Hi Ed, Matt Sheppeck here,
Thought I'd comment on the dice trick.
The outer digit pairs have a constant sum for each die:
7, 10, 9, 13, and 8. For a total of 47.
The middle digits are constant for each die as well:
8, 7, 6, 5, and 4. For a total of 30
So overall, the individual digits of the five topmost number on
the dice will always sum to 77, which is nice.
More importantly, these two traits lead to the following:
The first two digits of the four-digit final sum will have the
form L+3, while the last two digits will have the form R+0 (=
R).
Also, L+R=47.
(where L is the sum of the lefthand digits,
and R is the sum of the righthand digits.)
Since L+3+R=50, the four-digit final sum
will have the form "L+3,50-(L+3)" OR "50-R,R"
So one can quickly calculate the final sum by calculating the
sum of either the lefthand column or the righthand column.
(remembering to add 3 to the lefthand total, if it is used)and
then finding its complement with respect to 50.
On whether to Find L or R:
Roll: 186 + 179 + 168 + 459 + 147
L looks easier to find L = 1 + 1 + 1 + 4 + 1 = 8,
so, L+3= 8+3 = 11, 50-11=39, and the final sum is 1139.
Roll: 780 + 971 + 960 + 954 + 840
R looks easier to find R = 0 + 1 + 0 + 4 + 0 = 5
so, 50-5=45, and the final sum is 4505.
When they look equally easy to determine, it may be preferable
to Find L, since one can begin speaking the final result as they
are determining the last two digits. To conceal the method, and
sound impressive, I also recommend saying the result in long
form. (For example you find L+3=11, begin saying "one thousand,
one hundred and..." as you calculate 50-11=39, rather than
saying "eleven ahhhmm... 39."
Those are my thoughts. Now tell me I'm mising something that
makes the result still more 'immediate.'
- Matt
By the way, I originally thought it relied on the fact that the
numbers might be read differently upside down (186 as 981 or 960
as 096) but this must not happen or it WON'T work (an FYI for
those making the dice at home).
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Hi Ed,
The sum of the 10's digits will always add to 30 (8+7+6+5+4).
Notice that the sum of the 1's and 100's digits always add to 47
(7+10+9+13+8). So, the sum of the dice will be:
ABCD, where:
CD = the sum of the 1's
AB = the sum of the 100's + 3
= 47 - CD + 3
= 50 - CD
So, to get the last two digits, simply add the last digits from each
cube. To get the first two digits, subtract this number from 50.
Ex. 285,773,762,459,147
5+3+2+9+7 = 26
50 - 26 = 24
Answer = 2426
--
Nathan Stohler
--------------------------------------
Adding the sum of any five dice will give a four digit number between 1139 and 4505.
The trick is to add the last digit on each dice (this can be done fairly quickly)
e.g.
384
377
762
855
543
4+7+2+5+3 = 21
Twenty one is the last two digits of the sum.
The first two digits of the sum then can be calculated by subtracting this number from 50
e.g. 50-21 = 29
Therefore the sum of all four dice is 2921
...........
Alternatively (this is how I first analyzed the puzzle)
I noticed each dice had a unique second digit
e.g.
8
7
6
5
4
this will always contribute 300 to the total sum
80 + 70 +60 + 50 +40 = 300
the first and last digit on each dice also add up to a unique number
e.g.
384 3+4 =7
377 3+7 = 10
762 7+2 = 9
855 8+5 =13
543 5+3 = 8
by adding this combination you get
7 + 10 + 9 +13 +8 = 47
Now the total sum minus 300 must have the first two digits and the last two digits add up to 47
eg.
sum = ABCD and ABCD-300 = EFGH
EF +
GH
47
From the previous example EF will be
3+3+7+8+5 = 26
The last two digits of the final sum then are 47-26 = 21
putting these together results in 2621 - then add in the 300
The first two digits are then the 26 (equivalent to 2600) plus the 3 (for the 300 middle dice contribution) which is 29
again 2921 is the answer
After a few trials with this method it became obvious that the first two digits and the last two digits added up to 50 in each case and it was easier to add the last digit on each dice.
Bill Downs
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In the March 1978 issue of Games&Puzzles, J O P Sweeney wrote a letter about a dice trick he'd purchased many years earlier. The five dice are labeled 186-285-384-483-681-780, 179-278-377-773-872-971, 168-366-564-663-762-960, 459-558-657-756-855-954, 147-345-543-642-741-840. One rolls the five dice, and immediately announces the sum of the five topmost numbers. Of course, there's a trick to it. What is it?
Here is the answer.
Just add the last digits of 5 numbers. They will give you last 2 digts of the answer. First 2 digits can be found by subtracting last 2 digits from 50. So if last 2 digts are 39 then first 2 digits are 11 and the answer is 1139.
Nice one
My wife showed me this sight today.
Thanks.
Umesh
---------------------------------------------------
Dear Sir / Madam,
This is my answer to the puzzle of 6 July 2003 which was hosted at
http://www.mathpuzzle.com.
The first two digits of the sum = Sum of the first digits of all the five
topmost numbers + 3
The last two digits of the sum = Sum of the last digits of all the five
topmost numbers.
Thanks.
Sourav Banerjee.
--------------------------------------------------
Hi, here is what I found out for the dice-trick:
Start with 347, add the sum of the first digits to the 3 and subtract the same sum from 47, that 'll do. :)
kind regards,
Johan de Ruiter
the Netherlands
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Add the units digits. This sum matches the last two digits of the dice
sum. The 50-complement of the units sum gives the first two digits of
the dice sum.
Almost as easy: Add the hundreds digits. Add 3 to this total to get
the first two digits of the dice sum. 50 minus the first two digits
gives the last two digits. The advantage: you can begin to say the
dice sum before you think about the 50-complement.
Al Stanger
----------------------------------------------------
the trick: the sum of the five dice is always a four digit number, abcd,
where ab + cd = 50. as a result, all the performer must do is figure EITHER
ab or cd, and then subtract from 50 to get the solution.
i'm sure you'll get better explanations from number theorists for why
figuring either ab or cd is trivial to do, but here's a layperson's
explanation: the "tens" are always going to be 4, 5, 6, 7 and 8, and so
will ALWAYS sum to 30. therefore, summing the "ones" will yield cd every
time, and summing the "hundreds" (and then adding 3) will yield ab every
time.
so roll the dice, add five single digits and announce the solution one or
two seconds later. an impressive trick.
george tolley
-------------------------------------------------------
Add the rightmost digits of the dice showing, this is XX, subtract this
number from 50 to get YY, then the sum is YYXX
-Travis Taylor
-------------------------------------------------------
> The five dice are labeled 186-285-384-483-681-780, 179-278-377-773-872-971,
> 168-366-564-663-762-960, 459-558-657-756-855-954, 147-345-543-642-741-840.
> One rolls the five dice, and immediately announces the sum of the five
topmost
> numbers. Of course, there's a trick to it. What is it?
How:
Add up the last digits on each dice. These are the last two digits (pad
with 0 if necessary). Subtract this number from 50. These are the first
two digits. Putting the two together yields the total.
Example: 186,377,960,459,147
6+7+0+9+7 = 29, 50-29=21, so the total is 2129.
186+377+960+459+147 = 2129
Why:
Each die has the same center digit on all sides, and the first and last
digits on all sides have the same sum on each die: 7, 10, 9, 13, and 8
respectively. The center numbers sum to 30, and the digit sums sum to 47.
Therefore, if the last digits of a particular roll sum to x, the 3-digit
numbers sum to (47-x)*100 + 30*10 + x = (50-x)*100 + x.
--Stephen Kloder
------------------------------------------------------
The tens digit of each die is a constant, so that digit will always add up to 80 + 70 + 60 + 50 + 40 = 300.
The sum of the first and third digits on each die is also a constant, so the numbers on each die vary from each other by a multiple of 99.
To get the sum of the five dice, add up just the 5 "hundreds" digits. Call this number h. The total of the five 3-digit numbers is 100 * (h+3) + (47 - h). The sum is always a 4-digit number; the first two digits plus the last two digits always equals 50.
Don Rogers
------------------------------------------------------
Ed,
I just spotted the 1st+last digit on the numbers of a die is constant too...
(all 1st + all last = 47)
Therefore:
1) Add up the last digits of the 5 numbers. (quite trivial). This becomes
the last 2 digits of our result.
2) Subtract this number from 50, to produce the 1st 2 digits.
This will appear to be instant.
Alternatively, if the first digits are easier to add up, (eg, 1,1,1,1,4),
add these up, add 3 to produce the first 2 digits.
Subtract this 2-digit number from 50, to produce the last 2 digits again.
Cheers,
Alan O'Donnell.
---------------------------------------------------------
The numbers on each die have the same value mod 99
hence the sum of the dice always have to have
remainder 50 mod 99.
A number in the possible range: 1139 to 4505 must have
the form:
abcd where ab + cd = 50, hence if we know the leading
two digits, we know the remaining two digits. This is
since any number in this form satisfies;
n = 100 x + ( 50 - x ) (here x is the two digit number
ab).
Or n = 99 x + 50.
You quickly determine the first two digits by
observing that the carry from adding the five numbers
is always 3 (i.e., the last two digits of each number
add between 311 and 339 to the sum).
Hence the final rules are:
1) Add the leading digits of the five dice and add
three to that number to get a two digit number ab.
2) Subtract 50 - ab to get the two digit number cd.
3) The answer is the four digit number abcd.
Lyman Hurd
---------------------------------------------------
I'm still working on the explanation, but I found this method:
Take the most significant digit of each topmost face.
Sum them all
Add 3 and you'll have the 2 most significant digits of the result.
Substrat this from 50 and there are the 2 least significant digits of the result.
All possible sums are different (5*7=35 results)
RAMILO, S.A.
--------------------------------------------
Immediately??? Maybe for JHC; it'd take me a few seconds just to read the dice, let alone process them! ...Add the ones digits--these are the last two digits of the total; the first two digits are 50-that sum.
Nick Baxter
-----------------------------------------------
Hi Ed
The answer to the five dice problem is as follows
Calculate the sum of the digits in the units column (call it S)
Then calculate T = 50 - S
The sum of the three numbers on the upper face of the dice is merely the
concatenation of T and S
Eg
Suppose the five numbers rolled are 483, 278, 762, 657 and 543
S = 23 hence T = 27
And the total sum therefore is 2723 (which is correct)
Regards
Pete Kogel
-------------------------------------------------
The numbers on the dice differ by multiples of 99, so every time the
first two digits of the sum go up, the last two go down by the same
amount. So you only need to add the hundreds digits, add three for the
first two digits, and subtract that from 50 for the last two. Maybe it
can be further simplified, but I can do that in a couple of seconds.
--
Always carry a grapefruit, Treesong
---------------------------------------------------
The answer will be a four digit number. Sum the final digits of the
dice to get the final two digits of the answer; subtract this from 50
to get the first two digits.
I think it must be simpler than that, though, in order to call out the
answer "immediately"-- summing 5 digits in my head usually takes more
than a few seconds. But I can't see any more shortcuts...
--Doug Orleans