# [Enigma] Small problems

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• Jul 25th 2008, 07:44 AM
Moo
[Enigma] Small problems
Hi,

Here are some problems you can try to solve. I think it doesn't require much knowledge, but mostly tricks.

#1 : simple algebra
Less than 12 years ago, a mathematician, Vincent Urri, told his niece :
- It's curious ! I have in front of me 3 positive integers. Their product is equal to my age and their sum gives your age !
The niece, though she's good at calculations, wasn't able to guess what the 3 numbers were.

This year, Vincent Urri has just turned 100 years old. How old is his niece today ?

#2 : a little of number theory
1. What is the remainder in the division of http://www.gnux.be/latex/data/fdf0a6...56e932b2b2.png by 7 ?
2. What is the remainder in the division of http://www.gnux.be/latex/data/fdf0a6...56e932b2b2.png by 11 ?
3. What is the remainder in the division of http://www.gnux.be/latex/data/fd826b...49accfb088.png (where there are 502 times the pattern 2008) by 37 ?

#3 : simple geometry
The Swedish flag Attachment 7284 has a yellow cross of constant width in a blue background. The yellow area represents one third of the flag's area.
If the flag is 1m long and 66cm wide, what is the width of the yellow stripes ?

#4 : aŁg€bra
Teddy left his 9-digit code of his safe box in his safe box !
Luckily, he recalls that the code doesn't have any 0, that the numbers are all different and that :
- the number formed by the first 2 digits is a multiple of 2
- the number formed by the 2nd and the 3rd digits is a multiple of 3
- the number formed by the 3rd and the 4th digits is a multiple of 4
- and so on... until the number formed by the 8th and the 9th digits, multiple of 9.

With this information, he finds out 2 possibilities. What are they ?

Enjoy (Sun)
• Jul 25th 2008, 09:17 AM
galactus
For #2 part 1. Find the remainder when 2008^2008 is divided by 7.

The mod 7 residues of the powers of 2008 are either 1 or 6.

2008^0=1(mod 7)
2008^1=6(mod 7)
2008^2=1(mod 7)
2008^3=6(mod 7)
2008^4=1(mod 7)

and so on................

That whittles it down.

Since 2008/4=502, the remainder is 1.
• Jul 25th 2008, 09:26 AM
galactus
For the first one, is the niece now 26?.

The problem says, "less than 12 years ago". So, when Vincent was 90, then the product is 2*9*5=90.

2+9+5=16

That was 10 years ago, so the niece is now 26 yrs old.
• Jul 25th 2008, 09:38 AM
masters
Quote:

Originally Posted by galactus
For the first one, is the niece now 26?.

The problem says, "less than 12 years ago". So, when Vincent was 90, then the product is 2*9*5=90.

2+9+5=16

That was 10 years ago, so the niece is now 26 yrs old.

The product is 90 and the sum is 14. Therefore, she could be 24 after 10 years.
• Jul 25th 2008, 09:42 AM
galactus
That appears to work as well. I reckon we're looking for only one that works though.
• Jul 25th 2008, 09:49 AM
masters
Quote:

Originally Posted by galactus
That appears to work as well.

The statement "less than 12 years ago" could mean he was 96. I am assuming all three numbers are different.

96 = 3*4*8 with a sum of 15
96 = 2*6*8 with a sum of 16

which would put her at 19 or 20 now.

I must be overthinking this one.....(Wondering)
• Jul 25th 2008, 09:52 AM
galactus
Maybe this is nothing, but does 'told HIS niece' and 'how old is HER niece now?' have anything to do with it?. There appears to be confusion about the uncle/aunt's gender.
• Jul 25th 2008, 09:55 AM
Moo
Quote:

Originally Posted by galactus
Maybe this is nothing, but does 'told HIS niece' and 'how old is HER niece now?' have anything to do with it?. There appears to be confusion about the uncle/aunt's gender.

It's a mistake a French person would often make (Rofl)
There's no trap in it.

And there are 3 solutions (Wink)

The thing is, you have to take care of the information "The niece, though she's good at calculations, wasn't able to guess what the 3 numbers were."

None of your solutions are correct, and I need a complete demonstration :p
• Jul 25th 2008, 09:58 AM
galactus
OK, I thought maybe it was one of those red herring problems.
• Jul 25th 2008, 10:25 AM
masters
Quote:

Originally Posted by Moo
Hi,

#3 : simple geometry
The Swedish flag Attachment 7284 has a yellow cross of constant width in a blue background. The yellow area represents one third of the flag's area.
If the flag is 1m long and 66cm wide, what is the width of the yellow stripes ?

Enjoy (Sun)

Looks like the the flag area is 1m X .66m = .66 square meters.

The yellow cross is one-third of that which is .22 square meters.

Let W = the width of the yellow cross

1 * W = W square meters for the horizontal stripe

.66 * W = .66W square meters for the vertical stripe

Subtract the intersection (W *W= W square) which was counted twice.

Then we have: W + .66W - W^2 = .22

This can't be right. Too much work.
• Jul 25th 2008, 10:52 AM
Moo
Quote:

Originally Posted by masters
Looks like the the flag area is 1m X .66m = .66 square meters.

The yellow cross is one-third of that which is .22 square meters.

Let W = the width of the yellow cross

1 * W = W square meters for the horizontal stripe

.66 * W = .66W square meters for the vertical stripe

Subtract the intersection (W *W= W square) which was counted twice.

Then we have: W + .66W - W^2 = .22

This can't be right. Too much work.

Well, it's right lol. Is this too much work ? :eek:

but i must admit this problem isn't very *interesting*

now, do the others (Evilgrin), especially review the 1st problem :s
• Jul 25th 2008, 12:25 PM
masters
Quote:

Originally Posted by Moo

now, do the others (Evilgrin), especially review the 1st problem :s

The guy had to have been between 88 and 100 years old at the time he talked to his niece.

Between 100 and 88, there are only 3 numbers that can be factored into 3 single digit factors:

90 = 2*5*9
90 = 3*5*6

96 = 2*6*8
96 = 3*4*8

98 = 2*7*7

But you say none of the combinations is the correct solution.
• Jul 25th 2008, 10:46 PM
Moo
Quote:

Originally Posted by masters
The guy had to have been between 88 and 100 years old at the time he talked to his niece.

Between 100 and 88, there are only 3 numbers that can be factored into 3 single digit factors:

90 = 2*5*9
90 = 3*5*6

96 = 2*6*8
96 = 3*4*8

98 = 2*7*7

But you say none of the combinations is the correct solution.

What about 1 as a factor ?

Since the girl can't know what the numbers are, *major hint* you have to find the triplets whose product is equal and whose sum is equal (Tongueout)
• Jul 26th 2008, 03:42 AM
flyingsquirrel
Quote:

Originally Posted by Moo
#4 : aŁg€bra
Teddy left his 9-digit code of his safe box in his safe box !
Luckily, he recalls that the code doesn't have any 0, that the numbers are all different and that :
- the number formed by the first 2 digits is a multiple of 2
- the number formed by the 2nd and the 3rd digits is a multiple of 3
- the number formed by the 3rd and the 4th digits is a multiple of 4
- and so on... until the number formed by the 8th and the 9th digits, multiple of 9.

With this information, he finds out 2 possibilities. What are they ?

*781 254 963* and *187 254 963*

I solved the problem this way :

If the 9 digits of the code are a_1, a_2, ... , a_9 then the number a_8a_9 is a multiple of 9. As a_7a_8 is a multiple of 8, a_8 has to be even hence a_8a_9 is one of the following numbers : 27, 45, 63, 81.

If a_8a_9 = 27 then a_7a_8 is a multiple of 8 which last digit is 2 hence it is either 72 (impossible since a_9=7) either 32. With a_7a_8=32, one has necessarily a_6a_7=63. (only multiple of 7 which ends by 3 and has two digits) As a_4a_5 is a multiple of 5, a_5 is 0 (impossible) or 5. Since there is no multiple of 6 which starts by 5 and ends by a_6=3 the case a_8a_9=27 is impossible.

If a_8a_9=45 then ...

This is really tedious, is there a better method to find the answer ?
• Jul 26th 2008, 03:49 AM
Moo
Quote:

Originally Posted by flyingsquirrel
*781 254 963* and *187 254 963*

I solved the problem this way :

If the 9 digits of the code are a_1, a_2, ... , a_9 then the number a_8a_9 is a multiple of 9. As a_7a_8 is a multiple of 8, a_8 has to be even hence a_8a_9 is one of the following numbers : 27, 45, 63, 81.

If a_8a_9 = 27 then a_7a_8 is a multiple of 8 which last digit is 2 hence it is either 72 (impossible since a_9=7) either 32. With a_7a_8=32, one has necessarily a_6a_7=63. (only multiple of 7 which ends by 3 and has two digits) As a_4a_5 is a multiple of 5, a_5 is 0 (impossible) or 5. Since there is no multiple of 6 which starts by 5 and ends by a_6=3 the case a_8a_9=27 is impossible.

If a_8a_9=45 then ...

This is really tedious, is there a better method to find the answer ?

Unfortunately no :(
These are the correct answers.
Next time I promise, I'll post more interesting problems (Worried)
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