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Old 2007-07-05, 16:27   #1
mfgoode
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Question Missing Digits!



Heres one straight from the book so no Oohs! and Aahs.!

But dont use the computers! Just base your answer on strict math principles!
and I dont just want the answer only but also the method to eliminate guess work.

Well here it is:

Two digits of the 8 digit number 273*49*5 were erased.
The number is divisible by 9 and 11. Find the missing digits.
* stands for missing digit.

Try your hand on it!

Mally
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Old 2007-07-05, 16:44   #2
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Working mod 9:

2+7+3+a+4+9+b+5 = 0 => a+b = -30 = 6 (mod 9)

Working mod 11:
-2+7-3+a-4+9-b+5 = 0 => a-b = -12 = -1 (mod 11)

So:
a = b-1
a+a+1=6 (mod 9) -> a=7

The number must be: 27374985
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Old 2007-07-05, 16:52   #3
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let the first unknown diigt be a and the second b.
divisibility by 9 gives us : 3+a+b=0 mod 9
divisibility by 11 gives us : 5+9+a+7=b+4+3+2 mod 11 => a+21 = b+9 mod 11 => a+1=b mod 11 => Since a and b are not negative and less than 10 => a+1=b
substituting in the first equation 3+a+a+1=0 mod 9 => 2*a+4=18 (since 0 would give a negative a and 9 a fractional a) and thus a=7, b=8
the final number is 27374985

But I was to long in editing.

Last fiddled with by S485122 on 2007-07-05 at 16:53 Reason: to slow...
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Old 2007-07-05, 17:08   #4
mfgoode
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Thumbs up Quick on the draw!

:surprised

You are quick on the draw Alpertron and S485122 Hats off to you!
The next time I will give a harder one with you in mind

Mally
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Old 2007-07-05, 17:18   #5
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Quote:
Originally Posted by mfgoode View Post
:surprised

You are quick on the draw Alpertron and S485122 Hats off to you!
The next time I will give a harder one with you in mind

Mally
That one went over my head Mally.
Sounds like it was solved?

D.
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Old 2007-07-05, 17:41   #6
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This puzzle must be resolved by Mally. Please do not give any hints.

The number 823***546 is multiple of the numbers 7, 11 and 13. You will need to write the missing three digits without using the computer.
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Old 2007-07-05, 22:49   #7
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Quote:
Originally Posted by alpertron View Post
The number 823***546 is multiple of the numbers 7, 11 and 13. You will need to write the missing three digits without using the computer.
Let x be the unknown 3-digit number. We have:
823000546 + x*1000 = 0 (mod 1001)
implying that
x = 823000546 = 368 (mod 1001).
Therefore, x=368.
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Old 2007-07-05, 23:08   #8
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The puzzle is trivial, but I wanted Mally to solve it.
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Old 2007-07-06, 16:36   #9
mfgoode
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Quote:
Originally Posted by maxal View Post
Let x be the unknown 3-digit number. We have:
823000546 + x*1000 = 0 (mod 1001)
implying that
x = 823000546 = 368 (mod 1001).
Therefore, x=368.
:surprised

Thank you maxal for coming to my rescue in the face of an open challenge by Alpertron directed specifically to me and no other!

Well history is replete with such rivalry so its no surprise.

Well he has meant well and Im grateful to him for posing this similar problem to mine with a twist.

To be honest, I had no clue as to how to tackle it until I saw your concise and elegant solution.

Still there is a wide gap between your two lines. I understand the first but cannot arrive at the logic how you imply 368 from the last line.

I will be grateful if you give the full working as it is one for my book. Thanks!

Regards,

Mally
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Old 2007-07-06, 16:50   #10
mfgoode
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Post A Special!

Quote:
Originally Posted by alpertron View Post
This puzzle must be resolved by Mally. Please do not give any hints.
.
Thank you Alpertron for posing this problem to me.

Whereas I welcome all types of problems the ones I give are with a motive of enlightening, instructing, and for discussing and NOT for ridiculing or embarrassing anyone.

Let us therefore exchange views in the Spirit of Truth which is like the seven seas of old lying openly before us.

I know my problems have been controversial in the past but let the dead past bury the dead and move to a new order.

As Roger Penrose says "Controversy is an important part in the development of Science" [Road to Reality]

You your self put an end to the last one for which I am very grateful.

Thank you,

Mally
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Old 2007-07-06, 16:55   #11
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I wanted you to solve it because it is very similar to the problem you presented.

If a number is multiple of 7, 11 and 13, since all these numbers are relatively primes, you get that the original number must be multiple of 7*11*13 = 1001.

Now, given that 1000 = -1 (mod 1001) we get:

n = 823***546 = 823 * (1000^2) + x * 1000 + 546 = 823 - x + 546 = 0 (mod 1001)

This means that x = 823 + 546 = 1369 = 368 (mod 1001)

So the number n must be 823368546.
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