| From the problems, I think your daughter must be 8-9 years old. Here is
my suggestion as to how I would suggest things to someone of that age
and faced with these problems:
Problem 1:
First have her compute the 11 and three more i.e. 14
Then get her to devide that into two portions of the same size i.e. 7
Then let her think about things where 14 of something are 2 of another,
but really mean the same. If she cannot think of anything, point to the
way we count time in days, weeks, months etc. One answer to this
problem could be that 11 (days) and 3 more (days) equal 2 (weeks).
Problem 2:
First let her see that the two first digits of the first number, and of
the answer differ. This means that A+A must add up to something larger
than 10, otherwise the first digit of the answer would stay the same.
Get her to add the two same single digits together (i.e 1+1, 2+2, etc)
and observe that only when A is equal or larger than 5 this happens.
Then have her see that if you add two single digit > 5, that the first
digit of the answer is always 1, and then look at the equation again.
Have her see that this means that M = A + 1. From the two obeservations
combined have her set up the 5 remaining additions (e.g. 45+5, 56+6,
67+7, 78 + 8, and 89+9) and calculate the result. The answer should
than be obvious.
Hope this helps. Having no children of my own, I will take these
problems to my nephew of 8 and see if this works.
Wildrik
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| re .0
Knowing your daughter's age would seem to be crucial to choosing an
appropriate means of guiding her to the answer.
FWIW, I would do the second problem as follows.
(10M+A) + A = 10A+M
=> 9M = 8A
=> M=A=0 or M=8 and A=9
This solution requires an understanding of the base 10 positional
notation, some very simple algebraic manipulation and either spotting
the solution from 9M = 8A (not too hard) or knowing a little about prime
numbers.
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