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Your Comment: August 21, 12am
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2 comments
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John email hidden
Netherlands
wrote more than a year ago
Jerry I enjoyed you problems, very interesting page, as is all the rest on your website.
But your reasoning on Problem 4 is wrong.. I first assumed this as possible combinations of multiple true options. So i got to 3 possibilities. But 3 was not in the answers. Then I assumed singles too, makes it 8 because the last one cannot be single. Then I guessed you must have counted all false as a consistent too.
Then you answer was disappointing.. you present the solution:
2 + 2 + 1 + 2 + 2 = 9, but it does not compute that way.
Take: a + b + c + d + e from this line of components..
in a: one of the 2 is a single, in b too, and c is also a clear single.
Then your fault:
for d and e: next to single 4 and single 5, you count at each the combination with 6, but as soon as you have counted 45 or 46, it is 456, so you cannot count that anymore as 56 later.
So in your reasoning it should have become 2+2+1+2+1 = 8
Last point:
My conclusion was only 3 combinations exist: 1+5, 2+4, 4+5+6 .
and when counting singles too, there are 5 more, as single 6 is not an option since it exists only in combination 456.
As there was no 3 nor 8 in the answers, I had guessed the all 6 false coun ( truncated for length...)
0
0
John email hidden
Netherlands
wrote more than a year ago
Jerry I enjoyed you problems, very interesting page, as is all the rest on your website.
But your reasoning on Problem 4 is wrong.. I first assumed this as possible combinations of multiple true options. So i got to 3 possibilities. But 3 was not in the answers. Then I assumed singles too, makes it 8 because the last one cannot be single. Then I guessed you must have counted all false as a consistent too.
Then you answer was disappointing.. you present the solution:
2 + 2 + 1 + 2 + 2 = 9, but it does not compute that way.
Take: a + b + c + d + e from this line of components..
in a: one of the 2 is a single, in b too, and c is also a clear single.
Then your fault:
for d and e: next to single 4 and single 5, you count at each the combination with 6, but as soon as you have counted 45 or 46, it is 456, so you cannot count that anymore as 56 later.
So in your reasoning it should have become 2+2+1+2+1 = 8
Last point:
My conclusion was only 3 combinations exist: 1+5, 2+4, 4+5+6 .
and when counting singles too, there are 5 more, as single 6 is not an option since it exists only in combination 456.
As there was no 3 nor 8 in the answers, I had guessed the all 6 false coun ( truncated for length...)