## Summer!

cinelli
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### Summer!

I think I had better set this alphametic before the summer is over. I know it is wishful thinking but perhaps

`'   100   MORE  SOLVE    OUR SUMMER-------PROBLEM`

Cinelli

Gengulphus
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### Re: Summer!

cinelli wrote:I think I had better set this alphametic before the summer is over. I know it is wishful thinking but perhaps

`'   100   MORE  SOLVE    OUR SUMMER-------PROBLEM`

Spoiler...

To start with, there are 10 distinct letters, which must all be different, and therefore the 100 should not be regarded as 'reserving' the digits 1 and 0. Then the no-leading-zeros convention says that PROBLEM >= 1000000, and clearly PROBLEM = 100+MORE+SOLVE+OUR+SUMMER <= 100+9999+99999+999+999999 = 1111096, so P must be 1 and R must be 0 (since it's distinct from P).

Next consider the sum 100+MORE+SOLVE+OUR+UMMER. It must equal cOBLEM, where c is the carry into the top two digits, so 10 = PR = S+c. So c cannot be 0, and since we know that cOBLEM = 100+MORE+SOLVE+OUR+UMMER <= 100+9999+99999+999+99999 = 211096, c must be 1 or 2. But if c = 2, then O must be 0 or 1 to avoid exceeding 211096, and since both 0 and 1 are both already taken, c must be 1, from which it follows that S = 10-c = 9. Substituting P=1, R=0, S=9 into the sum leaves us with:

`    100   MO0E  9OLVE    OU0 9UMME0-------10OBLEM`

Looking at the rightmost column, it's clear that M must be even and not 0. So E is:

* Either one of 1-4, in which case there is no carry into the column to its left, so that column shows that V+U+E has last digit E, or that V+U has last digit 0. The only possibility is that V+U = 10 and V+U+E = 1E, i.e. there is a carry of 1 from that column to the column to its left.

* Or one of 6-9, in which case there is a carry of 1 into the column to its left, so that column shows that V+U+E+1 has last digit E, or that V+U+1 has last digit 0. The only possibility is that V+U = 9 and V+U+E+1 = 1E, i.e. there is a carry of 1 from that column to the column to its left.

So either way, there is a carry of 1 into the third rightmost column, which therefore shows that 1+O+L+O+M+1 has last digit L, so that 2*O+M+2 has last digit 0, so 2*O+M has last digit 8. It is the sum of three digits that are each <= 9, so it cannot be 28 or more and must be 8 or 18. Bearing in mind the fact that M must be even, nonzero and the bottom digit of twice E, we have the following possibilities:

`M  O  E------------2  3  1 or 64  2  2 or 76  1  3 or 88  0  3 or 82  8  1 or 64  7  2 or 76  6  3 or 88  5  4 or 9`

where I've struck out the ones ruled out by duplicate digits.

The carry out from the third rightmost column is the upper digit of 2*O+M+2. Calling that carry out c (not the same c as before), the column to its left shows that 2*M+O+c has last digit B. So for each of the remaining five possibilities, we can calculate the value of B:

`M  O  E  c  2*M+O+c  B  remaining digits----------------------------------------2  3  6  1   8       8  4,5,74  2  7  1  11       12  8  6  2  14       4  3,5,74  7  2  2  17       78  5  4  2  23       3  2,6,7`

where I've struck out the possibilities where B duplicates one of the three already-taken values 0, 1 and 9 or one of M, O or E, and listed the three digits remaining for L, U and V for each possibility we're left with. Recalling from above that if E is 1-4, V+U must equal 10 and if E is 6-9, V+U must equal 9, only the first possibility allows U and V to be chosen correctly. So the solution has M=2, O=3, E=6, B=8, U and V equal to 4 or 5 in one order or the other, and therefore L=7. It also has no carry out from the 4th rightmost column (since 2*M+O+c = 8 < 10) and so the top three columns show that 9+9U = 10O = 103, so U=4 and V=5, rather than the other way around. This completes the solution:

`    100   2306  93756    340 942260-------1038762`

Gengulphus

OLTB
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### Re: Summer!

cinelli wrote:I think I had better set this alphametic before the summer is over. I know it is wishful thinking but perhaps

`'   100   MORE  SOLVE    OUR SUMMER-------PROBLEM`

Cinelli

I've been looking at this all day (between working!) and I think I know the first number! I'm not a Maths graduate (just to 'O' level standard) but really enjoy trying to do these quizzes set by cinelli so I hope my theory is right. If so, I'll try and work out the next number!

cinelli and Gengulphus have given me pointers in the past so am I right in thinking that...

The 100 at the top of the question is a bit of a red herring as there aren't any letters to exchange and it's there to make the sum work? If so, I leave that alone.

Then, as Gengulphus has explained in the past, the first number of the answer (in this case the 'P' from PROBLEM) can't be a 0 so has to be between 1 and 9. The highest number that PROBLEM could be is 100+9999+99999+999+999999 which = 1,111,096. As the 'P' in PROBLEM can't be a 0 or higher than 1, my theory is that P must = 1. Actually, when I look at the answer, this must mean that the 'R' can't be a 1 as 'P' is and as the number 1,111,096 is the highest possible number, would I be right in thinking that 'R' can only be a 0?

That's all I've done so far and will try to work out some more, but not too sure on what the best next step should be.

Thanks again, cheers, OLTB.

cinelli
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### Re: Summer!

`'   100   MORE  SOLVE    OUR SUMMER-------PROBLEM-------11dcba`

Well solved, Gengulphus. My solution isn’t very different from yours. I feel happier if I can write out a series of simultaneous equations. I am including all the carry digits, a, b, c and d. This method may not work as well for all alphametics but here, it is quite quick. By inspection, P=1, R=0 and S=9. Working from the right

(1) 2*E = 10*a + M
(2) V + U + a = 10*b
(3) 1 + 2*O + M + b = 10*c
(4) 2*M + O + c = B + 10*d
(5) 9 + U + d = 10 + O

You may think this is complicating the issue but we can deduce that

From (1), M is even.
From (2), even if U and V are as large as possible, V + U + a cannot be
as high as 20, so b = 1.
Note that L does not appear.

We can rearrange these equations as follows:

(1) M = 2*E - 10*a
(2) V = 10*b - U - a
(3) O = (10*c - M - 2)/2
(4) B = 2*M + O + c - 10*d
(5) U = O + 1 - d

and if we use these equations in the order (1), (3), (4), (5), (2), the only digits we can choose in a systematic trial and error process are E and c. For example if E = 2 and c = 2, then it follows that M = 4, a = 0, O = 7, d = 1 and B = 7. The process stops here as we have a duplication of 7. Work systematically in tabular fashion as follows:

`E a M c O d B U V2 0 4 1 2x      2 7 1 7x3 0 6 1 1x      2 6x4 0 8 1 0x      2 5 2 3 4x5 1 0x6 1 2 1 3 0 8 4 5      2 8 1 4 8x7 1 4 1 2 1 1x      2 7x8 1 6 1 1x      2 6x`

x indicates that we stop as we have a duplication. Only one line proceeds without error, E = 6, M = 2, etc. The only digit not assigned is 7 so this must be L. The final sum with carries is

`'   100   2306  93756    340 942260-------1038762-------110111`

Cinelli

eepee
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### Re: Summer!

Many years ago I was inspired to write a program because a colleague had come across this alphametic in some magazine.

I THINK it was:-

SPRING
BRING
RAINS
GREEN
---------
PLAINS

Please not I have not just done it to prove I have the right words.

The program was written in ASM under CP/M! The person that suggested it thought we would make a lot of money by say, picking the word TESCO and creating alphametics with that word, that could then be sold to the said company! I don't think he ever sold any to anybody!

Regards,
ep

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### Re: Summer!

eepee wrote:I THINK it was:-

SPRING
BRING
RAINS
GREEN
---------
PLAINS

Hmm, I think it was more likely:

SPRING
RAINS
BRING
GREEN
---------
PLAINS GS

eepee
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### Re: Summer!

Don't tell me I am going to have to do it again after all these years!

Guess if it doesn't work first time I would have to try other variations such as BRING / BRINGS and RAIN / RAINS

Funnily enough, I know there is a box of floppy discs in the attic. It would not suprise me if one had the software I wrote for allegedly commercial purposes - it would prove this alphametic within five runs!

Now where did I put my Superbrain or my Amstrad PC??????????????

Maybe not.

Regards,
ep

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### Re: Summer!

GoSeigen wrote:Hmm, I think it was more likely:

SPRING
RAINS
BRING
GREEN
---------
PLAINS Have only recently seen this - the thread looked to be finished after cinelli posted his solution and a week had gone by without further contributions, so I'd dropped it from my bookmarks, and only looked at it again while searching for something else... Anyway, I've had a go at solving it, and provided I haven't made any mistakes, it has a unique solution and so is a valid alphametic. Here's a spoiler:

`SPRING RAINS BRING GREEN------PLAINS`

There is a rather clear 'divide and conquer' point between the first two columns and the remaining four. Specifically, if I look at the latter, I get the addition:

` RING AINS RING REEN-----cAINS`

where 'c' is the carry out of those four columns into the first two. (I've made 'c' a lower-case letter as a visual reminder that it's not like the other letters - in particular, it can freely equal any of them. Note also that leading zeros are allowed in this addition.) Those remaining columns then produce the addition:

`SP R B G c--PL`

We can clearly subtract AINS from both top and bottom of the first addition, reducing it to:

` RING RING REEN-----c0000`

That addition is clearly less than 9999+9999+9999 = 29997 and greater than 0000+0000+0000 = 00000, so c must be 1 or 2. Furthermore, for any particular value of R, it must lie between R000+R000+R000 = 3000*R and R999+R999+R999 = 3000*R + 2997, from which one can deduce that if c=1, then R=3 and ING+ING+EEN = 1000, and if c=2, then R=6 and ING+ING+EEN = 2000. In each case, it happens that ING+ING+EEN = c000:

` ING ING EEN-----c000`

Now look at each possible value of G. For each of them, the rightmost column allows us to deduce the value of N, then the column to its left allows us to deduce the value of E (taking care not to ignore the carry between the two columns). That in turn allows us to deduce the possible values of I from the leftmost column (again taking care not to ignore the carry) and that allows us to deduce what c000 is, and hence the values of c and R. I'll use 'x's to indicate possibilities ruled out by duplicate digits, while '-'s indicate those ruled out because I+I is required to be odd:

`G  N  E  I         c000  R---------------------------0  0x1  8  3  -2  6  7  -3  4  1  4x or 9   2000  64  2  5  2x or 7   2000  65  0  9  0x or 5x6  8  2  3 or 8x   1000  3x7  6  6x8  4  0  -9  2  4  -`

So we only have two remaining possibilities for the RING+RING+REEN = c0000 addition:

` 6943    6724 6943    6724 6114    6552-----   -----20000   20000`

Now for the two original leftmost columns, i.e. the SP+R+B+G+c = PL addition. We know that R=6, c=2 and G=3 or 4 respectively for the two possibilities, so it reduces to SP+B+11 = PL or SP+B+12 = PL respectively. That makes PL between 11 and 20 greater than SP, so P must be either 1 or 2 greater than S. The digits we have used so far are {1,3,4,6,9} or {2,4,5,6,7} respectively and S and P are both nonzero leading digits, so they must both come from {2,5,7,8} or {1,3,8,9} respectively. The only (S,P) combinations that satisfy all those conditions are (5,7) and (7,8) for the first possibility, and (1,3) and (8,9) for the second. Looking at each of those:

(S,P) = (5,7) for the first possibility leads to 57+B+11 = 7L, or 68+B = 70+L, or B=L+2 with B and L in {0,2,8}, which works for B=2, L=0, and then A (the tenth letter in the original alphametic) must be 8, the one remaining unused digit.

(S,P) = (7,8) for the first possibility leads to 78+B+11 = 8L, or 89+B = 80+L, or B+9=L with B and L in {0,2,5}, which doesn't work.

(S,P) = (1,3) for the second possibility leads to 13+B+12 = 3L, or 25+B = 30+L, or B=L+5 with B and L in {0,8,9}, which doesn't work.

(S,P) = (8,9) for the second possibility leads to 89+B+12 = 9L, or 101+B = 90+L, or B+11=L with B and L in {0,1,3}, which doesn't work.

So the unique solution to this alphametic is:

`576943 68945 26943 36114------708945`

Gengulphus

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### Re: Summer!

Remembering back to the project many years ago, there are three others that come to mind althought the SPRING one was the crowning glory.

For a first attempt:-
TWO
TWO
--------
Is a good starter but it does have two answers.

I think the following two are correct - but memory might be failing!

CROSS
----------

and

LYNDON
B
-----------
JOHNSON which is a multiplication

.......... and I promise this is my last word on the subject!

Regards,
ep

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### Re: Summer!

eepee wrote:Remembering back to the project many years ago, there are three others that come to mind althought the SPRING one was the crowning glory.

For a first attempt:-
TWO
TWO
--------
Is a good starter but it does have two answers.

Seems to have more than 2 answers, I've found three fairly easily...

GS

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### Re: Summer!

I seem to remember that one has to specify a certain number.

Can't remember which.

Regards,
ep

GoSeigen
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### Re: Summer!

I seem to remember that one has to specify a certain number.

Can't remember which.

Regards,
ep

Fun though, daughter enjoyed solving that one... with a bit of prompting from dad...

GS