On 2020-02-25 10:28, Tom Duff wrote:
I should say, I excluded 0 as an addend, because then you get a=a+0+0 for every a with a zero in it, which strikes me as not interesting.
Adding 0's to a previous solution is also uninteresting. I assume you meant to cut those out, too, but you (mistakenly?) included all forms of 150* = 50* + 50* + 50* (but almost no other examples [e.g. 182, 1820, 18200, ...], although you do include 195000=5000+95000+95000 and 19500, and 195 = 5 + 95 + 95) To answer Eric's question, it is impossible to have 3 distinct values for A, B, C. Also, the leading digit of D must be 1, and the 2nd digit must be either 8 or 9 or (in special cases when D=A+A+A), 5. (Boring, grinding argument follows:) In this case, for D to be an n digit number, then it has to be the sum of at 2 n-1 digit numbers, or at least 3 n-1 digit numbers to have a leading digit higher than 1. There are only 2 distinct n-1 digit numbers possible (the first n-1 and the last n-1 digits of D) so let's restrict ourselves to D's with a leading 1 [actually, D *always* must have a leading 1]. Let's focus on, x, the 2nd digit of D. We'd need 1x... + x... + the n-2 digit number, which can start with at most 9. The 2nd digit, x, of D must be x = (x + 1 + carry from the lower digits) mod 10, but 1 + <carry> cannot = 10 (<carry> can't be > 2). So no such 2nd digit x exists. So no 3 distinct A, B, C exist. For the non-distinct case, either D = A + A + A, or D = 2*A + B In the latter case, A must be the last n-1 digits of D. (If the leading digit is either 1 or 2, then A can't be the first n-1 digits -- the 2nd digit x of D would have to be x = 4 + x + <carry> or x = 2 + x + <carry>. <carry> + either 2 or 4 can never = 10, so A must be the last n-1 digits of D, not the first). By the same sort of (by now boring) reasoning, it is easy to see that the leading digit of D must always be 1. (For leading digit 2, we'd need both x = 2*x + 2 + <carry> mod 10, and 2*x + 2 + <carry> >= 20, so 2*x + 2 + <carry> = 20 + x, or x + 2 + <carry> = 20, which is impossible (x is a single digit, and <carry> <= 2)) So leading digit of D is 1, and all solutions are of the form D = 2*A + B, where x is the leading digit of A and the 2nd digit of D. 2*x + 1 + <carry> = 10 + x, or 2*x + <carry> = 10 + x. (The former is if B is the n-1 leading digits of D, the latter is if B is any shorter set of digits of D). So x must be either 8 or 9 when D = 2*A + B The only time A = B = C is if the final digit of D is 0 or 5 (A cannot be the first n-1 digits of D, else 3 * the leading digit would have to equal itself). If any example exists with A and D ending in 0, D = A + A + A, then D/10 A/10 must also be an example. So all such examples have to have a single digit solution. 5 is the only such example. No 2 digit value for A ending in 5 is possible, because the 2nd digit x, must be = to the last digit of 3*x + 1. So 1500... = 500.. + 500.. + 500.. are the only such examples of A + A + A
On Tue, Feb 25, 2020 at 10:26 AM Tom Duff <td@pixar.com> wrote:
I ran a search out to a billion & found 109 examples. There were no cases of A, B and C distinct, and they pretty much all fell into a few simple patterns. Here's the list: 15=5+5+5 19=1+9+9 150=50+50+50 182=18+82+82 191=9+91+91 195=5+95+95 199=1+99+99 1500=500+500+500 1819=181+819+819 1950=50+950+950 1981=19+981+981 1991=9+991+991 1995=5+995+995 1999=1+999+999 15000=5000+5000+5000 18182=1818+8182+8182 19091=909+9091+9091 19500=500+9500+9500 19802=198+9802+9802 19901=99+9901+9901 19950=50+9950+9950 19981=19+9981+9981 19991=9+9991+9991 19995=5+9995+9995 19999=1+9999+9999 150000=50000+50000+50000 181819=18181+81819+81819 195000=5000+95000+95000 198020=1980+98020+98020 199010=990+99010+99010 199091=909+99091+99091 199500=500+99500+99500 199801=199+99801+99801 199901=99+99901+99901 199950=50+99950+99950 199981=19+99981+99981 199991=9+99991+99991 199995=5+99995+99995 199999=1+99999+99999 1500000=500000+500000+500000 1818182=181818+818182+818182 1909091=90909+909091+909091 1950000=50000+950000+950000 1980199=19801+980199+980199 1995000=5000+995000+995000 1998002=1998+998002+998002 1999001=999+999001+999001 1999010=990+999010+999010 1999091=909+999091+999091 1999500=500+999500+999500 1999801=199+999801+999801 1999901=99+999901+999901 1999950=50+999950+999950 1999981=19+999981+999981 1999991=9+999991+999991 1999995=5+999995+999995 1999999=1+999999+999999 15000000=5000000+5000000+5000000 18181819=1818181+8181819+8181819 19500000=500000+9500000+9500000 19801981=198019+9801981+9801981 19900991=99009+9900991+9900991 19909091=90909+9909091+9909091 19950000=50000+9950000+9950000 19980020=19980+9980020+9980020 19990010=9990+9990010+9990010 19995000=5000+9995000+9995000 19998001=1999+9998001+9998001 19999001=999+9999001+9999001 19999010=990+9999010+9999010 19999091=909+9999091+9999091 19999500=500+9999500+9999500 19999801=199+9999801+9999801 19999901=99+9999901+9999901 19999950=50+9999950+9999950 19999981=19+9999981+9999981 19999991=9+9999991+9999991 19999995=5+9999995+9999995 19999999=1+9999999+9999999 150000000=50000000+50000000+50000000 181818182=18181818+81818182+81818182 190909091=9090909+90909091+90909091 195000000=5000000+95000000+95000000 198019802=1980198+98019802+98019802 199009901=990099+99009901+99009901 199500000=500000+99500000+99500000 199800200=199800+99800200+99800200 199900100=99900+99900100+99900100 199900991=99009+99900991+99900991 199909091=90909+99909091+99909091 199950000=50000+99950000+99950000 199980002=19998+99980002+99980002 199990001=9999+99990001+99990001 199990010=9990+99990010+99990010 199995000=5000+99995000+99995000 199998001=1999+99998001+99998001 199999001=999+99999001+99999001 199999010=990+99999010+99999010 199999091=909+99999091+99999091 199999500=500+99999500+99999500 199999801=199+99999801+99999801 199999901=99+99999901+99999901 199999950=50+99999950+99999950 199999981=19+99999981+99999981 199999991=9+99999991+99999991 199999995=5+99999995+99999995 199999999=1+99999999+99999999
On Mon, Feb 24, 2020 at 3:58 AM Éric Angelini <bk263401@skynet.be> wrote:
Hello Math-Fun, do you know any other solutions? (especially where A, B and C are distinct)
1818 = A 8182 = B + 8182 = C ------ = 18182 = D
Best, É.
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