Reconstructed Division
Replace the "x"s with the correct numbers:
posted by Zaux at
5:12 AM
Thursday, February 11, 2010
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20 Comments:
Hi Zaux. Should there be a 0 at the very end? i.e. is the division exact?
Chris .... no remainder
Struggling to find a working answer...
Is the divisor one of:
19,33,37 or 39 ??
Cam
Something seems wrong with the picture given i.e.
xxx
-xx1
given that the 0 should be carried down it should be
xx0
-xx1
----
x
which must be 9
and
9x
-3x
----
0
can't work.
Maybe intentionally part of the trick or maybe not?
Cam
Am I on the right track?
9XXX
------
11|10XX0X
99
--
XX4
XX1
---
3X
3X
--
The alignment is not pretty, but line for line can compare to the original image.
I'm going off an assumption the first digit of the dividend is not 0. and as Chris stated above, no remainder.
Cam ...
the divisor is one of the ones you mentioned
I re-checked ... the problem is illustrated correctly ... and the math works ... no remainder
forget it.. that didn't work out.
If there's no remainer, then the X above the 3 must also be a 3 and the X above the 1 must be a 4. But that X must also be a 0, beacuse it's just dropped down from above..
Damn.. Where's my brother when I need him!
It's not in Base5, is it?
Euclid,
I discarded 11 as a divisor since
11*X (where X is a single digit number) can not produce a 3 digit number ending in one.
19,33,37 and 39 are the only two digiti numbers that have the following properties:
-they can produce a 3 digit number ending in 1 if multiplied by a single digit number.
-they can produce a two digit number beginning with 3 if multiplied by a single digit number
Again to get XX-3x=0
I agree the above should be
XX4
XX1
but it should be a 0 rather than a 4 which gets carried.
maybe this question is done using base 4 arithmetic??
Cam
Or Base4, rather. If the 0 drops down and you borrow one, then subtrance one, you get 3. So the remainder could then work out.
standard base 10
jeez! ...this may help ... as I stated, my illustration of the problem is identical yo the published source ... and I did the math to make sure it works ... but I just assumed the illustration was correct ... there's is an error in the published problem ... I will correct it ... and I am so sorry FOR THEIR ERROR. i apologize for making everyone work on an impossiblity :-)
so you can work it out, the "1" should ber a "7"
looks like we cross-posted Euclid
Cam
Now that it's possible...
Answer:
102003/ 33=3091
Cam
exactly Cam ... and man I really do apologize ...I just assumed too much .. that the illustratioon was correct
If anyone wants to check out the published error ... it's in a book called Mathematical Quickies by Charles Trigg ... puzzle no. 173, page 48.
Method
13,31,33,39 are the only numbers that have the following properties:
-they can produce a 3 digit number ending in 7 if multiplied by a single digit number.
-they can produce a two digit number beginning with 3 if multiplied by a single digit number
Analyse for 33:
last digit must be 1
second last digit must be 9
third last digit is 0 as it skip a subtraction
only way to have a 3 digit number reduce to a 1 digit number after beind subtacted by x*33 is x=3 and the 3 digit number is witin 9 of 3*33=99 i.e. max 108
33*3091=102003
Cam
Zaux,
No need for apologies, it's not your fault.
We appreciate the amount of puzzles you post. If one or two of those puzzles should have an imperfect source, it's no big deal. The folks on here have been pretty savvy in identifying when a solution is no good or the problem is unsolvable as stated. We work through these things.
It just goes to show, that even the original sources, despite having fact checkers and editors (who get paid to make sure the puzzles and answers are correct), are still written by people, who can't be expected to be perfect.
Cam
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