What is the remainder when 104^358 + 7^29 is divided by 5?

Question

What is the remainder when  104^{358}+7^{29}  is divided by 5?

(A) 4
(B) 3
(C) 2
(D) 1
(E) 0

Fdambro294 · Answer

Cyclicity of Base with Units Digit of 4:

(4)^even Integer——-> Units Digit = 6

(4)^odd Integer ——-> Units Digit = 4

(104)^358———> Units Digit = 6

(7)^29 ———>

(7)^1 - Units Digit of 7
(7)^2 - Units Digit of 9
(7)^3 - Units Digit of 3
(7)^4 - Units Digit of 1

The pattern will keep on repeating for every block of 4 consecutive Integers

Thus, the Units Digit of (7)^29—-> 7

Therefore, the Units Digit of the Expression will be:

(......6) + (......7) = .......3

Concept: when dividing by 5:

If the Units Digit is from 1 thru 4, then the Remainder = Units Digit

If the Units Digit is from 6 thru 9, then the Remainder = (Units Digit) - (5)

In this case, the expression has a Units Digit of 3

Therefor the remainder when Divided by 5 is

Rem = 3

-B-

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CrackverbalGMAT · Answer

Concept:   When the remainder is 1 less than the divisor, then the remainder is -1

Each 104 when divided by 5 gives a remainder of 4 or -1

7^2 = 49 , when divided by 5 gives a remainder of 4 or -1

We have  R  + R  = R  + R

=  R  + R  = R  + R  = 1 + 2 = 3

Option B

Arun Kumar

Deyoz · Answer

Hi Arun , 
can you please clarify how you have simplified or conceptualized the last step

CrackverbalGMAT · Answer

Concept: When the remainder is 1 less than the divisor, then the remainder is -1

Each 104 when divided by 5 gives a remainder of 4 or -1

7^2 = 49. 49 when divided by 5 gives a remainder of 4 or -1

(-1) raised to an even number = 1 and (-1) raised to an odd number = -1

Now (-1)^358 = 1 and 1 divided by 5 gives a remainder of 1 (when the numerator is lesser than the divisor i.e the denominator, the remainder is the numerator)

Similarly (-1)^14 = 1 and 1 * 7 = 7. 7 divided by 5 gives a remainder of 2. therefore the sum of the remainders = 1 + 2 = 3

Hope this helps

Arun Kumar

danbrown9445 · Answer

104^{358}+7^{29} = (105-1)^{358}+7^{29}

(105-1)^{358} 
Remainder when  (-1)^{358}  divided by 5 is  (-1)^{358}=1  (105 divided by 5 will leave 0 remainder)

7^{29} 
Remainder when 7 divided by 5 is 2.
Cyclicity of 2 is 4 (2,4,8,6)
29 divided by 4 leaves remainder 1.
So used 1st value in cyclicity list.

Add remainders 1 + 2 = 3

BrushMyQuant · Answer

We know to find what is the remainder when  104^{358}+7^{29}  is divided by 5

Lets solve the problem using two methods:

Method 1: Units' Digit Cycle

Theory: Remainder of a number by 5 is same as the Remainder of the unit's digit of the number by 5

(  Watch this Video   to Learn How to find Remainders of Numbers by 5)

Units' digit of  104^{358} , will be same units' digit of  4^{358}

We can do this by finding the pattern / cycle of unit's digit of power of 4 and then generalizing it.

Unit's digit of  4^1  = 4
Unit's digit of  4^2  = 6
Unit's digit of  4^3  = 4

So, unit's digit of power of 4 repeats after every  2^{nd}  number.
=> If power is odd then units' digit is 4
=> If power is even then units' digit is 6

=> Unit's digits of  104^{358}  = 6

We can do this by finding the pattern / cycle of unit's digit of power of 7 and then generalizing it.

Unit's digit of  7^1  = 7
Unit's digit of  7^2  = 9
Unit's digit of  7^3  = 3
Unit's digit of  7^4  = 1
Unit's digit of  7^5  = 7

So, unit's digit of power of 7 repeats after every  4^{th}  number.
=> Remainder of 29 by 4 = 1
=> Units' digit of  7^{29}  = Units' digit of  7^1  = 7

=> Unit's digits of  104^{358}  + Unit's digits of  7^{29}  = 6 + 7 = 13
=> Remainder = Remainder of 13 by 5 = 3

Method 2: Binomial Theorem

We solve these problems by using Binomial Theorem, where we split the number into two parts, one part is a multiple of the divisor(5) and a big number, other part is a small number.

=>  104^{358}+7^{29}  =  (105 - 1)^{358}+(5 +2)^{29}

Watch this video to   MASTER BINOMIAL Theorem

Now, when we expand this expression then all the terms except the last term will be a multiple of 5. 
=> All terms except the last term will give 0 as remainder then divided by 5
=> Problem is reduced to what is the remainder when the last term is divided by 5
=> What is the remainder when  358C358 * 105^0 * (-1)^ {358}  +  29C29 * 5^0 * (2)^ {29}  is divided by 5 = Remainder of 1 by 5 + Remainder of  2^{29}  by 5

Unit's digit of  2^1  = 2
Unit's digit of  2^2  = 4
Unit's digit of  2^3  = 8
Unit's digit of  2^4  = 6
Unit's digit of  2^5  = 2
Unit's digit of  2^6  = 4

So, unit's digit of power of 2 repeats after every  4^{th}  number.
=> Remainder of 29 by 4 = 1
=> Units' digit of  2^{29}  = Units' digit of  2^1  = 2

=> Remainder of 1 by 5 + Remainder of  2^{29}  by 5 = 1 + 2 = 3

So,  Answer will be B 
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

What is the remainder when 104^358 + 7^29 is divided by 5?

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What is the remainder when 104^358 + 7^29 is divided by 5?

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