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The remainder when `9^103` is divided by 25 is equal toA. 5B. 6C. 4D. none of these

Answer» Correct Answer - c
We have,
`9^(103)`
=` 9xx(81)^(51)`
= ` 9(80 + 1)^(51)`
`= 9{""^(51) C_(0) (80)^(51)+ ""^(51)C_(1) (80)^(50) + ""^(51)C_(2)(80)^(49) +...+""^(51)C_(50)(80) +1}`
`9 [ 25 k + (51 xx80 + 1)] `
`= 25 k lambda+ 36729`
`25 lambda + 25 xx1469 + 4`
` 25(lambda + 14 69) + 4` Hence, the remainder is 4,


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