What is the remainder when 333^222 is divided by 7?

Question

A. 3
B. 2
C. 5
D. 7
E. 1

Bunuel · Accepted Answer

What is the remainder when 333^222 is divided by 7?
A. 3
B. 2
C. 5
D. 7
E. 1

\(333^{222}=(329+4)^{222}=(7*47+4)^{222}\). Now, if we expand this, all terms but the last one will have 7*47 as a multiple and thus will be divisible by 7. The last term will be \(4^{222}=2^{444}\).

So we should find the remainder when \(2^{444}\) is divided by 7.

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

The remainder repeats in blocks of three: {2-4-1}. So, the remainder of \(2^{444}\) divided by 7 would be the same as \(2^3\) divided by 7 (444 is a multiple of 3). \(2^3\) divided by 7 yields remainder of 1.

Answer: E.

Or, when finding the remainder of \(2^{444}\) divided by 7 we can use the same trick: \(2^{444}=(2^3)^{148}=8^{148}=(7+1)^{148}\). When we expand this, all terms but the last one will have 7 as a multiple and thus will be divisible by 7. The last term will be \(1^{148}=1\). 1 divided by 7 gives the remainder of 1.

Answer: E.

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Theory on remainders problems: https://gmatclub.com/forum/remainders-144665.html

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Hope it helps.

Bunuel · Answer

The units digit does not determine the remainder when dividing by 7. For example, 9 divided by 7 gives the remainder of 2, 19 divided by 7 gives the remainder of 5, 29 divided by 7 gives the remainder of 1, ...

TGC · Answer

(333)^222

Rem(333/7) = 4

=> (4) ^222

=> (16) ^111

Rem(16/7) = 2

(2)^111 = 2^100 * 2^11

Now let us observe the Rem(2^10)/7

2^10 = 1024 => Rem(1024/7) = 2

REM(2^100 * 2^11 ) be 7 = REM((2^10)^10 * 2^11) by 7 => REM((2)^10 * 2^10 * 2) by 7

=> Rem( 2* 2 *2 ) by 7

=> 8/7

=> 1

(E)

iNumbv · Answer

What is the remainder when 333^222 is divided by 7? 
A. 3
B. 2
C. 5
D. 7
E. 1

333^{222}=(329+4)^{222}=(7*47+4)^{222} . Now if we expand this, all terms but the last one will have 7*47 as a multiple and thus will be divisible by 7. The last term will be  4^{222}=2^{444} . So we should find the remainder when  2^{444}  is divided by 7.

2^1 divided by 7 yields remainder of 2;
2^2 divided by 7 yields remainder of 4;
2^3 divided by 7 yields remainder of 1;

2^4 divided by 7 yields remainder of 2;
2^5 divided by 7 yields remainder of 4;
2^6 divided by 7 yields remainder of 1;
...

The remainder repeats in blocks of three: {2-4-1}. So, the remainder of  2^{444}  divided by 7 would be the same as  2^3  divided by 7 (444 is a multiple of 3).  2^3  divided by 7 yields remainder of 1.

Answer: E.

Hey can you explain to me how you get a remainder of 2 when you divide 2^1/7?

mau5 · Answer

This might help : remainders-144665.html If x and y are positive integers, there exist unique integers q and r, called the quotient and remainder, respectively, such that y =divisor*quotient+remainder = xq + r and 0\leq{r}

agourav · Answer

A simple one line solution to this problem can be this:
Rem(333^222)/7 = Rem(4^222)/7 = Rem(64^74)/7=Rem((63+1)^74)/7 = 1

Bunuel · Answer

For more on this kind of questions check  Units digits, exponents, remainders problems collection .

gmatcracker2407 · Answer

Hi,
 I am confused between 2 approaches for these kinds of problems
Approach 1: Binomial Theorem. 
Approach 2: Find the unit's digit of the exponent and then find the remainder.

Unit's digit of 333^222 = unit's digit of 3^222. Then divide that by 7. 
Cyclicity of 3 = 4 {3,9,7,1}. 222/3 has a remainder of 2. 3^2 has a unit's digit of 9. 9/7 has a remainder of  2

In Approach 2, i don't always get the same ans as by using Approach 1. Which approach is preferred for these kinds of problems?

Thanks.

DesiGmat · Answer

Hi All,

I used the following approach.

(333^222)/7

(333/7) = Remainder is 4

4^222 can be written as 2^444 which can be written as (2^3)^148

now what we have to do find is

((2^3)^148)/7

we can write the above expression as

((7+1)^148)/7

now apply remainder theorem.

Hence Remainder is 1.

Option E is correct

Shree9975 · Answer

Hi  VeritasPrepKarishma  :
 Bunuel 
Can you please solve my doubt,
From 4^222 , 222 Is basically 55m+2. Since 4 has a cyclicity of { 4, 6} , the unit's digit here will be 6.
When you divide this by 7, the remainder will be 6.
But answer says remainder will be 1.

Can you please help.

chetan2u · Answer

Hi,
the units digit cannot determine the remainder except in the case of 2,5,10 etc...
6 will have remainder 6 but 16 will have 2 and so on..
the right way would be 4^222=(4^3)^74...
now 4^3=64 and the remainder will be 1 when divided by 7..
so ans will be1^74=1
1 is the remainder..
hope it helps

brs1cob · Answer

333 = (3*111) ^222
111/7 = (-1)^222 =1

now only 3^222
(3^2)^111
(9/7)^111 = 2^111 
(2^3)^37

(8/7)^111 =1

ans is 1

KarishmaParmar · Answer

I have no idea why this is a 95 % difficulty level question. Just know that- Each term in the expression (x+y)^n is divisible by x except for the last term which is y^n 333^222 / 7 Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that. (333)^222 = (336-3)^222 Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222 So now essentially our task is to find the remainder when 3^222 is divided by 7 Again the same routine 3^222 = (3^3)^74 = 27^12 or, (28-1)^12 All terms are divisible but the last i.e (-1)^12 12 being a positive power this is equal to 1 Answer E Am I missing something ? :roll:

ENGRTOMBA2018 · Answer

Knowing binomial theorem expansion is a great help. Additionally, for the last part you can use cyclicity to aid you in finding the remainder. Rem of 3^1/7 = 3 Rem of 3^2/7 = 2 Rem of 3^3/7 = 6 Rem of 3^4/7 = 4 Rem of 3^5/7 = 5 Rem of 3^6/7 = 1 ... and repeat thus the cyclicity of 3^n when divided by 7 = 6. 222/6 = 37 (exactly). Thus the remainder will be = 1. Read this cyclicity-on-the-gmat-213019.html for more on remainders and cyclicity. Hope this helps. P.S.: If a question seems "easy" to you need not necessarily be the same for some or in the case of this question, for the majority. 95% difficulty is not a manually inputted value but is calculated on the basis of number of incorrect attempts at this question.

jetmat · Answer

I've read all the previous posts and it seems one step is missing (certainly too obvious for most of you guys) but for the others, check it out (I've eluded the parenthesis to simplify):
333^222=3^222 x 111^222

111^222 / 7 is found by actually dividing 111 by 7;
=>111=77+41=77+42-1
111 =77+35+6=77+35+7-1
=>with the remainder -1 comes -1^222=1, because 222 is even

3^222 =9^111 gives the following remainders when divided by 7:
9=7-2 => 2^111
8=7+1=> 1^37 (111 is a multiple of 3 as the sum of its digits is a multiple of 3, you don't need to find the quotient 37, you only to have 111=3x, to get 2^3=8)
=> Thus the remainder of this part is 1

Finally, the remainder for the whole number is 1x1 =1

Hope it'll help

EgmatQuantExpert · Answer

Given

•  333^{222}

To Find

• The remainder when it is divided by 7.

Approach and Working Out

• N^6, when divided by 7, leaves a remainder of 1. (Fermat’s theorem)
 o When N is coprime to 7.

• 222 is a multiple of 6 and hence we can write it as ( 333^6 ) ^{37} .
 o The net remainder is 1.

Correct Answer: Option E

sidchandan · Answer

Bunuel  - why isn't the below approach correct?

333^222 is divided by 7?

333^222 is the same as 3^222 and thus 3^222 will have units digit 9 (222/4=55 R2, so from within 3-9-7-1, 9 will be the units digit.)
then 9/7 = R2. 2 is the answer.

chetan2u · Answer

The units digit is the remainder when we are dividing something by 10.
So, the remainder will be 9 when we divide by 10.

But the same is not applicable for division by 7.

For example
 63^{222}  will be divisible by 7 even though the units digit is 3.

hadimadi · Answer

Hi,

this is a method I learnt from Miss Eva Jager from her posts on GMAT Club.

It usually makes sense to work with multiples in these kind of questions:

The closest multiple of 7 to 333 is 329:

(329+4)^222=(M7+4)^222

Now, the question is:

What is the remainder when 4^222 is divided by 7?

Here, we can simply check for cyclicity:

4^1 -> R4
4^2 -> R2
4^3 -> R1
4^4 -> R4
4^5 -> R2
...

Cyclicity is at 3, and 3 divides 222 -> (E)

What is the remainder when 333^222 is divided by 7?

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What is the remainder when 333^222 is divided by 7?

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