integer cubic roots, ending on 5

# integer cubic roots, ending on 5

 The cubic Fives. Since the MCWC 2006 I organize at regular times a “Rekenwonder”-weekend. Gert Mttring is always present. In our room there is a huge flap-over on which we write less huge numbers, with which we do all kinds of calculations. There is a very intensive interchange of ideas. During the October weekend Gert said “It is typical that there are no questions in cubic roots with 5 numbers” “Yes, I understand, there are too many possibilities” “Indeed, but would it be too difficult to find something?”. Gert introduced to notions which I will use furthermore: The “question number” the number to find the root of, abbreviated as QN, and the “answer number”, by consequence AN . We agreed to think about and you’ll find the result of our thinking here under. As a pensioned man I have the luxury of much leasure time, so I had the time for thinking. So the first thing to do was making a table, which is to be found in the Excel file, the addition. There I found a typical “jump”, the difference between 2 succeeding numbers. This table is crucial. I marked the jumps in red. Ralf Laue was so kind to mail me 150 questions with “cubic fives”, so that I could test the working method thoroughly. After having made some examples Gert gave the necessary finishing touch and very important directions. He too filtered out that in some circumstances there is a difference of 50.000 in the answer found. This is due to the 10.000 in the AN. When we make a survey about thee table we can simplify the jump numbers by dividing by 125, Then we get resp. 460, 140, 300, 140 and 460 × 125. Important is that when calculating modulo 33 there is no common divisor with one of the jumps so that there can not be confusions. As general working method we advise in a cubic root to find the first digits and the last ones and finally the middle ones. The jumps of 37.500 mean that after 8 times you get the same result in the middle digits: 0 or 8 and 1 or 9. These differences can be filtered out either with estimation or more exactly be calculated with the 3a^2 method. As the proof of the pudding is in the eating some “Laue numbers”. These numbers are generated by Ralf Laue, for testing my method of working. Q(uestion) N(umber): cubic root of 22260|3687|6344|7625. A(nswer) N(umber): first digits 28. See table: last digits evidently 05 and an odd number of 100, see table. Middle digits AN: 22260-21952=308/28^2×3= 308/2352=± 0,13. With modulo 33. So first make for yourself a table modulo 33 exponent 3. So 2^3=8(33), 13^3= 19(33) etc. For me this was very simple as I work since my 10th year with the 11 test. 33 is even better, it has no doubles and 99 has many doubles. QN: from right to left the groups of 4 digits are respectively 2, 8, 24 and 18, overall 19(33). So the AN is necessarily 13(33). As AN 28 + 05 = 0(33) so the hundreds have to be 13(33) and be odd. Choices: 13, 46 and 79. As the even numbers drop out, the middle digits are either 13 or 79. As 22260 is close to 21952 and see the check of 308/2352 the AN will be 28.13.05. QN cubic root of 39|1315|6471|5773|7875. AN: first digits 73. Last digits 35, even hundreds necessarily 2(4), see table. Hundreds: 391315-389017=2298/73^2×3, rounded 16000 = 14, AN 73 14 35 Modulo 33: 21+31+3+28+6= 23(33). AN = 23(33). 73+35= 9. Middle digits have to be 23-9=14(33). So AN = 73 14 35. Finally: QN cubic root of 74|3738|9803|9876|5625. AN : first digits 90, last digits 25. Hundreds estimated 743738-729000=14738/3×90^2= ± 0,605, 0(4) and even. (33) = 15+9+2+9+8=10. So AN = 10(33). 90+25=16(33), so answer has to be 16 + X = 10(33), so X=27. The possibilities are by consequence: 27, 60 and 93. Middle digits have to be 60, being 27(33) and 0(4). So the final answer is 90 60 25. Remarks, additions or questions are welcome. Success!! Gert Mittring and Willem Bouman.