The Remainder: Newsletter Challenge (12/07/10)

I know the answer, but if I give the answer, no one else will have a chance to try. (Hint: it's a prime number between 20 and 40.) The remainder is .34482759...

Well since there are only 4 primes between 20 and 40 that's a pretty strong hint.

I agree with #1, however there are two answers - the obvious one is 1

since MOD 1 of any number is 0......................

Each of the numbers 9638, 2591, and 8739 is a multiple of the 'mystery divisor' (call it X) plus 10. So the remainder is 10/X.

If you take the inverse of the remainder 1/.34482759 you get 1/10th of the 'mystery divisor' X (or very close, considering there is some rounding-off error).

But if you take the inverse of the remainder for the trivial solution "1" -- i.e., if you take the inverse of 0 -- you get 'infinity'. Which doesn't really tell you anything about what your divisor actually was.

So yes, 1 is a trivial solution, but trivial solutions usually aren't very interesting.

I propose that one of the solutions to this puzzle stated in #2 is a fully valid

answer. Long division of the quotients by the (answer) leaves a remainder of 10

in each case. Long division of the quotients by one also results in remainders

of 0, which is an integer, in each case. Qed: the answer ONE (1) is as fully valid as

the second answer, which should be apparent from the second Hint : (10/r= answer)

IMO USB's point wasn't that you didn't have a VALID solution... only that said solution was trivial.

Three makes it unanimous....

29 is the answer

Easier still

trial and error proved that

Mathematics + Semantics = Politics.

I don't know how that got a GA, it is a "brute force" solution.

FOR i = 1 TO 2591
a = (9638 / i)
a = a - FIX(a)
a = FIX(a * i)
b = (2591 / i)
b = b - FIX(b)
b = FIX(b * i)
c = (8739 / i)
c = c - FIX(c)
c = FIX(c * i)
IF a = b AND b = c THEN PRINT a, b, c, i
NEXT

The answer is Jack Benny's age minus 10.

divider 29

remainder 10

GA from me.

Let == suffice for congruent

Then we have 9638 == x mod y

2591 == x mod y

8739 == x mod y

So (9638-2591)=7047 == 0 mod y Prime factors of 7047 is 3,3,3,3,3,29

Also (9638-8739)=899==0 mod y Prime factors for 899 is 29,31

So the only common factor is 29 and 29 is therefore the dividor

The answer is simple. When the integer is 1, all the remainders are all zero.

Hey Guys and Gals,

I've been well under the weather for the past few months and have not kept up with the rest of you. Mia culpa.

Good grief, Charlie Brown, what have I missed?

Anyway, this morning I decided to look at the latest challange question and immediately my high school Math teacher, Larry Smith, came to mind. His solution to any problem was to go for the simplest answer. Like PrinEngr in answer #18, I came up with one as the logical answer for the reason PrinEngr states, when any number is divided by one, the remainder is always zero. Hence, a very simple answer to an otherwise convoluted process. Kudos to PrinEngr for thinking outside the box!

At least I was not alone in my reasoning!

Kenneth Leigh and I am far from a rocket scientist. Just a simple, backwoods lawyer.

The numbers divided by 29!

Apparently there are a lot of us who used a spreadsheet to figure it out. What does this say about the state of mathematical reasoning today? At least my spreadsheet was elegant.

I don't know if it is the easiest way to do it, but I formulated 3 equations:

9638/x = c1 +y

2591/x = c2 + y

8739/x = c3 + y

You can subtract second from first, second from third, and third from first to get 3 new equations without y:

7047/x = c1 - c2

6148/x = c3 - c2

899/x = c1 -c3

The answer is the LCD of 7047,6148, and 899, which by brute force factoring into primes, is 29.

So, 29 is my answer.

-ebolton

Let 9638 = M*A + R and 8739 = M*B + R, where M is the integer by which we are dividing and R is the remainder which we want to be the same for both divisions. Subtracting one equation from the other, we get 899 = M*(A-B). M, A, and B are all integers so M must be a factor of 899 which has factors 29 and 31. M, the divisor, must be 29 or 31. The remainders for M=31 are 28, 18, and 28 and for M=29, the remainders are all 10.

Thanks,

Jim

Let 9638 = M*A + R and 8739 = M*B + R, where M is the integer by which we are dividing and R is the remainder which we want to be the same for both divisions. Subtracting one equation from the other, we get 899 = M*(A-B). M, A, and B are all integers so M must be a factor of 899 which has factors 29 and 31. M, the divisor, must be 29 or 31. The remainders for M=31 are 28, 18, and 28 and for M=29, the remainders are all 10.

Thanks,

Jim

Very good answer -- I completely forgot to consider negative numbers in my post.

the question is:

(9638-r)/n=a

(2591-r)/n=b

(8739-r)/n=c

or

9638-2591=7047=n(a-b)

9638-8739=899=n(a-c)

2591-8739=-6148=n(b-c)

all integer factors of 899 are: 1, 29, 31

only 2 of these integer numbers are factors of the other 2 equations: 1 and 29

if the solution 1 is to simple (trivial) there is only the integer 29 as used divisor

all integer factors of 899 are: 1, 29, 31

and the fourth integer factor is 899 (trivial, trivialer, trivialst)

'I let the machine do it for me.

'Copy and paste into an excel macro, then run the macro

'Lot's of extra fluff here, I like visual...

'Hey, where's my tabulation spacing??? Dang Programmers.

Private Sub FindAnswer()

Range("A1").Value = 9638
Range("A2").Value = 2591
Range("A3").Value = 8739

For I = 2 To 2592

Range("B1").Value = I
Range("B2").Value = I
Range("B3").Value = I

Range("C1").Value = Range("A1").Value / Range("B1").Value
Range("C2").Value = Range("A2").Value / Range("B2").Value
Range("C3").Value = Range("A3").Value / Range("B3").Value

Range("D1").Value = Range("A1").Value Mod Range("B1").Value
Range("D2").Value = Range("A2").Value Mod Range("B2").Value
Range("D3").Value = Range("A3").Value Mod Range("B3").Value

If (Range("D1").Value = Range("D2").Value) And (Range("D2").Value = Range("D3").Value) Then

MsgBox ("Found Answer")
Exit Sub

End If

Next I

End Sub

<?

$nums = array(9638, 2591, 8739);

for($i=;$i<1000;$i++){

foreach($nums as $index => $num){
$n[$index]=fmod($num,$i);
}

if ($n[0]==$n[1] && $n[1] == $n[2]){
echo $i."\r\n";
print_r($n);

break;
}


}

?>

29
Array
(
[0] => 10
[1] => 10
[2] => 10
)

If the unknown divisor is D and the remainder is R, and the lower case pronumerals a, b, c, d etc are integers, then 9638 = aD +R; 2591 = bD + R; 8739 = cD + R. So 9638 - 8739 = 899 = (a-c)D, i.e. 899 is a multiple of D.

Now factorising a number is a notorious brute force problem (as far as I know), but 899 happens to be 900 - 1 = (30+1)x(30-1) = 31 x 29. So D must be either 29 or 31. Both can be tried easily and I find that D must be 29.

More generally, we can write 9638 - 899x10 = 648, so if we divide out by D, we see that 648 = dD + R; and similarly, 2591 - 899x2 = 793 so 793 = eD + R.

So 793 - 648 = 145 is a multiple of D. We are now working with much smaller numbers, so we can easily see that 145 = 5x29, and since 5 is not a factor of 899, D must be 29.

Or, we can continue the same way: 899 = 6x145 + 29. So 29 must be a multiple of D. Since 29 is known to be a prime number, D must be 29 (Or, trivially, 1).

Simple programming solutions also exist. I do not believe that using them is cheating.

also you can say: 899 = (a+b)*(a-b) = a²-b² = 900 - 1 = 30²-1² = (30+1)*(30-1) = 31*29

When 9638, 2591, and 8739 are divided by a certain integer, the remainder is the same for each. What is the integer used as the divisor?

Calc the difference between each of the 3 numbers -

ie 899, 6148 & 7047

Factor them

899 = 29 * 31

6148 = 29 * 212

7047 = 29 * 243

Divisor is 29

9639

Well this puzzle has been solved, resolved, & programmed in several ways. The answers are 29, 1 & their negative values by general agreement. While solvable quickly using "Brute Force" programming techniques, other solutions illustrated the wonderful way numbers interact.

But now we have considerable time left over including the upcomming Holiday, when many may have spare time for contemplation. So here is another puzzle for those interested inremainder problems.

A rancher has been accumulating stock toward reaching the maximum capacity of his place of 500 head. One day he decides to count how many he had acquired, so had them all rounded up. He had a number of pens available so they ran half into the first pen and half into the second pen. But there was on o'l cow left over. So they took a fraction of the cows from the first & a fraction of the cows from the second and put them in the next pen so there was an equal number in each of the three pens - but - there was one o'l cow left over. Dividing the cows equally into 4, 5 & 6 pens always left one o'l cow left over. Finally they moved the cows so that there was equal number in seven pens - and - ii viola !! there wasn't any cow left over. So how many cows did the Rancher have???.

Not to give away the answer too quickly, but I'll start the ball rolling by noting that LCM(2,3,4,5,6) = 60.

7*a=n

6*b=n-1

5*c=n-1

4*d=n-1

4*d=5*c=6*b=7*a-1

a,b,c,d,n are integers

the number n has a last digit with 1 or 6

4,5,6 results in numbers like 60+m*60, m=0,1,2,...

7 results in number like 21+l*70 or 56*l*70 with l=0,1,2,... because 4,5,6 the result must be something with 21+l*70 = 60+m*60+1

the shortest solution is 4*70=40+4*60=280

Sorry. but 280 cows can be divided evenly into 2, 4, or 5 pens.

21+l*70=60+m*60+1

if l=m l=40/(70-60)=4

the value is 301

301-1=300 is dividable by 2, 3, 4, 5, 6

301 is dividable by 7

301=280+21

other solutions are existing for

l=4+i*6 and m=l+i where i=0,1,2,3,...

The first two lines may be true, but what motivates them?; where do they come from?

see post 38

The answer to he cows puzzle depends on finding a number so that when it is divided by 2,3,4,5,6 there is only 1 left over, when divided into 7 pens there are none left over - nice try - but not correct as has been pointed out. Condider how the cows could be moved from pen to pen........ (Hint : "a fraction of the cows from pen 1" etc...)

Next hint: 60 = 4(mod 7).

29

29 is not an integer, it has two integers.

I guess not many people remember their formula for finding the GCD (greatest common divisor)

(Step 1 - observation about remainders)

Because the challenge numbers all have the same remainder when divided by the mystery number, the difference between any two will have a zero remainder when divided by the mystery number.. if our numbers are x,y,z, we want to use (e.g.) (x-y) and (z-y) and find their GCD.

(Step 2: finding the GCD);

Finding the GCD of two numbers is an iterative process.

Divide the smaller into the larger, returning a remainder. Discard the larger number, now we have the prior divisor and the remainder of the division.

If the remainder was zero, the divisor was the GCD. if not repeat the process (the prior divisor is now the "larger number" and the remainder is the "smaller"

e.g. What is the GCD of 77 and 21?

77/21 = 3 r 14

21/14 = 1 r 7

14/7 = 2 r 0

7 is the GCD.

(Step 3 - apply the technique)

find the GCD of (9638-2591) and (8739-2591)

Share and Enjoy!

RufusVS

If memory serves, that's called "Euclid's algorithm."

Fantastic technique. I was never taught it and this is the first time I've seen it.

Thanks for sharing!

Three businessmen rent a hotel room for a night. The hotel manager tells them it will be $30 for one night. So each man hands over $10 and proceed to the room. The manager realizing he has overcharged the three men hands the bell hop $5 to refund them. The bell hop decides that $5 doesn't evenly divide by the 3 patrons, so he pockets $2 (of the $5) and gives a $1 refund to each guest.

From the businessmen's point of view they have each paid $9. So… (3 x $9 = $27) plus the $2 in the bell hop's pocket equals $29 dollars. So what happened to the other dollar?

Nothing. The addition/subtraction used in this story is meaningless.

I've always liked this one... perhaps it needs to be retitled "Lehman Brothers Accounting 101"

Nothing. Money is just redistributed.

Manager had $30. Now manager has $25, bell hop $2 and 3 businessman $1 each.

Another way of looking at it is 3 x $9 = $27 = $25 for manager and $2 for bell hop.

25+3=28

2 for the bell hop

When 9638, 2591, and 8739 are divided by 1 & 29, the remainder is the same for each.

-1, 1, -29, 29

-1, 1, -29, and 29.

From Groucho

Answer + 1, - 1 and + 29 and - 29

Divizor- 29, Remainder-10

7

integers 2,4,19,38 and 76 all give the same remainder when used as the divisor to 96382591 and 8739.

Regards

Ken

My out-of-the box chuckle at this one died when I realized the groupings were 4 rather than 3.

Clever in another universe, maybe...

29

29

The divisor is 29, for which the remainder is 10 in all cases.

I'd like to say that my fancy three equations and three unknowns math solved this, but I counted 5 unknowns, and then used a spreadsheet to calculate the MOD, and filter out only the remainders that matched the average of all remainders.

This yielded the answer, but I'd like to see the math proof.

Huh? What does "the average of all remainders" have to do with anything?

For what it is worth the original solution results from dividing the cows into 6 pens, then removing "x" cows from each and running them into pen 7. Then put the odd cow in pen 7 also. So now you have (N - 1)/6 - x left in pen 6 & (6 * x) + 1 in pen 7. Since the number now in pen 7 equals the number left in 6 the following equation gives the solution to the puzzle.

(N -1)/6 -x = 6x + 1 Simplifying & rearranging gives :

N = 42x + 7 Solution by trial is :

x .....N .......N/x

4 175 43.75

5 217 43.40

6 259 43.17

7 301 43.00 Answer !

Best wishes to all for a Happy Holiday !!!!

where is pen 5?

Pen 5 is between Pen 4 & Pen 6 . All the pens (1 - 7) now have 43 cows in each. Hope this clarifies the solution.

This month's Challenge Question:

When 9638, 2591, and 8739 are divided by a certain integer, the remainder is the same for each. What is the integer used as the divisor?

Divisor is 29. Remainder is 10 for each no.

Ans 3 . Michael Piner Wilmington NC

Ignoring '1' as a trivialsolution, the differences between the numbers (a-b, a-c, b-c) will contain one (or possibly more) common factor(s) F of the solution and set upper bounds to the solution, assuming that the smallest difference must be an integer multiple of the result.

The remainder R is given by

R= a-F*integer(a/F),

This can be checked with the other numbers b & c. The highest common factor of (a-R), (b-R) and (c-R) is the result. Alternatively, the result can be viewed as a product F*F1, where we already know F. F1 is The highest common factor of (a-R)/F, (b-R)/F and (c-R)/F.

and what are these numbers F, F1, R