Question

What are the last two digits of x, if x = (11¹+ 11²+ 11³+ ... + 11⁵⁵)?​

Answer 1

Given :- What are the last two digits of x, if x = (11¹+ 11²+ 11³+ ... + 11⁵⁵) ?

Answer :-

→ x = (11¹+ 11²+ 11³+ ... + 11⁵⁵)

in order to find last two digits we have to find the remainder when x divide by 100 .

so,

→ (x/100) = (11¹+ 11²+ 11³+ ... + 11⁵⁵) /100

→ (x/100) = [(11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 + 51)]

→ (x/100) = 5(11 + 21 + 31 + 41 _____ 91) + (11 + 21 + 31 + 41 + 51)

→ (x/100) = 5 * 459 + 155

→ (x/100) = 5 * 59 + 55

→ (x/100) = 295 + 55

→ (x/100) = 95 + 55

→ (x/100) = 150 = 50 (Ans.)

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