Page 154 - Kukanja Gabrijelčič, Mojca, and Maruška Seničar Željeznov, eds. 2018. Teaching Gifted and Talented Children in A New Educational Era. Koper: University of Primorska Press.
P. 154
ianne Nolte
‘subtraction.’ Is this always possible? Can you find out which final numbers
result in 1 and which in 0? And can you say, why?
– 1, 2, 3, 4 is one starting part of the natural numbers with the final number
4.
– 1, 2, 3, 4, 5, 6, 7 is one starting part of the natural numbers with the final
number 7.
If you do it in a clever way, you can construct tasks which result either in 1
or in 0.
Conditions:
– You can only use addition or subtraction.
– You must take every number exactly once.
– The numbers have to be a starting part of the natural numbers.
Result. Four successive numbers can be termed as a, a−1, a−2, a−3. It follows
for a > 3 e.g. that:
a − (a − 1) + (a − 3) − (a − 2) = 0 and a + (a − 3) − (a − 1) − (a − 2) = 0,
for a < 4: 3 − 2 − 1 = 0 and 2 − 1 = 1.
With small final numbers it is possible for the children to find results by trial
and error. When the figures get larger the students must set up a conjecture.
After three steps Simon recognized after calculating with the final number
66:
66 − 65 + 63 − 64 + 62 − 61 + 59 − 60 + 58 − 57 + 55 − 56 . . .
‘If you calculate this way you only have to puzzle over the numbers 1, 2, or
3 at the end.’ Even children at primary grade level are capable of generalize
their ideas. Nevertheless, this is a problem which can be interesting also for
prospective mathematics teachers.
Problem of the month’s at the University of Duisburg-Essen (December 2014,
see https://www.uni-due.de/didmath/problemdesmonatsarchiv.php). Given
is a natural number n. We are looking for arithmetic problems which must
follow these rules: You must use all numbers between 1 and n. You can only
use + and –. The result of the task must be 0 or 1.
Examples for these kinds of arithmetic problems are:
7+3−1−6+4−5−2 = 0
5+2−4−3+1 = 1
152
‘subtraction.’ Is this always possible? Can you find out which final numbers
result in 1 and which in 0? And can you say, why?
– 1, 2, 3, 4 is one starting part of the natural numbers with the final number
4.
– 1, 2, 3, 4, 5, 6, 7 is one starting part of the natural numbers with the final
number 7.
If you do it in a clever way, you can construct tasks which result either in 1
or in 0.
Conditions:
– You can only use addition or subtraction.
– You must take every number exactly once.
– The numbers have to be a starting part of the natural numbers.
Result. Four successive numbers can be termed as a, a−1, a−2, a−3. It follows
for a > 3 e.g. that:
a − (a − 1) + (a − 3) − (a − 2) = 0 and a + (a − 3) − (a − 1) − (a − 2) = 0,
for a < 4: 3 − 2 − 1 = 0 and 2 − 1 = 1.
With small final numbers it is possible for the children to find results by trial
and error. When the figures get larger the students must set up a conjecture.
After three steps Simon recognized after calculating with the final number
66:
66 − 65 + 63 − 64 + 62 − 61 + 59 − 60 + 58 − 57 + 55 − 56 . . .
‘If you calculate this way you only have to puzzle over the numbers 1, 2, or
3 at the end.’ Even children at primary grade level are capable of generalize
their ideas. Nevertheless, this is a problem which can be interesting also for
prospective mathematics teachers.
Problem of the month’s at the University of Duisburg-Essen (December 2014,
see https://www.uni-due.de/didmath/problemdesmonatsarchiv.php). Given
is a natural number n. We are looking for arithmetic problems which must
follow these rules: You must use all numbers between 1 and n. You can only
use + and –. The result of the task must be 0 or 1.
Examples for these kinds of arithmetic problems are:
7+3−1−6+4−5−2 = 0
5+2−4−3+1 = 1
152
