Latest TCS Aptitude Question SOLUTION: What is the remainder when 6^17+17^6 is divided by 7? a)1 b)6 c)0 d)3

6^17+17^6 = (7-1)^17 + (21-4)^6 = (7-1)^17 + (7*3 -4)^6....[eqn1]

if the [eqn1] is expanded then every term of the expansion except [(-1)^17 + (-4)^6] will have 7 as one of its factors.
Just think a little bit about the binomial expansion of both [(7-1)^17] and [(7*3 -4)^6] ,then u can readily point out that only the last term of both the expansions , i.e , [(-1)^17] & [(-4)^6] respectively,will don't have 7 as one of its factors.

So, we have to calculate the remainder when [(-1)^17 + (-4)^6] is divided by 7.
Now, clearly (-1)^17 = -1
and, (-4)^6 = 4^6 = 2^12 = (2^3)^4 = (7+1)^4.
Now, a same reasoning related to binomial expansion mentioned previously explains why, when (7+1)^4 is divided by 7 will leave a remainder 1.
So, (7+1)^4 will be of the for (7*A + 1); where A is some +ve integer, to know whose value isn't important in this case.
So, when
[(-1)^17 + (-4)^6] will be divided by 7
or, [-1 + (7+1)^4] will be divided by 7
or, when [-1 + 7*A +1] will be divided by 7
or, when [ 7*A ] will be divided by 7 , clearly therefore the remainder will be zero, i.e, 0.

So, the answer is OPTION 3)0.

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6^17 mod 7=(7-1)^17 mod 7=(-1)^17 mod 7=-1
17^6 mod 7=(7*2+3)^6 mod 7= (3)^6 mod 7=729 mod 7=1
the remainder when 6^17+17^6 is divided by 7 is
1-1=0(ans)

6^2 = 36
6^3 = 216
Thus 6 raise to anything will give you 6 in the decimal's place.

Similarly 17^6 will give 9 in the decimal place.

Adding the decimal place values of both i.e 6+9 = 15

Remainder = 1. Ans (a)

we can calculate this type of Questions with calculator.

6^17mod 7=((6^2 mod7)^6*6mod7)mod 7=((1^6)*6)mod 7=1*6mod 7=6 mod 7=6
17^6 mod 7=((17^2mod 7)^3)mod 7=2^3mod 7=8 mod 7=1
so,1+6=7 mod 7=1
so,answer is (a)1

remainder of 6^17 r1=6
remaindr of 17^6 r2=(17^2)^3=(289)^3=(7*41+2)^3=2
power should be oddexcept an case 1.
total remainder=r1+r2=8
now 8>7 again 8 is divide by 7

then we get remainder=1ans

short trick-

6^17+17^6

take6(6^n where n is odd)+17^2(17^n where n is even)=295 divide by 7

remainder=1apply this method another qus..

ANSWER C) 0 a congruent b (modm) means a-b is divisible by m

6 CONGRUENT -1(MOD 7) THEREFORE 6¹⁷ CONGRUENT (-1)¹⁷(MOD 7) = -1(MOD 7)

17 CONGRUENT 3(MOD 7) THEREFORE 17⁶ CONGRUENT 3⁶(MOD 7) = 1(MOD 7)

HENCE 6¹⁷ + 17⁶ CONGRUENT -1+1(MOD 7)= 0(MOD 7)

REMAINDER IS 0 .

Remainder of (6^17) = 6............r1
Remainder of (17^6) =(17^2)^3=(289)^3=(7*41+2)^3= 2^3 =8.......r2

total remainder= r1 + r2 = 14
now 14 is divide by 7 = 0 (Ans)

6 raised to power anything remainder will be 6
again 17^6 is calculated in units place as (7^6) which will be (7^(6/4)) ie 7^2 its unit digit is 9
as cyclicity of 7 as in units palce is
7^1=7
7^2=9
7^3=3
7^4=1
so (6+9)%7=1 ans

answer is 6..
when 6,6^2,6^3....
is divided by 7 gives the pattern of 6,1
i.e,if power is odd it gives 6 and if it is even it is 1
here 17 is odd that is 6+17^3 is divisible by 17 hence 0 is the remainder
so in total 6+0=6mod17=6 is the answer
hence option b is correct

in this problem we use unit digit method
6 power of any value we get the unit digit as 6 adding with 17^6 value is 9 total is 15 divided by 7 remaining is 1

6^17=...6(6 is the unit digit).....1
17^6=...9(1 is the unit digit).....2

(6^17)+(17^6)=6+9=15(which gives remainder as 1)
So remainder is 0.........

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2.418097655x10^12

0. As the series is absolutely divisible by 7

6^anything the unit digit will be 6 so its remainder is 6 then for 2nd case by dividing the (17^2)^3 we get (7*41+2)^3 then find unit digit for 2^3 is 8
its remainder now add two remainder and divide it by 7 so you vl get ans as 0

(6^17)/7 by reminder method
1*7=7
6-7=-1
(-1^17)=-1 (-1 power any odd number means = -1)

(17^6)=7
its formula like (n^(b-1))/b=1
(17^(7-1))/7=1

then -1+1=0
answer is 0

using -1 rule we get the value for (6)^17/7 as 6 because (a)^x/(a+1) gives remainder "a" when x is odd and gives remainder 1 when x is even..
Similarly, using prime rule (a)^(p-1)/p gives remainder 1 where p is a prime number.. So value of (17)^6/7 gives remainder 1

Now combining both both we get 6+1=7 which is completely divisible by 7
Hence the remainder is 0.. :)

6^17 + 17^6
1.For 6^17
6^1 %7=6
6^2 %7=1
6^3 %7=6
6^4 %7=1
6^5 %7=6
6^6 %7=1
6,1 repeat for 2 times. So 17/2 = 1
So 6^17 % 7 = 6
2.For 17^6
1st 17 % 7 = 3
Therefore we go with 3 instead of 17
3^6 = 729
729 % 7 = 1
So 17^6 % 7= 1
Finally
(6^17 + 17^6) % 7= (6+1) % 7= 7 % 7 =0
Ans c)0

6^17+!7^6=6+9=15/7=1

((-1)^17+(2)3)7=-1+1=0

first find unit place digit so 6 to the power any thing always 6
similarilly 17^6=9
so 6+9=15/7=1 ans is a.