(i)Find the remainder when 7 98 is divided by 5
(ii)Using binomial theorem prove that 6 n – 5n always leaves the remainder 1 when divided by 25.
(iii)Find the last digit, last two digits and last three digits of the number (27) 27 .
Text Solution
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(i) 4
(iii) 3, 03, 803
Sol. (i) 7 98 = (50 –1) 49 = 49 C 0 (50) 49 – 49 C 1 (50) 48 + .....– 49 C 49 ⇒ Remainder= 5 – 1 = 4
(ii) 6 n – 5n = (5 + 1) n – 5n = 5 n + n C 1 .5 n – 1 + ......+ n C n – 2 . 5 2 + n C n – 1 . 5 + 1 – 5n = 25 λ + 1
(iii) (27) 27 = 3 81 = 3.(9) 40
= 3(10 – 1) 40 = 3(10 40 – 40 C 1 .10 39 + ..... + 40 C 38 .10 2 – 40 C 39 .10 +1 ) = 3(1000 λ – 400 + 1 )
Last 3 digits of this number = 803.
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