# »  Solutions to puzzles in my National Review Online Diary

## December 2009

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In my December diary I posed the following brain-teaser from the 2009 China Girls Mathematical Olympiad:

Show that there are only finitely many triples (abc) of positive integers satisfying the equation  abc = 2009(a + b + c)

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Solution

Wolog (sorry: that's math-speak for "without loss of generality") we may assume  a ≥ b ≥ c.  So   b ⁄ a ≤ 1   and   c ⁄ a ≤ 1.

But the original equality can be rewritten as

bc = 2009 (1 + b ⁄ a + c ⁄ a)

We have just shown that the parenthesis on the right-hand side is  ≤ 3.  Therefore  bc ≤ 6027.

Since we are dealing with whole numbers here, it follows that only a finite number of (bc) is possible.

And since the original equality is linear in a, it can be rewritten to show a uniquely in terms of b and c.  It follows that only a finite number of a is possible.  Q.E.D.

[Actually a = 2009 (b + c) ⁄ (bc − 2009), but I wouldn't give extra points for writing this out.  For proof purposes, it's sufficient to note that there is some expression giving a single a for each (bc).  Note also that from this expression for a it follows that bc ≥ 2010, so we have both lower and upper bounds for bc:   2010 ≤ bc ≤ 6027.  The lower bound is attained several times, but the highest value I could find for bc was 4,920, corresponding to the first row of the table below.]

What is the actual number of possible (a,b,c)? A fast bit of VB coding came up with 258, listed below … but I don't guarantee this sort of thing. And if anyone can tell me how to get the number 258 (or whatever it is) with pure reasoning, I'd be most interested to hear.

I note before leaving this problem how mathematical intuition can sometimes lead one astray. Seeing that sum and that product, my mind immediately flew to the geometric and arithmetic means, the first of which, by a well-known theorem, cannot exceed the second. I then wasted several minutes — nothing to me, but perhaps fatal in an exam situation — fiddling with that theorem. All it gives you is lower bounds for  abc  and  a + b + c,  thus:

Since, by the aforementioned theorem, (abc) 1 ⁄ 3 ≤ (a + b + c) ⁄ 3, I can

•  substitute for a + b + c:

2009 × 3 ×(abc) 1 ⁄ 3 ≤ abc

which boils down to  abc ≥ 6027 3 ⁄ 2

•  substitute for  abc:

2009 (a + b + c) ≤ (a + b + c)3 ⁄ 27

which boils down to  a + b + c ≥ 542431 ⁄ 2

So  abc ≥ 467,899 and a + b + c ≥ 233.  Which is kind of interesting, but no help towards solving the problem.

a     b     c
98 82 60
119 112 41
123 86 49
123 105 42
131 82 49
139 98 41
147 94 41
154 91 41
164 71 49
175 84 41
196 79 41
196 123 29
196 164 24
246 59 49
259 70 41
259 164 21
273 82 35
287 56 49
287 105 28
287 154 21
294 67 41
301 287 14
328 77 35
328 266 14
343 64 41
350 287 13
364 63 41
364 246 14
410 51 49
451 50 49
451 217 14
490 59 41
504 123 21
511 205 14
539 58 41
574 56 42
574 77 31
574 86 28
574 119 21
574 126 20
574 196 14
574 273 11
574 371 9
581 574 7
588 164 16
630 533 7
637 123 20
656 49 47
656 84 28
656 294 10
656 343 9
679 56 41
693 287 10
770 82 28
779 49 46
779 112 21
779 196 13
784 55 41
812 574 6
861 64 35
861 84 27
861 175 14
861 434 7
931 54 41
973 410 7
1,036 287 9
1,066 168 14
1,148 112 20
1,148 166 14
1,148 385 7
1,148 896 4
1,211 861 4
1,246 164 14
1,271 49 44
1,271 833 4
1,316 369 7
1,351 574 5
1,421 779 4
1,435 77 28
1,435 104 21
1,435 161 14
1,435 266 9
1,435 560 5
1,519 52 41
1,640 98 22
1,764 82 26
1,764 205 11
1,886 49 43
1,886 686 4
2,009 68 31
2,009 135 16
2,009 202 11
2,009 336 7
2,009 403 6
2,009 671 4
2,009 1,006 3
2,011 2,009 2
2,065 287 8
2,156 656 4
2,254 51 41
2,255 490 5
2,255 1,813 2
2,296 154 14
2,296 329 7
2,296 644 4
2,345 328 7
2,499 246 9
2,499 1,681 2
2,583 1,645 2
2,829 196 11
2,891 164 13
3,024 861 3
3,157 151 14
3,353 1,435 2
3,430 123 17
3,444 99 21
3,675 369 6
3,724 451 5
3,731 49 42
3,731 238 9
3,731 581 4
4,018 71 29
4,018 86 24
4,018 212 10
4,018 447 5
4,018 1,340 2
4,019 4,018 1
4,046 574 4
4,305 308 7
4,354 3,731 1
4,459 50 41
4,756 1,274 2
4,756 3,479 1
4,823 3,444 1
4,879 98 21
4,879 3,416 1
5,166 148 14
5,243 82 25
5,292 3,239 1
5,488 1,230 2
5,740 189 11
5,740 1,218 2
6,027 3,014 1
6,314 301 7
6,314 546 4
6,314 2,947 1
6,601 147 14
6,699 2,870 1
6,888 231 9
6,888 427 5
7,421 539 4
7,503 2,744 1
8,036 536 4
8,036 2,679 1
8,050 1,148 2
8,575 2,624 1
8,771 533 4
8,897 58 35
9,044 2,583 1
9,184 146 14
9,471 70 29
9,758 1,120 2
10,250 2,499 1
10,535 205 10
10,619 2,478 1
11,858 2,419 1
12,054 46 44
12,054 1,096 2
12,054 2,411 1
12,341 294 7
14,063 521 4
14,063 2,344 1
14,309 343 6
14,350 293 7
14,760 1,078 2
15,211 145 14
15,744 2,303 1
16,079 2,296 1
16,646 518 4
17,150 697 3
17,444 1,066 2
18,424 2,255 1
18,491 2,254 1
19,516 119 17
20,335 410 5
20,377 203 10
21,238 2,219 1
21,707 2,214 1
22,099 101 20
22,099 2,210 1
25,830 48 42
26,117 184 11
26,404 56 36
27,265 63 32
27,265 1,043 2
28,249 686 3
28,413 290 7
28,700 72 28
29,479 2,156 1
29,561 511 4
30,996 84 24
32,144 65 31
32,144 2,143 1
33,579 96 21
34,153 1,035 2
34,839 2,132 1
37,310 112 18
38,171 509 4
39,319 406 5
40,467 2,114 1
40,754 126 16
42,271 1,029 2
42,476 289 7
43,214 2,107 1
45,346 144 14
50,225 106 19
50,323 1,025 2
51,254 2,091 1
51,660 168 12
56,252 155 13
59,696 2,079 1
62,279 1,021 2
62,279 2,076 1
66,871 224 9
68,306 506 4
74,620 252 8
84,419 2,058 1
84,665 288 7
97,293 1,015 2
98,154 336 6
98,154 2,051 1
100,499 2,050 1
117,383 2,044 1
136,612 403 5
136,612 2,039 1
145,796 504 4
184,828 1,010 2
193,725 672 3
194,299 2,030 1
194,873 67 30
271,215 2,024 1
289,870 1,008 2
290,444 2,023 1
299,341 134 15
339,521 503 4
405,818 2,019 1
423,899 201 10
578,879 2,016 1
675,024 1,006 2
675,024 2,015 1
685,069 335 6
809,627 2,014 1
817,663 402 5
1,348,039 2,012 1
1,352,057 670 3
2,021,054 2,011 1
2,023,063 1,005 2
4,040,099 2,010 1