Puzzles
Not Roman numerals
The letters \(I\), \(V\) and \(X\) each represent a different digit from 1 to 9. If
$$VI\times X=VVV,$$
what are \(I\), \(V\) and \(X\)?
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For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.
If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.
Now that we know that \(VI\) is 37, we see that
$$37\times X=333,$$
and so \(X=9\) and the final solution is
$$37\times9=333.$$
22 December
In base 2, 1/24 is
0.0000101010101010101010101010...
In base 3, 1/24 is
0.0010101010101010101010101010...
In base 4, 1/24 is
0.0022222222222222222222222222...
In base 5, 1/24 is
0.0101010101010101010101010101...
In base 6, 1/24 is
0.013.
Therefore base 6 is the lowest base in which 1/24 has a finite number of digits.
Today's number is the smallest base in which 1/10890 has a finite number of digits.
Note: 1/24 always represents 1 divided by twenty-four (ie the 24 is written in decimal).
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If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be
possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.
The prime factorisation of 10890 is 2×33×5×112. Therefore the smallest base in which 1/10890 has a
finite number of digits is 2×3×5×11=330.
15 December
Today's number is smallest three digit palindrome whose digits are all non-zero, and that is not divisible by any of its digits.
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The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).
The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is 323. This is not a multiple of 2 or 3 and so is the solution.
9 December
Today's number is the number of numbers between 10 and 1,000 that contain no 0, 1, 2 or 3.
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There are 6 available choices for each digit, so there are 62 and 63 such two- and three-digit numbers. 62+63=252.
4 December
Today's number is the number of 0s that 611! (611×610×...×2×1) ends in.
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:
$$219978\times4=879912$$
