Maths and IQ problems

ElansarGelmir · #31 15-04-2005

Quote:

Originally Posted by prince

Actually it can be followed by anything since I can create a pattern from any finite sequence.

But I think you're seeking for
211213

There might be another possible way
122113

ElansarGelmir · #32 15-04-2005

Quote:

Originally Posted by topdog

U, D, T, Q, C, S, S, O...

N?

Here's another one [URGH!!! Got to go back to my Chem]

A man wanted to marry a woman. She agreed to marry him, if and only if he can guess the age of all her 3 daughters. She gave him 2 clues:
1. The product of the age of her daughters is [edited: 36]
2. The sum of the age of her daughters equals to the address of the house next door

The man made some calculations, went out to find out the address of the house next door, but still could not arrive with the answer. He begged for one more clue. Pitying the man (I think she wanted to marry him too), she gave him one more clue:
3. Her oldest daughter loves strawberry

Instantly, the guy got the answer and they were happily married to each other. The question: How did he solve it?

littlebigone · #33 15-04-2005

i think the questions wrong.

He should be able to make the guess based on the first two clues.

There are only 5 unique combinations that i can find:
1,1,32
1,2,16
1,4,8
2,2,8
2,4,4

And they all give unique sums.

I think your first clue is wrong.

I think the puzzle is supposed to end up with som combination adding up to the same number which means he wouldn't be able to choose between the combinations. But then because she has an "oldest daughter", one of the combinations would have to have her having older twins. So by elimination, he guessed the correct ages.

littlebigone · #34 15-04-2005

should be 36
then you'll get two combinations which add up to 13
1,6,6 and 2,2,9
so using the same logic, we see that the daughters's ages are 2,2,9

someone · #35 15-04-2005

Yes. The question is wrong. the product of the 3 numbers should be 36. so that the factorizations can be
36, 1, 1 - 38
18,2,1 - 21
9,4,1 - 14
9,2,2 - 13
6,6,1 - 13
6,3,2 - 11
4,3,3 -10
All the 3 numbers sum up to distinct numbers except the 13. so the address of the next door must be 13. he is not able to guess until he knows there's an 'oldest' daughter. therefore, between 9,2,2 and 6,6,1, he chose 9,2,2.

someone · #36 15-04-2005

sorry.. i am late 8O

ElansarGelmir · #37 15-04-2005

Oops...

Yeap... sorry... was posting this question while i was attempting my chem lab report... pai seh

chiunlin · #38 16-04-2005

Here's another question:
A monk climbs up a hill following a path at starting from the bottom at 12pm. The next day, he climbs down the hill following the same path at 12pm. Prove that when the monk is going down the hill, the monk will reach a spot somewhere between the peak and the bottom where he was at at the same time at the day before.

SpRInG · #39 16-04-2005

chiunlin, dun understand ur question lar

bp_ffei · #40 16-04-2005

I think you should add assumptions, like he is travelling with constant speed etc.

Anyway, is it a maths question? I don't seem to see any trickery in the question...

littlebigone Slightly Senior Member Join Date: Apr 2003 Posts: 867	#34 15-04-2005 should be 36 then you'll get two combinations which add up to 13 1,6,6 and 2,2,9 so using the same logic, we see that the daughters's ages are 2,2,9 __________________ Life is what happens while you're busy making other plans ~John Lennon

someone Super Junior Member Join Date: Apr 2004 Posts: 30	#36 15-04-2005 sorry.. i am late 8O

ElansarGelmir Senior Member Join Date: Mar 2004 Posts: 2,072	#37 15-04-2005 Oops... Yeap... sorry... was posting this question while i was attempting my chem lab report... pai seh __________________ Money is not the most important thing in life. Love is. Fortunately, I LOVE MONEY!!!

chiunlin Slightly Senior Member Join Date: Jan 2004 Posts: 705	#38 16-04-2005 Here's another question: A monk climbs up a hill following a path at starting from the bottom at 12pm. The next day, he climbs down the hill following the same path at 12pm. Prove that when the monk is going down the hill, the monk will reach a spot somewhere between the peak and the bottom where he was at at the same time at the day before.

SpRInG Member Join Date: Mar 2004 Posts: 294	#39 16-04-2005 chiunlin, dun understand ur question lar

bp_ffei Member Join Date: Mar 2005 Posts: 359	#40 16-04-2005 I think you should add assumptions, like he is travelling with constant speed etc. Anyway, is it a maths question? I don't seem to see any trickery in the question... __________________ You are unique. Just like everyone else.