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

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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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24 Feb 2020, 02:55
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75% (hard)

Question Stats:

51% (01:54) correct 49% (02:28) wrong based on 85 sessions

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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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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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24 Feb 2020, 03:41
1
5
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
##### General Discussion
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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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Updated on: 20 Mar 2020, 04:47
1
2
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.

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

Originally posted by NHiromoto on 17 Mar 2020, 21:26.
Last edited by NHiromoto on 20 Mar 2020, 04:47, edited 1 time in total.
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Re: What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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19 Mar 2020, 01:37
NHiromoto wrote:
What is the remainder when $$5555^{2222}+2222^{5555}$$ is divided by 7?

Using the binomial theorem, $$5555^{2222}+2222^{5555}= 7n + 4^{2222} + 3^{5555}$$
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.
$$4^{2222} + 3^{5555} = m * (16 + 243)$$
Because 259 is divisible by 7, the remainder is 0.

can you explain the bold line please?

thanks
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Re: What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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20 Mar 2020, 04:49
1
NHiromoto wrote:
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.

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

thank you so much!!
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Re: What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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28 Mar 2020, 07:01
I understand nothing of this question. Why do we pick 4 and 3 for b in the binominal theorem and how do we get from 3 to 243?
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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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28 Mar 2020, 07:08
2
Bunuel wrote:
What is the remainder when $$5555^{2222} + 2222^{5555}$$ is divided by 7?

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

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

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

Remainder $$5555^{2222} + 2222^{5555}$$ divided by 7 = Remainder $$4^{2222} + 3^{5555}$$ divided by 7 = 0 {because 4+3=7}
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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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29 Mar 2020, 00:32
Bunuel wrote:
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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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
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What is the remainder when 5555^2222 + 2222^5555 is divided by 7?  [#permalink]

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30 May 2020, 22:23
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)
(2+5)/7 = 0 remainder that is the ans

if anyone have any query plz ask
What is the remainder when 5555^2222 + 2222^5555 is divided by 7?   [#permalink] 30 May 2020, 22:23