# Special Factors of 3-Digit Numbers

## Main Question or Discussion Point

Find all positive integers n (n will be called special)
such that if n divides a three digit number say ABC where A = hundreds, B = tens, C = ones. Then n also divides BCA and CAB.

So far I have worked out that 1 is special of course, so is 3 (factor test sum of digits will always be the same) and 9 as well as 111,222,333....999 and any number greater than 1000 as it will only divide into the 3 digit number 000.

Also worked out: n cannot be:
n - divides this - not this
2 - 132 - 321
4 - 492 - 249
5 - 335 - 353
6 - 336 - 363
7 - 861 - 618
8 - 984 - 849

I am told that there are two more such numbers n under 100. How do you find them? And please explan your reasoning. Thanks!

I just found another 27, but I dont know why it is one of the numbers. Please explain! thanks!

HallsofIvy
Homework Helper
Perhaps the fact that the three numbers you have found are 3, 9, and 27 will tell you something.

27 should not follow this rule. You can think on this. The number should be divisible by 99, 198, 297,,,,,,.

Sorry - Its BCA and CAB. Ithought of CBA.
Here ABC - BCA = 100(A - B) + 10(B - C) C - A.
Taking A, B and C as consecutive integers or in AP we get them to be having values like, 108, 216, and then 348. So what Halls mean need not happen coz we have already seen that 4 does not follow the rule. 81 as you intend does not follow the rule.

AKG
Homework Helper
Hint: If a and b are special, what of ab?
Hint: If n | ABC, then n | ABC0, i.e. n | 10 x ABC. If also n | BCA, then n | ABC0 - BCA. What is the prime factorization of ABC0 - BCA?

Yea, it can be shown true for 27, it is important to remember the form.

If we have 100a+10b+c==0 Mod 27, then c==-(19a+10b) Mod 27, so we throw this into 19b+10(-19a-10b)+b==19b-19b-189a==0Mod27. The other case works the same.

You can see that 486, 648, and 864 are all divisible by 27, but 468 is not.

If you go through the procedure above without regard for a modulus, you can find what that other number is.

Last edited:
AKG said:
Hint: If a and b are special, what of ab?
Hint: If n | ABC, then n | ABC0, i.e. n | 10 x ABC. If also n | BCA, then n | ABC0 - BCA. What is the prime factorization of ABC0 - BCA?
Are you saying that say 27 divides 135 and it also divides 351.
THerefore, add a 0 to 135 and 27 divides (1350-351)?

I'm not sure what you mean, but if it's this it's wrong...

While I worked this problem out by modulo arithmetic, I now see that AKG saw it on a somewhat simpler basis. (I was orginally thrown by: Hint: If a and b are special, what of ab? )

AKG is saying that it does not hurt to multiply by 10 and look at ABCO-BCA, dividing factors will go through that. Since 1350-351= 999, we check out the divisors of 999= 37x27.

AKG
Homework Helper
MGS said:
Are you saying that say 27 divides 135 and it also divides 351.
THerefore, add a 0 to 135 and 27 divides (1350-351)?

I'm not sure what you mean, but if it's this it's wrong...
Well, 27 does divide 135 and 351, and also divides (1350 - 351). However, I gave two hints, both of which are questions, so they can't really be wrong. Just try answering them.

robert Ihnot said:
While I worked this problem out by modulo arithmetic, I now see that AKG saw it on a somewhat simpler basis. (I was orginally thrown by: Hint: If a and b are special, what of ab? )

AKG is saying that it does not hurt to multiply by 10 and look at ABCO-BCA, dividing factors will go through that. Since 1350-351= 999, we check out the divisors of 999= 37x27.
Can you please show me the modulo method? Thanks a bunch!

If we have 100a+10b+c==0 Mod 27, then c==-(19a+10b) Mod 27,

The modulo method just consists of reducing the number by the modulus. In this case 100a = 3x27a+19a. Since we "cast out" multiples of 27, we are just left with 19a.

This might be easier to see if we took a small number like, say, 2. In that case all numbers are either even or odd. If even, then they are congruent to 0 and if odd, congruent to 1. This gives a arithmetic of 0+1 ==1 Mod 2, or odd + even equals odd. Furthermore 1+1==0 mod 2, or odd + odd = even number. The other cases are 0+0 ==0, and 1+0 ==1 Mod 2.

I am unsure what you are seeking. Yet,...

...where A, B, and C are non-zero leading, distinct digits, the following exist where n (such that n is greater than 1 and less than three digits) divides ABC, BCA and CAB:

2 - 246, 248, 264, 268, 284, 286, 426, 428, 462, 468, 482, 486, 624, 628, 642, 648, 682, 684, 824, 826, 842, 846, 862, 864

3 - 123, 126, 129, 132, 135, 138, 147, 153, 156, 159, 162, 165, 168, 174, 183, 186, 189, 192, 195, 198, 213, 216, 219, 231, 234, 237, 243, 246, 249, 258, 261, 264, 267, 273, 276, 279, 285, 291, 294, 297, 312, 315, 318, 321, 324, 327, 342, 345, 348, 351, 354, 357, 369, 372, 375, 378, 381, 384, 387, 396, 417, 423, 426, 429, 432, 435, 438, 453, 456, 459, 462, 465, 468, 471, 483, 486, 489, 492, 495, 498, 513, 516, 519, 528, 531, 534, 537, 543, 546, 549, 561, 564, 567, 573, 576, 579, 582, 591, 594, 597, 612, 615, 618, 621, 624, 627, 639, 642, 645, 648, 651, 654, 657, 672, 675, 678, 681, 684, 687, 693, 714, 723, 726, 729, 732, 735, 738, 741, 753, 756, 759, 762, 765, 768, 783, 786, 789, 792, 795, 798, 813, 816, 819, 825, 831, 834, 837, 843, 846, 849, 852, 861, 864, 867, 873, 876, 879, 891, 894, 897, 912, 915, 918, 921, 924, 927, 936, 942, 945, 948, 951, 954, 957, 963, 972, 975, 978, 981, 984, 987

6 - 246, 264, 426, 462, 468, 486, 624, 642, 648, 684, 846, 864

9 - 126, 135, 153, 162, 189, 198, 216, 234, 243, 261, 279, 297, 315, 324, 342, 351, 369, 378, 387, 396, 423, 432, 459, 468, 486, 495, 513, 531, 549, 567, 576, 594, 612, 621, 639, 648, 657, 675, 684, 693, 729, 738, 756, 765, 783, 792, 819, 837, 846, 864, 873, 891, 918, 927, 936, 945, 954, 963, 972, 981

18 - 468, 486, 648, 684, 846, 864

27 - 135, 162, 189, 216, 243, 297, 324, 351, 378, 432, 459, 486, 513, 567, 594, 621, 648, 675, 729, 756, 783, 837, 864, 891, 918, 945, 972

37 - 148, 185, 259, 296, 481, 518, 592, 629, 814, 851, 925, 962

54 - 486, 648, 864

6, 9, 18, 27 and 54 each are comprised of only the prime factors 2 and/or 3. The only other prime factor that has a separate solution with the stated restrictions is 37.

If A, B, and C do not need be distinct, other values of n do exist. In addition to supplemental solutions for the values of n already listed, the values for n that have solutions are: 4, 5, 7, 8, 12, 15, 21, 24, and 74. The prime factors remain 2, 3 and 37 with an additional prime of 5 for n as 5 and 15.

After re-reading the posts and the question, I can provide an answer to which factors are zero modulo to ABC, BCA and CAB. They are 1, 3, 9, 27 and 37.