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Seven Up

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Congratulations to Rachel Walker, Year 10, The Mount School York for her use of algebra here:

For two digit numbers that are seven times the sum of their digits (7-up numbers):

$\begin{eqnarray} \\ 10a + b &=& 7a + 7b \\ 3a&=&6b\\ a&=&2b. \end{eqnarray}$

The two digit 7-up numbers are: 21, 42, 63 and 84.

Using Rachel's method you can show that, for two digit K-up numbers, $2 \leq K \leq 9$ and so, for all these values of $K$, you can find the condition for solutions to exist and hence find the solutions.

Extending this method to three digit numbers:

$\begin{eqnarray} \\ 100a + 10b +c &=& 7(a + b + c) \\ 93a + 3b&=& 6c \\ 31a +b&=& 2c. \end{eqnarray}$

As the digits $a$, $b$ and $c$ are between 0 and 9 and $a \neq 0$, the right hand side of this expression can't be bigger than 18 whereas the left hand side is at least 31 which is impossible. This shows that there are no three digit seven-up numbers.

Because, as she says, there aren't any proper 7- up three digit numbers, Fiona Conroy, Year 10, The Mount School York, found some 'honorary' 7- up numbers: 105 = 7(10 + 5) 126 = 7(12 + 6) 147 = 7(14 + 7) 168 = 7(16 + 8) 189 = 7(18 + 9).

Fiona found the following set of K- up numbers for $K=37$:
111 = 37(1 + 1 + 1)
 
222 = 37(2 + 2 + 2)
 
333 = 37(3 + 3 + 3) 
 
444 = 37(4 + 4 + 4) 
 
555 = 37(5 + 5 + 5)
 
666 = 37(6 + 6 + 6)
 
777 = 37(7 + 7 + 7)
 
888 = 37(8 + 8 + 8)
 
999 = 37(9 + 9 + 9)
 

Christine Eaves, from the same school, found some interesting patterns:

190 = 19(1 + 9 + 0)
 
280 = 28(2 + 8 + 0)
 
370 = 37(3 + 7 + 0)
 
460 = 46(4 + 6 + 0)
 
570 = 57(5 + 7 + 0)
 
640 = 64(6 + 4 + 0)
 
730 = 73(7 + 3 + 0)
 
820 = 82(8 + 2 + 0)
 
910 = 91(9 + 1 + 0)
 

and another set:

198 = 11(1 + 9 + 8)
 
288 = 16(2 + 8 + 8)
 
378 = 21(3 + 7 + 8)
 
468 = 26(4 + 6 + 8)
 
558 = 31(5 + 5 + 8)
 
648 = 36(6 + 4 + 8)
 
738 = 41(7 + 3 + 8)
 
828 = 46(8 + 2 + 8)
 
918 = 51(9 + 1 + 8)
 

There are even easier cases if you go with the 100s:

100 = 100(1 + 0 + 0)
 
200 = 100(2 + 0 + 0)  etc
 

Altogether there are 180 K- up three digit numbers, and you can gather them in sets sharing such patterns. Can you use algebra to prove that $K$ cannot be larger than 100? Using this fact you might like to write a computer program to list all the three digit K-up numbers and to verify that there are 180 such numbers.