Brain Teaser No 1

Don,
I thought it was too easy! Apology accepted!
Steve D

Another simple one, with plenty of time over the weekend to respond...

A Trip to Salisbury

Tim and Tom drive to Salisbury for a visit to the Naim factory. On the way there, Tim drives the first 45 miles then Tom drives until they reach the factory.

They have a superb day at the factory, including the usual tour; three course cordon blue lunch, and free selection of 'gift'….Tom picked a 500 power amp whilst Tim chose a cds3, nothing outrageous. (Cherie would have cleaned the factory out!)

Tim then drove back except for the last 55 miles which Tom drives.

Who drove the furthest, and by how much?

Cheers

Don

If the distance to the Naim factory is x, then on the way there we have
Tim = 45 miles
Tom = (x-45) miles
On the way back we have
Tim = (x-55) miles
Tom = 55 miles
So in total we have
Tim = (x-10) miles
Tom = (x+10) miles
So Tom drives 20 miles further than Tim.
John

Another simple one

What is the smallest four digit number that cannot be formed from the sum of consecutive numbers.

For example 334+335+336 = 1005

So 1005 is not a valid answer.

I picked 334/335/336 because they seemed like powerful music to my ears......

Cheers

Don

Nine minutes

I just noticed that John took nine minutes to notice, read, solve, write and publish his answer to the Tim and Tom question.......

Probably some kind of record!!

Well done!

Cheers

Don

what it can't be...

Don,

Well, very quick inspection says it can't be odd, since for x>999, the pair (x/2,x/2+1) will total to x (using integer arithmetic).

Dan

Dan,

OK, it is an even number.

Cheers

Don

8192?

Not proven to my own satisfaction, so I might well change my mind....

Paul

Don,

I'm still looking for a solution to this one - but I'm afraid it might require one of those brute force spreadsheet solutions. My only progress is that the possible totals (t) for a sequence of m numbers ending in n is:

t = m(2n-m+1)/2

For the case you gave n = 336 and m = 3:
t = 3(672-3+1)/2 = 1005

So it's down to looking for a t such that the equation above has no integer solutions (n,m) with m<n. Not sure that this gets me anywhere though,

cheers,

Dan

p.s. I'm a little worried - this is supposed to be a 'simple one'

[This message was edited by d marsh on WEDNESDAY 24 September 2003 at 02:10.]

1024

1024 too.

(It's blindingly obvious that the sum of {1024} is 1024, I presume 'consecutive implies more than one....)

Paul

Minky and Paul,

Can you please show you proof.

TIA,

Dan

I did an exhaustive search with a computer, which I suppose qualifies as proof nowadays.

The impossibles are 1024, 2048, 4096 and 8092. It's likely that there's an elegant proof, given the nature of those numbers...

#include <iostream>
#include <bitset>

int main(int argc, char* argv[])
{
	std::bitset< 10000 > bs ;

	for ( int i = 1; i < 5001; ++i )
	{
		int t ( i ) ;
		int n ( i + 1  ) ;
		while ( t < 9999 )
		{
			t += n ;
			if ( t > 999 && t < 10000)
			{
				bs.set ( t ) ;
			}
			++n ;
		}
	}

	std::cout << "The following numbers between 1000 and 10000 cannot be constructed from summing consecutive numbers...\n" ;

	for ( i = 1000; i < 10000; ++i )
	{
		if ( !bs.test ( i ))
		{
			std::cout << i << "\n" ;
		}
	}

	return 0;
}

Paul

Or to put it another way (Cache object script) :

Test;
F I=1000:2:9998 D Find Q:'FOUND
Q
Find;
W I," "
F J=1:1:I/2 D Q:TOT=I
.S TOT=0,STRING=""
.F K=J+1:1:I/2 S TOT=TOT+K,STRING=STRING_"+"_K Q:TOT=I!(TOT>I)
S FOUND=TOT=I
I 'FOUND W "Gotcha !",! Q
S $E(STRING,1)=""
W "= ",STRING,!
Q

Which gives :

1000 = 28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+5
2
1002 = 78+79+80+81+82+83+84+85+86+87+88+89
1004 = 122+123+124+125+126+127+128+129
1006 = 250+251+252+253
1008 =16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+4
0+41+42+43+44+45+46+47
1010 = 41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60
1012 = 33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55
1014 = 7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+3
2+33+34+35+36+37+38+39+40+41+42+43+44+45
1016 = 56+57+58+59+60+61+62+63+64+65+66+67+68+69+70+71
1018 = 253+254+255+256
1020 = 6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31
+32+33+34+35+36+37+38+39+40+41+42+43+44+45
1022 = 23+24+25+26+27+28+29+30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+4
7+48+49+50
1024 Gotcha !

Not so much a proof, more of a deduction:-
Any odd integer has 2 consecutive numbers which sum to it.
Therefore the answer must be even.
An even number cannot have 2 consecutive numbers which sum to it, but may have more than 2.
If the integer is even, keep dividing by 2 until the result becomes odd. Then, consecutive numbers exist, and can be deduced.
If continuing to divide by 2 always produces an even result (until it reaches 2), then no consecutive numbers exist.
Therefore, an integer which cannot be summed by consecutive numbers must be a power of 2.
The first 4 digit power of 2 is 1024, or 2 to the power 10.
Steve D

Not so much a proof, more of a deduction:-

and a pretty good deduction at that....

Cheers

Don

quote:

---

and a pretty good deduction at that....

---

Seconded. Minky, what are 'Cache object script(s)' used for?

cheers,

Dan

Dan,

Cache is a "post relational" OO database and programming environment. Cache object script (COS) is it's native language. It's super fast and packed with cool stuff.

Have a look at www.intersys.com

Embellishment gone mad....

So, for those of you unlucky enough to miss the Heathrow hifi show, let me remind you that the BIG HIT of the show was the little SILVER push-on facia embellisher that Naim have produced for their main amplifier range.

Now whilst I was at the show, I forgot to ask whether Naim have produced an oversized embellisher to fit the 500 power amp, so I popped into my local dealer's at the weekend to find out. He didn't know! BUT.......he has heard rumours that Naim are producing a range of embellishers each with a single letter of the alphabet cut out, presumably so you can string them together and spell out your name or something like that. He said that although all the As cost the same and all the Bs cost the same, etc different letters sometimes cost different amounts with a surprisingly wide range of prices.

I wanted to spell out 'one' 'two' 'three' 'four' etc so as to number my equipment according to it's shelf. I noticed that the price of the letters needed to spell out 'four' cost me £4. Surprised by this coincidence I worked out the cost of spelling out each of the numbers 'one' to 'twelve' because I have 12 shelves. In ten out of the twelve cases the cost in pounds equalled the number being spelled out.

For which shelf numbers was the cost different from the number?

Cheers

Don

Oblongs !!

What is the area of the fourth rectangle?

The four smaller rectangles make a single large rectangle

Cheers

Don

242.6667

I would concur with Minky on 242 2/3. And I'd go so far as to say that the top left quadrant (with area 91) is a square.

How can I put this....


Minky and two-sheds

I could pretend that version one was intentional, especially since the top left turned out to be a square......

However, it was meant to be a simple brain teaser, that could be done in the head....

So I apologise for the careless goof....and publish the right picture below

But thanks for tackling the first version

Cheers

Don

Don,

Which is why I didn't offer any proof. You could do this by (with area required = p, x-axis of left side = x, y-axis of bottom left = z, y-axis of top right = t)

t = 91/x
x = 208/z
t = 91/208/z
p = 91z/208 * 288/z
p = 91*288/208 = 126

But given that p has the same relationship with 288 that 91 has with 208 why not just say :

p = 91 / 208 * 208 = 126

I get 126 also, via the area ratio method --

A = 288*91/208
= (12*12*2) * (7*13) / (13*8*2)
= (12 * 12 * 7) / 8
= 3 * 6 * 7
= 3 * 42
= 126

cheers,

Dan