What is the remainder when 5555^2222 + 2222^5555 is divided by 7?

Question

What is the remainder when  5555^{2222} + 2222^{5555}  is divided by 7?

A. 0
B. 1
C. 3
D. 4
E. 6

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danbrown9445 · Accepted Answer

5555^{2222} + 2222^{5555}

5555 leaves remainder 4 when divided by 7 & 2222 leaves remainder 3 when divided by 7

=4^{2222} + 3^{5555}  
 =4^{2220}*4^2 + 3^{5553}*3^2  
 =(4^3)^{740}*16 + (3^3)^{1851}*9

\frac{(4^3)}{7}  gives remainder 1 and  \frac{(3^3)}{7}  gives remainder -1

=64^{740}*16 + 27^{1851}*9 
 =1^{740}*16 + (-1)^{1851}*9 
 =16 + (-9) 
 =7

And 7/7 = 0 gives the answer

NHiromoto · Answer

What is the remainder when  5555^{2222}+2222^{5555}  is divided by 7?

In the binomial theorem,  (a + b)^n=(a^n) + nC1(a^{n-1}*b^1) + nC2(a^{n-2}*b^2) + .... + nC1(a^1*b^{n-1}) + b^n .(C stands for the combination)
When it comes to the expression given,  5555^{2222} = (7n + 4)^{2222} . (n is the quotient of 5555/7)
Let's say a=7n and b=4,  (7n + 4)^{2222} = 7n^{2222} + ..... + (2222C2221)(7n^1 * 4^{2221}) + 4^{2222}. 
Thus,  5555^{2222} = 7m + 4^{2222} .(7m is the simplified form of  7n^{2222} + ..... + (2222C2221)(7n^1 * 4^{2221}) )
Similarly,  2222^{5555} = (7r + 3)^{5555} .
Hence,  2222^{5555} = 7s + 3^{5555} .
So,  5555^{2222} + 2222^{5555} = 7m + 7s + 4^{2222} + 3^{5555}

Since we need only the remainder of the expression, we can ignore 7n.

Then,  4^{2222} + 3^{5555} = 16^{1111} + 243^{1111} 
Because  a^n + b^n  is divisible by a + b when n is odd.

4^{2222} + 3^{5555} = t * (16 + 243) 
Because 259 is divisible by 7, the remainder is 0.

The answer is A

Mansoor50 
Sorry, but I could not post a new comment.
So, I revised my post to express the binomial theorem explicitly.
Hope it helps.

lacktutor · Answer

What is the remainder when  5555^{2222}+2222^{5555}  is divided by 7?
-->  (7m+4)^{2222}+(7n+3)^{2222}= ... + 4^{2222}+ ...+ 3^{5555}

4^{2222}= (7-3)^{2222}= ...+3^{2222} 
-------------------------------------------

--> 3^{2222} + 3^{5555}= 3^{2222}*(1+ 3^{3333})= 3^{2222}*(1+ 27^{1111})= 3^{2222}*(1+ (28-1)^{1111})

-->  3^{2222}*(1+ (7x^{1111}+ ...-(1)^{1111})= 3^{2222}*(1+ (7x)^{1111}+ ...-1)= 3^{2222}*(7x)^{1111}

3^{2222}*(7x)^{1111}  can be divided by 7. The remainder will be 0

The answer choice A is Correct

GMATinsight · Answer

Concept : Remainder of when  5555^{2222}  is divided by 7 = Remainder when  4^{2222}  where Remainder   = 4

Concept : Remainder of when  2222^{5555}  is divided by 7 = Remainder when  3^{5555}  where Remainder   = 3

Remainder  5555^{2222} + 2222^{5555}  divided by 7 = Remainder  4^{2222} + 3^{5555}  divided by 7 = 0 {because 4+3=7}

Kinshook · Answer

Asked: What is the remainder when  5555^{2222} + 2222^{5555}  is divided by 7?

Remainder when 5555^{2222} + 2222^{5555} is divided by 7 
= Remainder when 4^{2222} + 3^{5555} is divided by 7
= Remainder when 4^2 + (-1)^{1851}*3^2 is divided by 7 since 4^3 = 64 = 63 + 1 and 3^3 = 27 = 28 -1
= Remainder when 16 - 9 = 7 is divided by 7
= 0

IMO A

aryan440 · Answer

step 1: 5555/7 
55/7 = 6 rem
65/7 = 2 rem
25/7 = 4rem
so 4^2 =16
16/7 = 2 rem-------------- (i)
Step 2: 2222/7
22/7 = 1 rem
12/7 = 5rem
52/7 =3 rem
now 3^5 = 9*9*3
again 9/7 = 2 rem
9/7 = 2 rem
2*2*3/7 = 5 rem----------- (ii)
add (i+ii)
(2+5)/7 = 0 remainder that is the ans

if anyone have any query plz ask

Puja01 · Answer

Why can't we just estimate the units digit and figure out the remainder based on the units digit?
Would be quicker no? Solved it very quickly and got the correct answer. Is it the approach correct or did I just get lucky?

Bunuel · Answer

The remainder when dividing by 7 is not determined by the units digit. For example, 11 divided by 7 gives a remainder of 4, but 21 divided by 7 gives 0. So using just the last digit doesn’t work here.

mvgkannan · Answer

Ok. The problem looks so huge when we see it initially. (Actually i didn't apply this shortcut so it took me almost 5 mins to solve this by identifying the cyclicity alone, and in a way i am putting up this explanation to help me memorize this technique).

\(5555^{2222} + 2222^{5555}\) has to be first reduced to simpler nos in the base. Understanding the modulo theorems, we know,

\((a + b) mod m ≡ [ a(mod m) + b(mod m)]mod m\)
Also,
\( a^b mod m ≡ [[a (mod m)]^ b]mod m\).

We get

\([5555^{2222} + 2222^{5555} ] (mod 7) ≡ [4^{2222} (mod 7) + 3^{5555} (mod 7)] mod 7\).

Now applying Eulers totient function, which say

If \( gcd⁡(a,n)=1 \) , then
\( a^{φ(n)}≡1 mod n\).

Since 7 is prime, φ(n) = n-1 = 7-1 = 6.

The Euler's Totient Function, denoted as φ(n), counts how many positive integers less than or equal to n are coprime to n.

φ(n)=number of integers k such that 1≤k≤n, gcd⁡(k,n)=1

Since 7 is prime, φ(n) = n-1 = 7-1 = 6, So we have to turn the powers into multiples of 6. So,
\([5555^{2222} + 2222^{5555} ] (mod 7) ≡ [4^{2222} (mod 7) + 3^{5555} (mod 7)] mod 7\)
can be rewritten as,

\([4^{2220} (mod 7) .4^2 (mod 7) + 3^{5550} (mod 7) .3^5 (mod 7)] mod 7\)

As,

\(3^{5550} (mod 7)= 4^{2220} (mod 7)=1\),

we get,
\([4^2 (mod 7) + 3^5 (mod 7)] mod 7\)
\([16 (mod 7) + 243 (mod 7)] mod 7\)
\([2 + 5 ] mod 7\)
=0.
Option A.

Bunuel

What is the remainder when \(5555^{2222} + 2222^{5555}\) is divided by 7?

A. 0
B. 1
C. 3
D. 4
E. 6

Are You Up For the Challenge: 700 Level Questions

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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?

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