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Mensa number puzzle for math geniuses (1 Viewer)

icecat4270

Footballguy
I found an old puzzle book published by Mensa and one of the puzzle types has me baffled about their answer.

The problem is:

Place 6 three digit numbers of 100 plus at the end of 685 so that six numbers of six digits are produced. When each number is divided by 111 six whole numbers can be found. Which numbers should be placed in the grid.

_ _ _

_ _ _

685 -> _ _ _

_ _ _

_ _ _

_ _ _

In other words find six numbers between 685100 and 685999 that are evenly divisible by 111.

The first spoiler is my method and answer. The second is the books answer.

The way I have solved it is taking 685000/111 and take the answer ignoring the remainder and adding one. I get 6171 and some remainder add one for 6172. Then I multiply it by 111 to get 685092 which is less than 685100 so it's not valid. adding 111 (or multiplying 6173*111) gets 685203 which is valid. I continue adding 111 getting 685314, 685425, 685536, 685647, 685758, 685869, 685980.

So the answer would be:

203

314

425

536

647

758

869

980
The answer in the book is:

314

425

536

647

758

869
There are several similar problems throughout the book and most are like this where I think there are more answers than they are showing.

Is my answer incorrect or is the book correct?

Is there a better method for solving this problem?

 

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