The product of the last two digits of is __________. [2025]

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Solution

Q.
The product of the last two digits of ${(1919)}^{1919}$ is __________. [2025]

Ans.
(63)

We have, ${(1919)}^{1919} = {(1920 - 1)}^{1919}$

   $= {}^{1919}C_{0} {(1920)}^{1919} - {}^{1919}C_{1} {(1920)}^{1918} + . . . + {}^{1919}C_{1918} (1920) - {}^{1919}C_{1919}$

   $= {(1920)}^{1919} - 1919 {(1920)}^{1918} + . . . + 1919 (1920) - 1$

   $= 100 λ + 1919 (1920) - 1$

[As all terms will have 100 as multiple except the last two terms]

   $= 100 λ + 3684480 - 1 = 100 λ + 3684479$

$∴$ Last two digits will be 79

So, required product = $7 \times 9 = 63$ .

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