\(5^{16} - 3^{16}\)

\(5^8*2 - 3^8*2\)

\((5^8 +3^8) (5^8 - 3^8)\)

now just break down 5^8 - 3^8.

apply \(a^2 -b^2.\)

at least we got : (\(5^8 +3^8) (5^4 +3^4) (5^2 +3^2) ( 5 +3 ) (5-3)\)

Now divide :

\((5^8 +3^8) (5^4 +3^4) (5^2 +3^2) ( 5 +3 ) (5-3) /16.\)

\((5^8 +3^8) (5^4 +3^4) (5^2 +3^2) *8*2 /16.\)

Reminder is 0.

A is the correct answer.

I solved using cyclicity

for any power to 5 we would always get 5 as units digit and for 3 ^ 16 we would get 1 as unit digits

so 5-1= 4 unit digit

16= 2^4

for an even Nr ending with 4 unit digit when divided by 2 would always give remainder as 0 , so IMO A ..

Solution

To find:
We are asked to find out,

• The remainder when \(5^{16} – 3^{16}\) is divided by 8

Approach and Working:

• \(5^{16} – 3^{16}\) can be written as \((5^8)^2 – (3^8)^2\)

We know that, \(a^2 – b^2 = (a + b) * (a - b)\)

• Thus, \((5^8)^2 – (3^8)^2 = (5^8 + 3^8) * (5^8 – 3^8) = (5^8 + 3^8) * (5^4 + 3^4) * (5^4 – 3^4) \)
\(= (5^8 + 3^8) * (5^4 + 3^4) * (5^2 + 3^2) * (5^2 - 3^2) = (5^8 + 3^8) * (5^4 + 3^4) * (5^2 + 3^2) * (5 + 3) * (5 – 3)\)

• So, \(5^{16} – 3^{16} = 16 * (5^8 + 3^8) * (5^4 + 3^4) * (5^2 + 3^2) * (5^2 - 3^2)\)

If you observe carefully, the above expression is a multiple of 8

• Therefore, the remainder will be 0

Hence the correct answer is Option A.

Answer: A


_________________

(5-3)^16/16= (2)^16/2^4.
The remainder is zero.

=== Please suggest if we can solve like this

I dont think we can.
its against the rules of indices

Posted from my mobile device

5^16-3^16

25^8 - 9^8

((16+9)^8 - 9^8)/16

Remiander

(9^8 - 9^8)/16

0

[size=80][b][i]Posted from my mobile device[/i][/b][/size]

Hey suchita2409,
I am afraid you can not solve it like that.

But, I have an alternate solution. I think you might prefer to solve the question from that method.

In this method, we are using the formula a^2 -b^2 = (a-b)(a+b)

5^16- 3^16 = (5^8- 3^8 )( 5^8 + 3^8)
= (5^4 - 3^4)(5^4 + 3^4)(5^8 + 3^8)
= (5^2 - 3^2)(5^2 + 3^2)(5^4 + 3^4)(5^8 + 3^8)

Now, (5^2 - 3^2) = 25 - 9 = 16
And, 16 is a multiple of 8.
S0, 5^16 - 3^16 is also a multiple of 8.

Hence, 5^16- 3^16 is divisible by 8. Thus, the remainder is 0.
_________________

Am I correct in solving the problem as below? And can this strategy be applied to similar problems??

5^2-3^2= 16; This divided by 16 -rem=0
5^4-3^4= 544; This divided by 16 -rem=0

so for all even powers of this expression, the remainder is 0 and hence for 5^16-3^16. Remainder =0.

Hi egmat,

Can this problem be solved without using the identity.
(5^16 - 3^16) / 8 = ( 5^16/8 ) - ( 3^16/8 )
Now we make use of cyclic pattern and compute remainder.

5/8 = remainder is always 5
3/8 = remainder is 1 as pattern repeats after 2
So remainder is = 5-1=4
This will be divisible by 8 as 4 ^16 hence the remainder is 0

Please comment on the approach whether it is correct or not

(5^2)^8-(3^2)^8
=(25)^8- (9)^8
=(25-9)× integer
=16× integer
Reminder 0

Posted from my mobile device

5=8(1)+5
25=8(3)+1
Both only contain 5 as prime factor yet have different remainders when divided by 8 (think this cyclicity continues 125=8(15)+5; 625=8(78)+1)

A side note for others, a^b isn't necessarily divisible by b
e.g. 4^3=64, 64=3(21)+1

Ciclicity tells us that the units digit of the subtraction will be 5-1=4.
Hence, the remainder must be even.
Answer (A).

5^16 is always odd
3^16 is always odd
it means 5^16- 3^16 is always even---(x)
IQ part:
if a even number (16) will divide any even number (from (x)) then how the remainder is any ODD number

Looking into choices , ANS must be (a) 0 ie Even number.

Ans A