Prove that, if ABA is divisible by 7 (with no remaining), then BAB is also divisible by 7 (with no remaining) as well.

Example 1; 161 = 7x23, 616 = 7x88, both 161 and 616 are divisible by 7 (with no remaining).

Example 2; 252 = 7x36, 525 = 7x75, both 252 and 525 are divisible by 7 (with no remaining).



Solution to the Problem:


ABA = 7x (since we know that ABA is divisible by 7)
BAB = y (we don’t know yet , whether y is divisible by 7 or not)

100A + 10B + A = 7x
100B + 10A + B = y

If we subtract two sides:
7x – y = 91A-91B

And then take y to left, and others to right:
y = 7x - 91A + 91B

Factor out 7:
y = 7 (x - 13A + 13B)
y/7 = x - 13A + 13B

So, y is divisible by 7


Dr. Hari Kishan sent in another way of looking at the problem:
"The three digit numbers divisible by 7 are 161, 252, 343, 434, 525, 595, 616, 686,
868 and 959 which satisfy the given condition, i.e. ABA and BAB are divisible by 7."

There is another number which satisfies the condition, but it is trivial (A = B):
777

Below is the list of all 128 3-digit numbers that are multiples of 7:
105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497, 504, 511, 518, 525, 532, 539, 546, 553, 560, 567, 574, 581, 588, 595, 602, 609, 616, 623, 630, 637, 644, 651, 658, 665, 672, 679, 686, 693, 700, 707, 714, 721, 728, 735, 742, 749, 756, 763, 770, 777, 784, 791, 798, 805, 812, 819, 826, 833, 840, 847, 854, 861, 868, 875, 882, 889, 896, 903, 910, 917, 924, 931, 938, 945, 952, 959, 966, 973, 980, 987, 994


Correctly solved by:

1. Dr. Hari Kishan D.N. College,
Meerut, Uttar Pradesh, India
2. Brijesh Dave Mumbai City, Maharashtra, India
3. Ritwik Chaudhuri Santiniketan, West Bengal, India
4. Kelly Stubblefield Mobile, Alabama