(ABCD) * 9 = DCBA , what is C ?
iris
A, B, C and D are different integars ranging from 0-9, and
( ABCD ) * 9 = DCBA
what is C ?
Please give me the answer as well as the method.
Thanx,
Iris.
Allan Adler
> A, B, C and D are different integars ranging from 0-9, and
> ( ABCD ) * 9 = DCBA
> what is C ?
I just happen to know that 33^2=1089 and 99^2=9801. This gives a solution
since 33^2 * 9 = 33^2 * 3^2 = (33*3)^2 = 99^2. However, to prove that
this is the only solution, and just in case one doesn't just happen
to know the answer, let us proceed systematically to solve this
alphametic.
(1) A=1.
Proof: 9*A is less than 10 (otherwise (ABCD)*9 would have 5 digits),
so A is 0 or 1. If A is 0 then so is D, since the last digit
of D*9 is A. So A is 1.
(2) D=9.
Proof: D is at least 9*A=9, by looking at the leftmost digits, and can't
be bigger than 9, so D is 9.
(3) B=0.
Proof: Since D = 9*A = 9, there can be no carry to the leftmost column
when one multiplies B by 9, so B is 0 or 1. B can't be 1,
since A is 1. So B is 0.
(4) C=8.
Proof: By now, ABCD looks like 10C9 and DCBA looks like 9C01.
If we actually multiply 10C9 by 9, we see that the unit's
digit of 9*C + 8 is 0, so 9*C has to end in 2, so C is 8.
Now solve the same problem in the base N, with 9 replaced by N-1.
Allan Adler
a...@zurich.ai.mit.edu
****************************************************************************
* *
* Disclaimer: I am a guest and *not* a member of the MIT Artificial *
* Intelligence Lab. My actions and comments do not reflect *
* in any way on MIT. Morever, I am nowhere near the Boston *
* metropolitan area. *
* *
****************************************************************************
Dan Smith
C = 8. This comes from ABCD = 1089 and DCBA = 9801 = 1089 * 9.
My method is somewhat unorthodox, but it worked so here it is.
DCBA has to be divisible by 891. This has to be true because:
- DCBA is a multiple of 9 (it is some number times 9)
- ABCD is also a multiple of 9 (any number whose digits add to a multiple of 9 must be divisible by 9, so if DCBA is divisible by 9, then so is ABCD).
- DCBA is a multiple of 81, because it is 9 times a multiple of 9.
- Any four digit number whose odd digits add up to the same as the even digits is divisible by 11. (Example: We know that 3872 is divisible by 11, because 3 + 7 = 8 + 2.)
- Any number that's divisible by 81 and by 11 must be divisible by 891, because 81 and 11 have no common factors.
99 * 9 = 891
198 * 9 = 1782
297 * 9 = 2673
396 * 9 = 3564
495 * 9 = 4455
594 * 9 = 5346
693 * 9 = 6237
792 * 9 = 7128
891 * 9 = 8019
990 * 9 = 8910
* 1089 * 9 = 9801 *
Through brute force (looking at the 11 possibilities above), I found the solution.
Cheers,
--Dan
iris wrote:
Hey,
A, B, C and D are different integars ranging from 0-9, and
( ABCD ) * 9 = DCBA
what is C ?
Please give me the answer as well as the method.
Thanx,
Iris.
Dan Smith
You got to the right answer, but some of your reasoning below needs some
expansion:
--Dan
>
>
> (1) A=1.
> Proof: 9*A is less than 10 (otherwise (ABCD)*9 would have 5 digits),
> so A is 0 or 1. If A is 0 then so is D, since the last digit
> of D*9 is A. So A is 1.
So why couldn't A and D both be 0 then?
>
> (2) D=9.
> Proof: D is at least 9*A=9, by looking at the leftmost digits, and can't
> be bigger than 9, so D is 9.
But what about roundoff? But D isn't necessarily 9*A, more precisely your above
arguement shows that D=(9*A) mod 10 (the last digit of 9*A)
Dan Smith
are distinct.
Apologies,
--Dan
Kurt Foster
<a...@nestle.ai.mit.edu> said:
. Now solve the same problem in the base N, with 9 replaced by N-1.
OK. Since A, B, C and D are distinct digits, we have N >= 4. Since
DCBA = (N-1)*ABCD, we have the base-N addition
DCBA + ABCD = ABCD0 (base N).
Now A, being the carry into the N^4 place, must be 1. Then, the ones
column of the addition gives A+D = N (i.e. 10 base N); thus D = N-1.
Now, the addition in the N's column is 1+C+B = D or 1+C+B = 1D = N+D.
But the latter possibility cannot occur, because it would make the
addition in the N^2 column 1+C+B also, which would again give D in the N^2
digit of the sum, rather than C (the digits are distinct). Therefore, the
N's column of the addition is 1+C+B = D, and the N^2 column is C+B = C,
whence B = 0. Now 1+C = D = N-1 (as already determined), and C = N-2.
Note that 1*N^3 + 0*N^2 + (N-2)*N + (N-1)*1 = (N-1)*(N+1)^2, so the
number is a perfect square precisely when the base N is one more than a
perfect square.
Ed Turco
>Hey,
> A, B, C and D are different integars ranging from 0-9, and
> ( ABCD ) * 9 = DCBA
>what is C ?
>Please give me the answer as well as the method.
>Thanx,
>Iris.
Zero
Igor Karaszik
iris wrote:
>
> Hey,
>
> A, B, C and D are different integars ranging from 0-9, and
> ( ABCD ) * 9 = DCBA
>
> what is C ?
>
> Please give me the answer as well as the method.
>
> Thanx,
> Iris.
I know ABCD is my base number
and DCBA = ABCD x 9
and ABCD0 = ABCD x 10
A B C D 0
- D C B A
------------
0 A B C D
|
|
A must be different of 0, so you must have a carried 1 from the right and you
can only carry 1 so
A = 1;
0 - A = D
0 - 1 = 9 (and you carry 1)
D = 9;
Now you see: C - C = B or C - C - 1 = B (supossing a carry)
in the first case B = 0;
and in the second case B = 9 (and you carry 1), but if you use B = 9 ABCD x 9
> 10000 so
B = 0;
finally D - B - 1 = C (applying the carried 1 from the 0-A)
9 - 0 - 1 = C
C = 8
The whole number: 1089
Bill Dubuque
>
> DCBA = ABCD * 9, what is C? (assume decimal digits A,B,C,D all differ)
Easily A=1 so D=9. Now 1000 <= 1BC9 <= 1111, so B=0 (else B=1=A)
Hence finally C=8 since, casting 9's, 9 divides A+B+C+D = 10+C. QED
Alternatively, by casting 9's and 11's derive that 99 divides ABCD.
But between 1000 and 1111 the only multiple of 99 is 990+99 = 1089. QED
Recall casting 9's and 11's:
9|ABCD <=> 9|A+B+C+D where X|Y means X divides Y
11|ABCD <=> 11|A-B+C-D
Thus 9|DCBA => 9|ABCD
ABCD|DCBA => 11|ABCD
-Bill Dubuque
Rob Johnson
"iris" <mij...@netteens.net> wrote:
> A, B, C and D are different integars ranging from 0-9, and
> ( ABCD ) * 9 = DCBA
>
>what is C ?
Here is a fairly straightforward method:
9000A + 900B + 90C + 9D = 1000D + 100C + 10B + A [1]
Subtracting yields
8999A + 890B = 10C + 991D [2]
reducing mod 10 gives
A + D = 0 mod 10 [3]
If A = 0, then D = 0 and then 89B = C, thus C = 0 and B = 0.
Then the digits are all the same, and they were specified to be
different.
If A = 1, then D = 9 and then
80 + 890B = 10C
8 + 89B = C [4]
If B were anything but 0, C would have more than one digit.
Therefore, C = 8. Thus, ABCD = 1089 and
1089 * 9 = 9801
Rob Johnson
robj...@idt.net
Allan Adler
This discussion reminds me of something I discovered in high school
and which I subsequently used as a topic in a Mickey Mouse Math course
I taught at Brandeis. I'm going to typeset it with TeX and place it at
a web site in the next few days.
AilashNG
Answer is C=0
Bill
> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>> Hey,
>> A, B, C and D are different integars ranging from 0-9, and
>> ( ABCD ) * 9 = DCBA
>>
>> what is C ?
>>
>>
>> Thanx,
>> Iris.
Bill
Bill
ABCD *9 = ABCD (10 -1) = ABCD*10 - ABCD=ABCD0 - ABCD
Have fun!
Bill
>
> Bill
>
>
Jim Burns
> Multiplication (A*B*C*D)*9=D*C*B*A
> Answer is C=0
>
A or B or C or D = 0
But, for multiplication,
(1) order is irrelevant, so writing DCBA likely
likely wouldn't be done
(2) it turns an interesting problem into a trivial problem
Jim Burns
> avilas...@gmail.com wrote:
>> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>>> Hey,
>>>
>>> A, B, C and D are different integars ranging from 0-9, and
>>> ( ABCD ) * 9 = DCBA
>>>
>>> what is C ?
>>>
>>> Please give me the answer as well as the method.
>
... unless A = 9 = 0, but A and D have to be different.
Also:
9000*A + 900*B + 90*C + 9*D = 1000*D + 100*C + 10*B + A
8999*A + 890*B = 10*C + 991*D
8999*A + 890*B = 10*C + 991*(9 - A)
9990*A + 890*B = 10*C + 8919
But this looks like there is no solution.
Jim Burns
>> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>>> Hey,
>>>
>>> A, B, C and D are different integars ranging from 0-9, and
>>> ( ABCD ) * 9 = DCBA
>>>
>>> what is C ?
>>>
>>> Please give me the answer as well as the method.
>
***fixed typo***
Peter Percival
> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>> Hey,
>>
>> A, B, C and D are different integars ranging from 0-9, and
>> ( ABCD ) * 9 = DCBA
>>
>> what is C ?
>>
>> Please give me the answer as well as the method.
>>
>> Iris.
>
ABCD = 1089 (or 0000).
--
Made weak by time and fate, but strong in will
To strive, to seek, to find, and not to yield.
Ulysses, Alfred, Lord Tennyson
Peter Percival
> avilas...@gmail.com wrote:
>> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>>> Hey,
>>>
>>> A, B, C and D are different integars ranging from 0-9, and
>>> ( ABCD ) * 9 = DCBA
>>>
>>> what is C ?
>>>
>>> Please give me the answer as well as the method.
>>>
>>> Thanx,
>>> Iris.
>>
> ABCD = 1089 (or 0000
Peter Percival
> On 8/3/2016 10:38 AM, Bill wrote:
>> avilas...@gmail.com wrote:
>>> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>>>> Hey,
>>>>
>>>> A, B, C and D are different integars ranging from 0-9, and
>>>> ( ABCD ) * 9 = DCBA
>>>>
>>>> what is C ?
>>>>
>>>> Please give me the answer as well as the method.
>
>> Well, (D*9) Mod 10 =A. That should narrow it down some.
>>
>
> That helps. Then A + D = 9
> ... unless A = D = 0, but A and D have to be different.
>
> ***fixed typo***
>
> Also:
>
> 9000*A + 900*B + 90*C + 9*D = 1000*D + 100*C + 10*B + A
>
> 8999*A + 890*B = 10*C + 991*D
>
> 8999*A + 890*B = 10*C + 991*(9 - A)
>
> 9990*A + 890*B = 10*C + 8919
>
> But this looks like there is no solution.
>
>
Jim Burns
> Jim Burns wrote:
>> On 8/3/2016 10:38 AM, Bill wrote:
>>> avilas...@gmail.com wrote:
>>>> On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>>>>> Hey,
>>>>>
>>>>> A, B, C and D are different integars ranging from 0-9, and
>>>>> ( ABCD ) * 9 = DCBA
>>>>>
>>>>> what is C ?
>>>>>
>>>>> Please give me the answer as well as the method.
>>
>>> Well, (D*9) Mod 10 =A. That should narrow it down some.
>>>
>>
>> That helps. Then A + D = 9
>
> 10?
But, I would just like to point out that
if 9 = 10 my solution is correct.
Barry Schwarz
>On Thursday, January 13, 2000 at 4:00:00 PM UTC+8, iris wrote:
>> Hey,
>>
>> A, B, C and D are different integars ranging from 0-9, and
>> ( ABCD ) * 9 = DCBA
>>
>> what is C ?
>>
>> Please give me the answer as well as the method.
not four. So A must be 0 or 1. If A is 0, the only way the product
could have a units digit of 0 is if D is also 0. Since A must not
equal D, A must be 1.
Since the product must now have a units digit of 1, D must be 9.
Consider B. We know the product is at least 9000. If B is 2 or
greater, the carry would force the product to five digits. B must be
0 or 1. A is 1 so B must be 0.
Consider C. 10C9 * 9 = 9C01. 9*9 produces a carry of 8. Therefore,
9*C must produce a units digit of 2. C must be 8.
--
Remove del for email
Bill
instead of 4...
This is fun, but we need a tougher problem!
> Have fun!
> Bill
>
>>
>> Bill
>>
>>
>