Fill in the digits instead of the Xs Clue1 There is no remainder Clue2 There is only one 7 in the position shown XXX)XXXXXX(7XX XXX)XXXX XXX)X_______ XXX)XXXXXX XXX)XXXXXX XXX)XX_________ XXX)XX0000
Can you explain the notation a little? At first I thought it was a long division problem with the first parenthesis separating the divisor and the dividend but the parenthesis just before the 7 and lack of a quotient breaks that pattern...
Your first instinct was right: it is a division problem, but the quotient is written to the right. I'm sure that studiot will confirm this.
studiot, Does that mean that there is no other seven anywhere a "X" is substituted, or only in the quotient? Ratch
Yes it is long division and I have worked it out for you, then replaced the digits with Xs If you prefer solve = 7XX none of the Xs are the digit 7
This is the trouble with trying to simplify to explain. I only posted that to confirm I was looking at a particular long division calculation. 330480/432 doesn't meet the original spec. 432)330480(76X XXX)3024 XXX)X_______ XXX)xx2808 XXX)xx2592 XXX)XX_________ XXX)XX0000 etc has the wrong number of digits in the wrong place for the second line of the calculation.
Well, I still don't understand the original notation, as I mentioned in my reply. Odd thing, most of the X's in the original post (#1) and in your follow up (#8) come across on my viewer (IE7) as ghosts, i.e., white on a light blue background. Initially, I dismissed it as an artifact. However, my e-mail shows: Can you state the problem in some wordprocessor or other program, then export as pdf or an image? John
studiot, There are 164 different solutions. The first and second in dividend-divisor order are 100452/132 = 761 and 100584/132 = 762. Davebee submitted the last or 164th solution in the series. I was going to wait a little longer to submit my entry. I have a confession to make. I used my programming skills to brutally extract the solutions with the help of my PC. I know that is not the right thing to do, but I just could not help myself. The PC sure figured out the solutions a lot faster than I could have done. If anyone wants some more solutions or all the solutions, let me know. Ratch
Ahh, I did the same thing, except maybe a bit sloppier, as I only ended up with about 150 solutions. I was planning to also show the 100504, 136, 739 result to show that my answer wasn't unique but I'm at work with too much to get done before I go home tonight.
Congratualtions davebee on being the first to publish. Thanks, too Ratch. Those who are interested in logic may realise that the fact that the first subtraction cannot be more than 999 (there are three xxxs in this position) means that the first digit on the left cannot be greater than 1 or less than 1 and therefore must be 1.
I had better make it clear that davebee's solution fails in the third line, i.e. the second subtraction. Oh, and Ratch, your two published solutions also fall at the same hurdle. Sorry. But perhaps I've now given too much away?
studiot, I don't understand what you are saying. Could you go through one of our faulty solutions point by point so we can understand? Ratch
OK we are dividing a six digit number ABCDEF by a three digit number, GHI to get a quotient JKL We are told J=7 Proceeding throught the long division: GHI into ABC won't go so bring down D and move one place right. GHI into ABCD goes 7 times. We are told the product 7 times GHI is only three digits ie not more than than 999. If G is 2 or greater then the product would have four digits. So G=1 This also tells us that A = 1 since the maximum product 1HI times 7HL is 199 times 799 = 159001 and we don't show leading zeros. So place 7 times 1HI under the 1BCDEF and do the subtraction. We are told this results in a two digit remainder so bring down E to make three digits. We are told this is not divisible by 1HI. (Davebee's solution would make it divisible 6 times.) So we place a zero in the quotient to note this fact This means K=0 Then we bring down the last digit, L to make a four digit number. We are told that 1HI divides this exactly with no remainder so this four digit number is equal to L times 1HI. This should tell us something about L????? You can't bring down two digits at a time, without placing the all important zero in the quotient, by the rules of long division. Does this help?
studiot, Let's take the last solution of the series 100584/132 = 762 . So then we have 1005 - 7*132 = 81 . bringing down the E = 8 gives 818, right? Wait a minute. Why or where does it say that particular partial division cannot have zero for a remainder? "6" appears a valid number as far as I can see. I observe no violation of the rules here. Anyway, I looked over my computer generated solutions and the only one with a zero for the second digit of the quotient is 100536/142 = 708 . It is the 124th solution of the series. Ratch
Because your calculation would look like my attachment. This is not the pattern in the original, which has four digits in the second subtraction and concludes there. And yes I've put the quotient on top rather than to the right.
Finally I get it - you need the zero in the quotient in order to allow pulling down both remaining digits in one operation... thanks for the puzzle!