Find all values of a,b,c such that a-b = c, a,b,c are three digit numbers and the digits are distinct.

Question

Hello everyone this is my first question on here.

What is the relation between a and b when a-b=c | a,b,c ∈ Z | 123 ≤ a,b,c ≤ 987 and a,b,c have distinct digits 1-9.

For example one possible value of a and b are: a = 459, b = 173, c = a-b = 286. This works as no digit repeats, it is a three digit number, 0 is not used and a is greater than b.

I've gone through every possibility with a computer which gives 336 possible combinations of a and b. Obviously though a-b=c and a-c=b are both true so every combination has a "duplicate".

I have not yet found a way of deriving these numbers with a mathematical rule, as it doesn't fit with any standard formulas. If anyone could find one i'd be very interested.

Here's the list of possible values of a (left-column) and b if it can be of aid:

• 459 : 173, 176, 183, 186, 273, 276, 283, 286
• 468 : 173, 175, 193, 195, 273, 275, 293, 295
• 486 : 127, 129, 157, 159, 327, 329, 357, 359
• 495 : 127, 128, 167, 168, 327, 328, 367, 368
• 549 : 162, 167, 182, 187, 362, 367, 382, 387
• 567 : 128, 129, 138, 139, 218, 219, 248, 249, 318, 319, 348, 349, 428, 429, 438, 439
• 576 : 182, 184, 192, 194, 382, 384, 392, 394
• 594 : 216, 218, 276, 278, 316, 318, 376, 378
• 639 : 152, 157, 182, 187, 452, 457, 482, 487
• 648 : 251, 257, 291, 297, 351, 357, 391, 397
• 657 : 218, 219, 238, 239, 418, 419, 438, 439
• 675 : 182, 183, 192, 193, 281, 284, 291, 294, 381, 384, 391, 394, 482, 483, 492, 493
• 693 : 215, 218, 275, 278, 415, 418, 475, 478
• 729 : 143, 146, 183, 186, 543, 546, 583, 586
• 738 : 142, 146, 192, 196, 542, 546, 592, 596
• 783 : 124, 129, 154, 159, 214, 219, 264, 269, 514, 519, 564, 569, 624, 629, 654, 659
• 792 : 134, 138, 154, 158, 634, 638, 654, 658
• 819 : 243, 246, 273, 276, 352, 357, 362, 367, 452, 457, 462, 467, 543, 546, 573, 576
• 837 : 142, 145, 192, 195, 241, 246, 291, 296, 541, 546, 591, 596, 642, 645, 692, 695
• 846 : 317, 319, 327, 329, 517, 519, 527, 529
• 864 : 125, 129, 135, 139, 271, 273, 291, 293, 571, 573, 591, 593, 725, 729, 735, 739
• 873 : 214, 219, 254, 259, 614, 619, 654, 659
• 891 : 234, 237, 254, 257, 324, 327, 364, 367, 524, 527, 564, 567, 634, 637, 654, 657
• 918 : 243, 245, 273, 275, 342, 346, 372, 376, 542, 546, 572, 576, 643, 645, 673, 675
• 927 : 341, 346, 381, 386, 541, 546, 581, 586
• 936 : 152, 154, 182, 184, 752, 754, 782, 784
• 945 : 162, 163, 182, 183, 317, 318, 327, 328, 617, 618, 627, 628, 762, 763, 782, 783
• 954 : 216, 218, 236, 238, 271, 273, 281, 283, 671, 673, 681, 683, 716, 718, 736, 738
• 963 : 215, 218, 245, 248, 715, 718, 745, 748
• 972 : 314, 318, 354, 358, 614, 618, 654, 658
• 981 : 235, 236, 245, 246, 324, 327, 354, 357, 624, 627, 654, 657, 735, 736, 745, 746

Also, I've noticed the sum of every "a" value is 18, this may be significant.

A sub question that I have aswell is why the number of possibilities is 336. I noticed it is equal to $8*7*6$ but why I don’t know.

Thankyou for taking the time to read this question.

Answer 1

Casting out nines is an old method for checking arithmetic. It's based on arithmetic modulo $9$. What makes the method tick is that if we add up the digits of a positive integer $n$, the sum is congruent to $n$ modulo $9$. For example the sum of the digits of $782$ is $17$ and one can check that $$782\equiv17\pmod{9}$$ Indeed $$782=9\cdot85+17$$ Of course, we can add up the digits of $17$ and see that to see that $$782\equiv17\equiv8\pmod{9}$$ and that's how casting out nines is done. If the sum of the digits is not a one-digit number, we add up the digits of the sum, repeating the process until we do arrive at a one-digit number. Leet's call this the "digital root" of the number we started with.

Now, if we want to check the addition of a column of figures, we add up the digital roots of all the summands, take the digital root of that sum, and it should equal the digital root of the total. Of course, we if we made a mistake, and the difference between the correct answer and the answer we erroneously got is divisible by $9,$ casting out nines won't detect it.

Before explaining how this applies to your question, let me take a moment to explain about the digital roots. Suppose $n=3472,$ for example. Then, $$ \begin{align} n&=3\cdot1000+4\cdot100+7\cdot10+2\\ &=3(999+1)+4(99+1)+7(9+1)+2\\ &\equiv3+4+7+2\pmod{9} \end{align}$$

Now, back to the question. Rewrite it as $a=b+c$ Each of the digits from $1$ to $9$ occurs once in the sum $a+b+c$ and the sum of the digits from $1$ to $9$ is $45$ so by looking at the digital roots, we have $$9|(a+b+c)=2a\iff 9|a$$ since $2$ and $9$ are relatively prime. Since $a$ has three distinct digits, the sum of the digits of $a$ can only be $9$ or $18.$

As I indicated in a comment, it's easy to exclude the possibility that the sum of the digits is $9$. The only possibilities are $1+2+6,\ 1+3+5\text{ and }2+3+4.$ In each case, it's easy to see that none of the $3$ digits could be the leading digit.