 7 | 9 | 3 |
| 8 | 4 | 2 |
| 9 | 1 | 9 |
So numbers ending in 0, 1, 5 and 6 will have squares and cubes that end in the same digit. There are 4×9=36 two-digit numbers then end in one of these digits.
Extension
How many two-digit numbers are there in binary whose square and cube end in the same digit?
How many two-digit numbers are there in ternary whose square and cube end in the same digit?
How many two-digit numbers are there in base \(n\) whose square and cube end in the same digit?
Digitless factor
Ted thinks of a three-digit number. He removes one of its digits to make a two-digit number.
Ted notices that his three-digit number is exactly 37 times his two-digit number. What was Ted's three-digit number?
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Ted's number was 925: \(925\div25=37\).
If Ted had removed the final digit of his number, then he would be looking for a solution of \(ABC = 37\times AB\). But \(ABC\)
is between 10 and 11 times \(AB\) (it is \(10\times AB + C\)) and so cannot be 37 times \(AB\). So Ted cannot have removed the final digit.
Therefore, Ted must have removed one of the first two digits: so two- and three- digit numbers have the same final digit (\(C\)).
The final digit of the three-digit number (\(C\)) will be the final digit of \(7\times C\) (7 times the final digit of the two digit number).
This is only possible if the final digit is \(C\) is 0 or 5.
This only leaves four possible solutions—10, 15, 20 and 25—as \(30\times37>1000\). Of these only \(925=37\times25\) works.
Extension
How many three-digit numbers are there that are a multiple of one of the two-digit numbers you can make by removing a digit?
Backwards fours
If A, B, C, D and E are all unique digits, what values would work with the following equation?
$$ABCCDE\times 4 = EDCCBA$$
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EDCCBA is a multiple of four, so A is even. A cannot be more than 2, as otherwise EDCCBA would have more digits. So A is 2.
E must therefore be 8 or 9 (as 4 times B is less than E) and 3 or 8 (as 4 times E ends in 2). Therefore E is 8.
Carrying on like this, we find:
$$219978\times4=879912$$
Cube multiples
Six different (strictly) positive integers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of 6.
What is the smallest possible sum of the six numbers?
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Consider the top, front, and right sides of the cube.
If the top number is \(a\) more than a multiple of six, then the front and right numbers must both be \(a\) less than a multiple of 6 (so that when added to the top number they make a multiple of 6). But when the front and right numbers are added, they make \(2a\) less than a multiple of 6; but this must also be a multiple of 6.
This is only possible if \(a=0\) or \(a=3\). So the numbers must be either all multiples of 6, or all 3 more than multiples of 6.
The smallest set of numbers that are all 3 more than multiples of 6 is 3,9,15,21,27,33. The sum of these is 108. The smallest set of numbers that are all multiples of 6 is the set with each number three more than these, so 108 is the smallest possible total.
Extension
Six numbers are written on the faces of a cube. The sum of the numbers on any two adjacent faces is a multiple of \(n\).
What is the smallest possible sum of the six numbers?
