17 41 5 41 is divided by 11 then what is remainder

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Read Solution (32) :

RAKESH (9 years ago)

(x^n + y^n) is always divisible by (x+y) if n is odd number.

17^41+5^41 is always divisible by 17+5 = 22 so also divisible by 11.

17^41+5^41 / 11 => rem = 0

Is this solution Helpfull? Yes (98) | No (1)

mithun (9 years ago)

17^41+5^41=17^41+5^41/11
=6^41/11+(-6)^41/11
=(6/11)+(-6/11)
=0

Is this solution Helpfull? Yes (11) | No (1)

miraj ansari (9 years ago)

17^41+5^41/11

here we can use the simple logic that 17+5=22 which is completely divisible by 11 and rem=0

so the overall rem.is 0 Ans...

Is this solution Helpfull? Yes (10) | No (6)

Rakesh (9 years ago)

The unit digit of 17 is repeating after 4 cycle.
17^1= unit digit is 7
17^2= unit digit is 9,,17^3= ---------- is 3, 17^4= 1, 17^5= again unit digit 7
so 17^41= unit digit is 7.
Similarly 5^41= Unit digit is 5
so 5+7=12....12/11= remainder is 1

Is this solution Helpfull? Yes (5) | No (13)

Nomendra Sinha (9 years ago)

yes 0 is ans. i have done calculation mistake
17^41=(11+6)^41---->6^41
5^41=(-6)^41
0/11 remainder 0

Is this solution Helpfull? Yes (4) | No

Nomendra Sinha (9 years ago)

Unit digit approach is not applicable in remainder theorem.
Exa. If unit digit is 7 and you want to divide by 3 then you get remainder 1 but its wrong because may be no. before 7 is 2 so 27/3 remainder 0 actually.

Is this solution Helpfull? Yes (4) | No

C.P.Shrikrishnan (9 years ago)

17 can be written as (11+6)^41/11 and 5 can be written as (11-6)^41/11
=(11)^41/11+(6)^41/11+(11)^41/11-(6)^41/11
=0+6+0-6
=0

Is this solution Helpfull? Yes (2) | No (3)

Shirsendu Bhattacharjee (9 years ago)

Since the powers are same.
We take powers common.
(17+5)^41=22^41
Dividing it by 11...
22^41/11,.. Since 22 is divisible by 11, hence remainder is 0.

Is this solution Helpfull? Yes (2) | No (2)

Pranoy Bej (9 years ago)

ANSWER==> 0
17^41+5^41/11=(17+5)(17^40+.......)/11= no remainder
another way (x^n+y^n) is completely divisible by (x+y) if n= even
so 17^41+5^41 is divisible by 17+5=22=2*11
hence remainder=0

Is this solution Helpfull? Yes (1) | No

vino vijayarangan (9 years ago)

17^41,we can write it as 77.
power=odd,keep 7 it as and multiply unit digits so 77
5^41=55
77+55=132/11=rem(0)

Is this solution Helpfull? Yes (1) | No

PRIYANKA GUHA (9 years ago)

17^41+5^41/11 = (11+6)^41+5^41/11 = 6^41+5^41/11
the unit digit of 6^41=6 and unit digit of 5^41=5
(6+5)/11 = 11/11 =0

Is this solution Helpfull? Yes (1) | No

manaswini devabhakthuni (9 years ago)

the remanider is 1

Is this solution Helpfull? Yes | No (6)

Nomendra Sinha (9 years ago)

17^41=(11+5)^41 11 is multiple so no remainder=5^41
So 2*5^41=2*(11*2+3)^20 *5 =10*3^20 =10*(11-2)^10= 10*2^10=10*(33-1)^2=10
So 10/11 remainder is -1 or 10

Is this solution Helpfull? Yes | No (4)

prasann mishra (9 years ago)

remainder 0

6+5 mod 11 = 0

Is this solution Helpfull? Yes | No (1)

farahan (9 years ago)

17^41 means we can tke 17 as 7 n 7 has cyclic 4 so
17^41=7 n 5^41=5 bcz 5 hve a cyclic 1.......
so rem=7+5/11= 1 so rem=1

Is this solution Helpfull? Yes | No (4)

prasanth (9 years ago)

the remainder is 1.

Is this solution Helpfull? Yes | No (6)

saranya.k (9 years ago)

17^41+5^41/11
first divide 17 by 11 the remainder is 6 and divide 5 by 11 the remainder is 5 therefore 6^41+5^41/5 the unit place of 6^41 is 6 and unit place of 5^41 is 5 therefore (6+5)/11 the remainder is so the remainder is 0

Is this solution Helpfull? Yes | No (3)

N.ANNUSHYA (9 years ago)

17^41+5^41

=(Rem(17/11)^41+Rem(5/11)^41)
=(6^41+5^41)
=(11^41)
11^41 is completely divisible by 11.Hence remainder is 0

Is this solution Helpfull? Yes | No (3)

payal sharma (9 years ago)

(17^41)/11= ((17-6)^41)= (-6^41)/11
(5^41)/11= ((5+6)^41)= (6^41)/11
(-6+6)^41/11 = 0.
ans:0

Is this solution Helpfull? Yes | No (2)

NATHIYA (9 years ago)

17mod41 + 5 mod41
6+9=15
divided by 11 we get remainder is 4
so ans is 4.

Is this solution Helpfull? Yes | No (4)

Vasanth (9 years ago)

Answer is 5

Is this solution Helpfull? Yes | No (5)

hari (9 years ago)

i) 5^anything = 5
ii) 17/11=> R 6 , 6^anytjing =6

so ans is 5+6 = 11 and then 11/11=1 so R = 0

Is this solution Helpfull? Yes | No (1)

pooja verma (9 years ago)

17=11+6;
6^41=6*6^40=6(36^20)=6*3^20;
5^41= 5*5^40=5(25)^20 = 5*3^20;
[3^20(6+5)]/11 =[11*3^20]/11;
rem=0