2. Due to
106=8× 13+2 2=2× 1
=8× 12+ 10 10=2×5
=8× 1 1+ 18 18=2×9=6×3
=8× 10+26 26=2× 13
=8×9+34 34=2× 17
=8×8+42 42=2×2 1=6×7= 14×3
=8×7+50 50=2×25= 10×5
=8×6+58 58=2×29
=8×5+66 66=2×33=6× 1 1=22×3
=8×4+74 74=2×37
=8×3+82 82=2×4 1
=8×2+90 90=2×45=6× 15
= 10×9= 18×5=30×3
=8× 1+98 98=2×49= 14×7
=8×0+ 106 106=2×53
Because 106 is an available fraction, from the above decomposition, to get 106, a should be 1, 3,5,7,9,1,13,/kloc. because
108=8× 13+4 4=4× 1
=8× 12+ 12 12=4×3
=8× 1 1+20 20=4×5
=8× 10+28 28=4×7
=8×9+36 36=4×9
=8×8+44 44=4× 1 1
=8×7+52 52=4× 13
=8×6+60 60=4× 15
=8×5+68 68=4× 17
=8×4+76 76=4× 19
=8×3+84 84=4×2 1
=8×2+92 92=4×23
=8× 1+ 100 100=4×25
Because 108 is an available fraction, to get 108 from the above decomposition, A should be 1, 3,5,7,9,1,13,/kloc.
Combined with the above analysis, to calculate these two numbers at the same time, A can only be 1, 3, 5, 7, 9, 1 1, 13, 15 and 65438. Because 83 = 8×10+3 3 = 3×1
=8×9+ 1 1 1 1= 1 1× 1
=8×8+ 19 19= 19× 1
=8×7+27 27=3× 19=27× 1
=8×6+35 35=5×7=35× 1
=8×5+43 43=43× 1
=8×4+5 1 5 1=3× 17=5 1× 1
=8×3+59 59=59× 1
=8×2+67 67=67× 1
=8× 1+75 75=3×25=5× 15=75× 1
Because 83 is an unattainable score, from the above breakdown, A can't be 1, 3, 5, 7, 9,1,15, 17, 19. According to the above analysis, A can only be 13 or 2 1.
Identify one
If a=2 1, 2 1 is odd, and 8 and 1 10 are even, then when 2 1 and 8 constitute1/kloc-0, 2/kloc-0. Because110-21× 2 = 68,68 ÷ 8 = 8 ... 4; 1 10-2 1×4=26,26÷8=3……2; 2 1× 6 > 1 10, so a≠2 1.
Let's see if a= 13 fits the question.
Because 103=8×8+3× 13.
104=8× 13
105=8×5+5× 13
106=8× 10+2× 13
107=8×2+7× 13
108=8×7+4× 13
109=8× 12+ 1× 13
1 10=8×4+6× 13
So a= 13 is the only solution that satisfies the meaning of the question.