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There were a number of correct solutions to this problem which either involved trial and improvement or the solution of a quadratic equation.
Correct solutions were received from Michael Brooker, Yatir Halvi (Maccabim-Reut High School), Ross Haines, Robert Haynes and Andre Lazanu. Well done!
Ross Haines first explained how number bases work:
We use base ten, which is this:
Thousands | Hundreds | Tens | Ones |
This can also be written as
10 cubeds | 10 squareds | 10s | 1s |
The second way is the way you write bases. It is the same format for all of them. For example, base 4 would be written as:
4 cubeds | 4 squareds | 4s | 1s |
To work a number from a base to base 10 you times the number by the column it is in. eg, using the number 321 in base four
4 cubeds | 4 squareds | 4s | 1s |
3 |
2 |
1 |
Which gives:
1 x 1 equals 1
2 x 4 equals 8
3 x ( 4 squared) equals 48
so 321 in base 4 would be 1+8+48=57 in base 10
Here is the trial and improvement solution submitted by Michael:
I worked it out like this: I decided to begin with bases much higher than 10, as 123 is much smaller than 3723. By trial and error I discovered that the answer was somewhere between 40 and 80.
So I tried translating "123" in Base 60 into Base 10. This time I did get 3723. Using powers I worked out the value of each base-60 digit in base 10 and added the figures together:
"3", the digit on the right, is the same in decimal
"2" x 60' gives 120
"1" x 60 ² gives 3600
Total = 3723
Yatir and Andrei found the solution by solving the quadratic equation 3x^2+2x+3=3723.
This works because the values of the columns in base x are:
x ^{2} | x | 1 | |||
So | |||||
1 | 2 | 3 | = 1x ^{2} + 2x + 3 | ||
= 3723 |
Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.