Brain Teaser No 1

quote:

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Originally posted by Don Atkinson:
Minky,

_Otherwise known as the "Hardy-Ramanujan Number"._

i'm almost afraid to ask who, or what, Hardy-Ramanujan is (are ?)

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The answer lies within

quote:

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Femant.......well, I think the margins of this post are slightly too small to explain who he was.....

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Probably. This IS a cool thread though. I am starting to remember why I liked Maths so much so many years ago.

Here's a whopper:

523^3 + 134^3 = 467^3 + 352^3 = 145,461,771

523 is the 99th prime, so I'm sure it's not a multiple.

cheers

Dan

Well done to Don for getting the answer correct and Minky for identifying the origin of the problem.
John

...and for the benefit of those who haven't identified a, b, c and d.......

1 and 12
9 and 10

cheers

Don

Has this one been done? --

What is the lowest pair of consecutive primes whose digits sum to the same number?

cheers

Dan

Dan,

Has this one been done? --

Not in this thread it hasn't........

Cheers

Don

quote:

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Originally posted by Vuk's son:
is the difference 18?

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Yes.

Dan

I believe the answer is 523 and 541.
John

I misread the question and did it for the PRODUCT.

Strictly speaking the lowest are :

101/103

But this is too easy.

The lowest that don't contain the digit zero are :

1913/1931

John,

Well done -- that's the pair I had in mind. I hope my post on the previous page didn't give it away.

cheers

Dan

Omer,

Did you notice that Minky's numbers also differ by 18.

Dan

While trying to get the answer I noticed something interesting. The sum of the digits of prime numbers is never a multiple of 3 (other than 3 itself) i.e. there are no sums equal to 6, 9, 12, 15 etc. I wonder why?
John

As an example here are all the prime number less than 100 with their sums:
2,2
3,3
5,5
7,7
11,2
13,4
17,8
19,10
23,5
29,11
31,4
37,10
41,5
43,7
47,11
53,8
59,14
61,7
67,13
71,8
73,10
79,16
83,11
89,17
97,16
John

John,

I just tried it for the first 100,000 primes and got this result :

3

I'm pretty sure that any number that is a multiple of 3 will have a digital sum that is also a multiple of 3 and vice-versa (try it), and any number that is a multiple of 3 (except for 3) ain't a prime.

'digital sum' is a multiple of 3, so,

a+b=3n

10a+b= (a+b) + 9a

since a+b is a multiple of 3, and 9a is obviously a multiple of 3, the number is also a multiple of 3, as is its digital sum. Extend to infinite digits as required.

Paul

three

When I went to school, (Dickens was halfway through writing Bleak House) we had a simple test to see if a (big) number was divisible by 3.

You added the digits in the number together, and if this sum was divisible by 3, then so was the number.

Seems like John has rediscovered this rule

Cheers

Don

When I heard one of the historic maths problems had been cracked I have to admit to being somewhat disappointed, I was hoping it was going to be Hilbert's problem number 8.
John

In order to give the other teams a chance at the 2007 Rugby World Cup a change is proposed to the laws of the games whereby the game is finished and a team wins when they score any of the following:
1) 10 converted tries (7 points)
2) 10 unconverted tries (5 points)
3) 10 penalties or drop goals (3 points)

Under the new rules, what is the lowest score that it is impossible for a team to achieve?
John

[This message was edited by John Channing on FRIDAY 28 November 2003 at 17:06.]

John,

I'm having a bit of difficulty grasping your rugby teaser. This is probably because I'm not a rugby follower so don't know the basic scoring system.

Given there are TWO teams playing, the lowest score that the LOOSING team couldn't possibly score would seem to be ONE (1).

The WINNING team would seem to be able to score a minimum of 30 (10 penalties) up to a maximum of 70 (10 converted tries)

However,unless there are mutually exclusive scoring combinations, it would seem that they couldn't have 31 as a winning score.

As I said, i'm not a rugby follower, so you might have to explain a few rules....

Cheers

Don

In my haste to create a topical question I have fairly clearly managed to fail to word it so as to exclude trivial solutions i.e. it is impossible to score 1, 2 or 4 points but none of these are the solution I was looking for. I will therefore withdraw this question.
John
Ps. the answer was 141 points

Don,
You can score any combination of tries, converted tries, penalties and drop goals. The equation I was looking at was:

3i + 5j + 7k

where i<=10 or i<10 if j=10 or k=10 etc.
which can't form a total of 141 (but everything else from 5 to 140 and 142).
John

Rotating Dates

1961 was a special year.

Having written the year on a sheet of paper, I noticed that if the sheet was rotated 180 degrees clockwise (or anti-clockwise because that's the same end-result) the date still reads as 1961.

Of course, when I write 1961 I don't put those upward strokes at the top of the 1 nor do I put the horizontal bar across the bottom.....I just write a simple vertical line

What was the most recent previous year that was a 'rotating date' ?

When will be the next year that this will be possible ?

Cheers

Don

I suppose it depends whether it needs to read as the same date when it has been rotated, if it does I would go for 1881. Obviously dates like 1918 or 1981 can be rotated 180 degrees and still look like a date, but obviously not the same one.
John

the same date

John, it has to read as the same date so 1961 and 1881 qualify. 1918 and 1981 don't.

and as to the future.....?

Cheers

Don

Don,

I wonder whether anyone will be around to spot this when it happens in 6009 ?