Question

find remainder when 2^97 is divided by 15

Answer 1

2⁹⁷ is divided by 15 .
First of all we have to find out last digit of 2⁹⁷
see , 2¹ = 2
2² = 4
2³ = 8
2⁴ = 16
2⁵ = 32 , after 4 terms last digit is repeating .
Hence, you should divide by 4 from given power .
means , if given power 97 then, you should write , 97 = 4 × 24 + 1
here remainder is 1 so, put the 1 in place of 97 now, you can get last digit .
so, last digit of 2⁹⁷ = last digit of 2⁴ˣ²⁴⁺¹ = last digit of 2¹ = 2

Hence , last digit of 2⁹⁷ = 2

Again , we try to understand what will remainder if we divide by 15 from 2ⁿ
2⁴/15 = 16/15 ⇒ remainder = 1
2⁵/15 = 32/15 ⇒remainder = 2
2⁶/15 = 64/15 ⇒remainder = 4
2⁷/15 = 128/15 ⇒remainder = 8
2⁸/15 = 256/15 ⇒remainder = 1 {repeating }

Hence, if we 2ⁿ is divided by 15 , remainder will be 1 , 2, 4, 8

Now, come to the point.
We have , 2⁹⁷/15
we get last digit = 2
It means remainder must be 2

Hence, remainder = 2

Answer 2

Answer:

Step-by-step explanation:

2⁹⁷ is divided by 15 . 2⁵ = 32 , after 4 terms last digit is repeating . Hence, you should divide by 4 from given power . here remainder is 1 so, put the 1 in place of 97 now, you can get last digit

as we know thw binomial theorem

$(a + b)^n = a^n + (nC_1)a^{n-1} b + (nC_{2} )a^{n-2} b^{2} + …$

now in ques,

$2^{97} = 16^{\frac{97}{4} } =(15+1)^{\frac{97}{4}}$

$(15 + 1)^\frac{97}{4} = 15^\frac{97}{4} + (\frac{97}{4}C_1)15^{\frac{97}{4}-1} 1 + (\frac{97}{4}C_{2} )15^{\frac{97}{4}-2} 1^{2} +..... +1^\frac{97}{4}$

on dividing

$\frac{(15 + 1)^\frac{97}{4}}{15} =\frac{ 15^\frac{97}{4} + (\frac{97}{4}C_1)15^{\frac{97}{4}-1} 1 + (\frac{97}{4}C_{2} )15^{\frac{97}{4}-2} 1^{2} +..... +1^\frac{97}{4}}{15}$

$=\frac{1^\frac{97}{4} }{15} = \frac{1}{15}$

remainder=1