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Summer!

cinelli
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Summer!

#159674

Postby cinelli » August 14th, 2018, 9:12 pm

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!

#159800

Postby Gengulphus » August 15th, 2018, 11:37 am

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 6
4 2 2 or 7
6 1 3 or 8
8 0 3 or 8
2 8 1 or 6
4 7 2 or 7
6 6 3 or 8
8 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 4,5,7
4 2 7 1 11 1
2 8 6 2 14 4 3,5,7
4 7 2 2 17 7
8 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!

#159891

Postby OLTB » August 15th, 2018, 4:53 pm

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!

#160969

Postby cinelli » August 21st, 2018, 10:39 am

Spoiler alert
'   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 V
2 0 4 1 2x
2 7 1 7x
3 0 6 1 1x
2 6x
4 0 8 1 0x
2 5 2 3 4x
5 1 0x
6 1 2 1 3 0 8 4 5
2 8 1 4 8x
7 1 4 1 2 1 1x
2 7x
8 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!

#163194

Postby eepee » August 30th, 2018, 4:01 pm

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

Obviously an addition.
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

GoSeigen
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Re: Summer!

#163230

Postby GoSeigen » August 30th, 2018, 8:05 pm

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!

#163352

Postby eepee » August 31st, 2018, 1:35 pm

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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