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Wise

cinelli
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Wise

#368024

Postby cinelli » December 20th, 2020, 2:00 pm

We haven't had an alphametic for a while.

.    AS
-----
AS)WISE
J*
---
O*
*E
--

The asterisks may stand for any digit but each distinct letter stands for a particular but different digit.

Cinelli

Gengulphus
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Re: Wise

#368561

Postby Gengulphus » December 21st, 2020, 11:13 pm

cinelli wrote:We haven't had an alphametic for a while.

.    AS
-----
AS)WISE
J*
---
O*
*E
--

The asterisks may stand for any digit but each distinct letter stands for a particular but different digit.

Spoiler...

Subtracting the 2-digit number J* from the 3-digit number WIS produces the 1-digit number O, so 100 <= WIS = J* + O <= 99+9 = 108. The fact that the division comes out exactly means that WISE is the square of AS, so it's a square in the range 1000-1089. Looking for squares in that range, 31^2 = 961 is too small, 32^2 = 1024 is in range, 33^2 = 1089 is (just) in range, 34^2 = 1156 is too big. So either WISE = 1024 and AS = 32 or WISE = 1089 and AS = 33 - but the latter is ruled out because it would make A and S be the same digit. So the solution has to be:

.    32
-----
32)1024
96
---
64
64
--

if it's anything - and a check shows that satisfies all the requirements about digits being the same or different, so it is indeed the solution.

Gengulphus

AsleepInYorkshire
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Re: Wise

#368570

Postby AsleepInYorkshire » December 21st, 2020, 11:41 pm

Gengulphus wrote:
cinelli wrote:We haven't had an alphametic for a while.

.    AS
-----
AS)WISE
J*
---
O*
*E
--

The asterisks may stand for any digit but each distinct letter stands for a particular but different digit.

Spoiler...

Subtracting the 2-digit number J* from the 3-digit number WIS produces the 1-digit number O, so 100 <= WIS = J* + O <= 99+9 = 108. The fact that the division comes out exactly means that WISE is the square of AS, so it's a square in the range 1000-1089. Looking for squares in that range, 31^2 = 961 is too small, 32^2 = 1024 is in range, 33^2 = 1089 is (just) in range, 34^2 = 1156 is too big. So either WISE = 1024 and AS = 32 or WISE = 1089 and AS = 33 - but the latter is ruled out because it would make A and S be the same digit. So the solution has to be:

.    32
-----
32)1024
96
---
64
64
--

if it's anything - and a check shows that satisfies all the requirements about digits being the same or different, so it is indeed the solution.

Gengulphus

Geng,

I love you to bits. But ... before I shoot myself (and probably miss) can you please put that in language a really simple sheep herding Yorkshireman can understand please :roll:

You lost me at "subtracting" :oops:

Merry Xmas Gengulphus

AiY

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Re: Wise

#368797

Postby Gengulphus » December 22nd, 2020, 2:44 pm

AsleepInYorkshire wrote:I love you to bits. But ... before I shoot myself (and probably miss) can you please put that in language a really simple sheep herding Yorkshireman can understand please :roll:

You lost me at "subtracting" :oops:

Think about how a long division is done - for example, if I divide 123456 by 789, I start with:

.
-------
789)123456

I get started by finding the leftmost place where I can subtract a nonzero multiple of the divisor 789 from an initial section of 123456. I can't do so from the initial section 1, nor from the initial section 12, nor from the initial section 123, because they're all smaller than 789, but I can from the initial section 1234. I then subtract the largest multiple of 789 that I can from that initial section, which in this case is 1*789 = 789, 2*789 = 1578 being larger than 1234, and I record the fact that it was 1*789 by putting 1 into the quotient:

.      1
-------
789)123456
789
---
445

So the long division so far basically contains the subtraction 1234-789 = 445 and the multiplication 1*789 = 789.

I then extend the remainder 445 by the next digit 5 of 123456, to produce:

.      1
-------
789)123456
789
---
4455

and repeat the process of subtracting the largest multiple of 789 that I can from the extended remainder 4455 and recording which multiple it is in the quotient - in this case, 5*789 = 3945 can be subtracted, but 6*789 = 4734 is too big:

.      15
-------
789)123456
789
---
4455
3945
----
510

and now the long division basically contains the subtractions 1234-789 = 445 and 4455-3945 = 510, and the multiplications 1*789 = 789 and 5*789 = 3945.

Finally I extend the remainder 510 with the last digit 6 of 123456, getting 5106 and again subtract the largest multiple of 789 I can, which is 6*789 = 4734, 7*789 = 5523 being too big:

.      156
-------
789)123456
789
---
4455
3945
----
5106
4734
----
372

and now the long division basically contains the subtractions 1234-789 = 445, 4455-3945 = 510 and 5106-4734 = 372, and the multiplications 1*789 = 789, 5*789 = 3945 and 6*789 = 4734.

The way to solve a long division alphametic is basically to read off the subtractions and multiplications it contains, plus some other information such as:

A) Each remainder after doing a subtraction but before extending it with another digit from the number we're dividing into must be less than the divisor, since otherwise we would have been able to subtract a bigger multiple of the divisor (in the above example of a long division, the three multiples of the divisor are 445, 510 and 372 and are all less that the divisor 789).

B) If the largest multiple of the divisor that can be subtracted is zero times the divisor, which is 0, we don't bother subtracting zero - we just record a quotient digit of 0 and extend the remainder with another digit of the number we're dividing into, or stop if there are no more such digits. This allows quotient digits of 0 to be spotted from the pattern of subtractions (but doesn't happen in the above example of a long division).

C) The lengths of the numbers tell us what range they're in - a number with one digit must be in the range 0-9 (and usually 1-9 because numbers that are 0 are generally left out of long divisions), a number with two digits must be in the range 10-99, a number with three digits must be in the range 100-999, etc. This can give useful information - e.g. if the divisor has three digits and one of the subtractions subtracts a 4-digit number, we must be subtracting at least twice the divisor, and the corresponding quotient digit must be at least 2. If we know that the number's digits are different because different letters are used for them, we can narrow those ranges a bit - a 3-digit number "***" can be anywhere in the range 100-999, but a 3-digit number "CAT" has to be in the range 102-987. If we actually get to know what one of the digits is, especially the first digit, we can narrow the range more - e.g. if we know that C represents the digit 1, a 3-digit number "CAT" has to be in the range 102-198 (and a 4-digit multiple of it has to be at least 6 times it).

D) The quotient times the divisor, plus the final remainder, must equal the number we're dividing into - in the above example of a long division, 156*789 + 372 = 123456.

So cinelli's alphametic:

.    AS
-----
AS)WISE
J*
---
O*
*E
--

gives us various clues:

Subtractions: WIS - J* = O, O* - *E = 0 (number zero, not letter O)

Multiplications: A x AS = J*, S x AS = *E

Remainders less than divisor: O < AS, 0 < AS

Zero digits in the quotient: neither A nor S is 0

Ranges: AS is in the range 10-98; WISE is in the range 1023-9876 (and so WIS is in the range 102-987); J*, O* and *E are in the range 10-99 (and so J, O and the * at the start of *E are in the range 1-9)

Overall: WISE (number being divided into) = AS (divisor) * AS (quotient) + 0 (final remainder - number zero, not letter O)

It's then a matter of trying to find a way to deduce what each digit is by deductions from those clues, plus the standard ones that different letters represent different digits, preferably in as simple, straightforward and direct a way as possible. My spoiler says how I did it - and it turned out to be much less complex, indirect and contorted than most alphametics!

Gengulphus


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