Puzzles
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:
$$219978\times4=879912$$
11 December
Two more than today's number is the reverse of two times today's number.
4 December
Pick a three digit number whose digits are all different.
Sort the digits into ascending and descending order to form two new numbers. Find the difference between these numbers.
Repeat this process until the number stops changing. The final result is today's number.
1 December
Today's number is the smallest three digit number such that the sum of its digits is equal to the product of its digits.
XYZ
Which digits \(X\), \(Y\) and \(Z\) fill this sum?
$$
\begin{array}{cccc}
&X&Z&Y\\
+&X&Y&Z\\
\hline
&Y&Z&X
\end{array}
$$
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Both the units and tens columns contain \(Y+Z\). The results are different (\(X\) and \(Z\)), so \(Y+Z=X+10\) and \(Z=X+1\) (because the 1 carries into the next column).
Therefore, \(Y+X+1 = X+10\), so \(Y=9\).
From the hundreds column, we see that \(X+X+1=Y\), so \(X=4\) and \(Z=5\).
Extension
Which digits \(X\), \(Y\) and \(Z\) fill this sum?
$$
\begin{array}{cccc}
&X&Z&Y\\
+&X&Y&Z\\
\hline
&Z&Y&X
\end{array}
$$
