Problem of the Month (October 2017)

The pair of numbers 692 and 1118 have a very interesting property. Note that 6922 + 11182 = 1728788. And note that 6923 + 11183 = 1728788920 starts with exactly the same digits! What other pairs of positive integers x and y have the property that x3 + y3 starts with the digits of x2 + y2? More generally, what pairs of positive integers x and y have the property that xn + yn starts with the digits of xm + ym, where n>m>1? Even more generally, what collections of numbers {x, y, ... , z, m, n} have the property that xn + yn + . . . + zn starts with the digits of xm + ym + . . . + zm? What about other bases?

Bryce Herdt points out that for any m and x<y (as long as y is not a power of 10), there is always some n>m that works. This is because as n grows, xn << yn, so xn will eventually contribute no digits to the sum. By taking logarithms, it's easy to see that yn will eventually begin with any arbitrary digits, for example the digits of xm+ym.

George Sicherman sent several solutions for 10 ≤ n ≤20, and some solutions in base 8.

Bryce Herdt sent some small solutions for 20 ≤ n ≤30.

Joe DeVincentis shared his technique for finding large solutions. Consider the m=2 and n=3 case as an example. Assume x < y. We want x3 + y3 ≈ 10d (x3 + y2), or y2 (y–10d) ≈ x2 (10d–x), where d is the number of digits of agreement. Since the right hand side has a maximum, we can take its derivative and find the maximum occurs around x = (2/3)10d. By fixing d, looping y to start at 10d+1, and searching for x values slightly less than 10d or slightly more than 10d/2, he only needs to check about 10d cases for each d.

David Broadhurst also sent some huge solutions he found by efficiently solving the above cubic using cosines of a third of an angle.

There is a trivial infinite family for xn + yn starting with the digits of xm + ym: x=y=10k.

The other known cases with 1 < m < n ≤ 10 are shown below:

m=2 n=3
xm + ymxn + ynAuthor
6922 + 11182 = 17287886923 + 11183 = 1728788920
223772222 + 1036200452 = 11237853790239309223772223 + 1036200453 = 1123785379023930900736173JD
2237722202 + 10362004502 = 11237853790239309002237722203 + 10362004503 = 1123785379023930900736173000JD
2249002772 + 10364926272 = 11248971004200378582249002773 + 10364926273 = 1124897100420037858898449816JD
421681727112 + 1087027311462 = 13594438548384283562837 421681727113 + 1087027311463 = 1359443854838428356283751827361567DB
704875606952 + 1117432313772 = 17455045971305066199154 704875606953 + 1117432313773 = 1745504597130506619915466723682008DB
1182808641702 + 10120437212252 = 1038222856499747505289525 1182808641703 + 10120437212253 = 1038222856499747505289525051686978625DB
9207263483102 + 10598298250782 = 1970976266597131517562184 9207263483103 + 10598298250783 = 1970976266597131517562184310094565552DB

m=2 n=4
xm + ymxn + ynAuthor
72 + 112 = 17074 + 114 = 17042
52 + 322 = 104954 + 324 = 1049201
287357182 + 335798832 = 1953350031269213 287357184 + 335798834 = 1953350031269213637900200303297JD
236392852 + 3171003612 = 101111454741641546 236392854 + 3171003614 = 10111145474164154604331367054963666JD
Infinite Family: (9)7(9)2 + 1(0)1(9)2 = 1(9)68(0)2 (9)7(9)4 + 1(0)1(9)4 = 1(9)68(0)271(9)68(0)2JD

m=2 n=5
xm + ymxn + ynAuthor
42 + 52 = 4145 + 55 = 4149
14812 + 47812 = 2505132214815 + 47815 = 2505132230829059302JD
11872 + 464262 = 2156782445 11875 + 464265 = 215678244558228574108083JD
1397752 + 21574442 = 4674101663761 1397755 + 21574445 = 46741016637617978188741861977599JD
1811992 + 46439432 = 21599039664850 1811995 + 46439435 = 2159903966485079912370886208033942JD
33391552 + 103249812 = 117755188764386 33391555 + 103249815 = 117755188764386934627241251719620776JD
117147182 + 230057202 = 666497770537924 117147185 + 230057205 = 6664977705379247707157854914065625568JD

m=2 n=6
xm + ymxn + ynAuthor
42 + 182 = 34046 + 186 = 34016320
72 + 322 = 107376 + 326 = 1073859473
1662 + 3332 = 1384451666 + 3336 = 1384456392420985
182922 + 335322 = 1458992288 182926 + 335326 = 1458992288564093012918509568JD
3605522 + 10289032 = 1188639128113 3605526 + 10289036 = 1188639128113897585733510774092112593JD
284256502 + 1018798082 = 11187512856039364 284256506 + 1018798086 = 1118751285603936459848547744496942724819666580544JD

m=2 n=7
xm + ymxn + ynAuthor
42 + 42 = 3247 + 47 = 32768
772 + 4012 = 166730777 + 4017 = 1667303986766629654
20632 + 26762 = 1141694520637 + 26767 = 1141694525437148055180143
347732 + 661082 = 5579429193 347737 + 661087 = 5579429193390324289414066886893149JD
24405452 + 25699282 = 12560789822209 24405457 + 25699287 = 1256078982220917238676625775706391870219346337JD

m=2 n=8
xm + ymxn + ynAuthor
82 + 222 = 54888 + 228 = 54892650752
2632 + 4842 = 3034252638 + 4848 = 3034251506629606158017
1704640932 + 2284175602 = 81232588718666249 1704640938 + 2284175608 = 8123258871866624959113474177492797247323078342136557672056058634401JD

m=2 n=9
xm + ymxn + ynAuthor
12 + 22 = 519 + 29 = 513
72 + 1392 = 1937079 + 1399 = 19370159742464385266
1952 + 3852 = 1862501959 + 3859 = 186250825801426164062500
128532 + 378612 = 1598654930 128539 + 378619 = 159865493047585719627933107945922498339674JD
2229142 + 2829222 = 129735509480 2229149 + 2829229 = 12973550948006507932374877844699524995029023435776JD
4784152 + 37362642 = 14188549589921 4784159 + 37362649 = 141885495899215201431194446630473684470081169302554323726159JD

m=2 n=10
xm + ymxn + ynAuthor
9082 + 24112 = 663738590810 + 241110 = 6637385314425205826968902633702425GS

m=3 n=8
xm + ymxn + ynAuthor
13 + 43 = 6518 + 48 = 65537JD

Phil Mason looked for solutions for xm + ym + zm dividing xn + yn + zn with 1 ≤ x ≤ y ≤ z < 1000 and 1 < m < n < 10. Here are his solutions for {m,n} = {2,3}, {2,4}, {2,5}, {2,6}, {2,7}, {2,8}, {2,9}, and 3 solutions for m=3.

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 10/1/17.