279000=2.79×105, 27900010=2.7910×1050=28578.67126×1050=2.857867126×1054 approximately.
To get a more accurate figure 2.79=2+0.79, so 2.7910=
(2+0.79)10 which can be expanded binomially:
210 + 10×29×0.79+45×28×0.792+120×27×0.793+210×26×0.794+252×25×0.795+
0.7910+10×2×0.799+45×22×0.798+120×23×0.797+210×24×0.796,
1024+15.8×256.151710881+112.338×64.2430874555+473.31744×16.38950081+1308.722722×4.6241+2481.33828.
Each of these products can be hand-calculated and summed to give an exact answer for 2.7910.
The problem is simplified by considering only the fractional part of 2.7910 because we already know from the approximate solution what the integer part is (28578). For example:
15.8×256.151710881=(15+0.8)(256+0.151710881)=(3840+15×0.151710881+0.8×256+0.8×0.151710881)=
3840+2.275663215+204.8+0.1213687048. We can ignore all the integer parts and just sum the fractions:
1.19703192, making the fraction part 0.19703192. Similarly for all the other products. Then simply sum all these fractional contributions. The result should match the approximation and continue beyond the approximate decimal places. Any error will be apparent if there is a large discrepancy in the calculations.