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 |