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What is the remainder when 333^222 is divided by 7?
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Updated on: 21 Jul 2013, 03:30
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What is the remainder when 333^222 is divided by 7? A. 3 B. 2 C. 5 D. 7 E. 1
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Originally posted by jonyg on 21 Jul 2013, 02:16.
Last edited by Bunuel on 21 Jul 2013, 03:30, edited 2 times in total.
Renamed the topic, edited the question and the tags.




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What is the remainder when 333^222 is divided by 7?
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21 Jul 2013, 03:27
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: {241}. 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. Similar question to practice: http://gmatclub.com/forum/whatisther ... 34778.htmlhttp://gmatclub.com/forum/when5125is ... 30220.htmlhttp://gmatclub.com/forum/whatisther ... 26493.htmlhttp://gmatclub.com/forum/whatisther ... 00316.htmlhttp://gmatclub.com/forum/whatisther ... 99724.htmlTheory on remainders problems: http://gmatclub.com/forum/remainders144665.htmlDS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=198PS remainders problems to practice: http://gmatclub.com/forum/search.php?se ... tag_id=199Hope it helps.
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Re: What is the remainder when 333^222 is divided by 7?
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28 Jul 2013, 04:05
(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)



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Re: What is the remainder when 333^222 is divided by 7?
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31 Jul 2013, 04:27
333^222/7 3^222 *(111)^222 /7 =>(111)^222/7=> in terms of remainder (6)^222/7 or (1)^222/7 which leaves 1 now the other part (3^2)^111/7 => (2)^111/7 =>(8)^27/7=>1^27 and this part is also one .
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Re: what is the reminder when 333^222 is divided by 7?
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01 Aug 2013, 13:37
Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1
official answer=>e
source random internet 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: {241}. 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?



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Re: what is the reminder when 333^222 is divided by 7?
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02 Aug 2013, 04:28
iNumbv wrote: Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1
official answer=>e
source random internet Hey can you explain to me how you get a remainder of 2 when you divide 2^1/7? This might help : remainders144665.htmlIf 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 \(0\leq{r}<x.\) For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since 15 = 6*2 + 3. Notice that \(0\leq{r}<x\) means that remainder is a nonnegative integer and always less than divisor. As for your query, we can write \(2 = 0*7+2\), where 7 is the divisor, and 2 is the remainder. Hope this helps.
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Re: What is the remainder when 333^222 is divided by 7?
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02 Aug 2013, 05:18
Hi, My solution is as follows when 333/7 gives reminder 4 thus we have to find out 4^222 now 7 is a prime no so according to fermants littile therom (4^6)/7=1 now we have to see if 222 is divisble by 6 thus 222=6*37 hence 4^6k/7 =1 hence answer is 1 ie e



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Re: What is the remainder when 333^222 is divided by 7?
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30 Aug 2013, 11:58
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



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Re: What is the remainder when 333^222 is divided by 7?
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09 Mar 2014, 13:14
Bumping for review and further discussion.For more on this kind of questions check Units digits, exponents, remainders problems collection.
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Re: What is the remainder when 333^222 is divided by 7?
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29 Apr 2014, 14:36
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.



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Re: What is the remainder when 333^222 is divided by 7?
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29 Apr 2014, 20:39
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



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Re: What is the remainder when 333^222 is divided by 7?
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30 Apr 2014, 07:44
gmatcracker2407 wrote: 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. 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, ...
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Re: What is the remainder when 333^222 is divided by 7?
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25 Jul 2015, 05:43
Hi VeritasPrepKarishma : BunuelCan 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.



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Re: What is the remainder when 333^222 is divided by 7?
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25 Jul 2015, 06:01
Shree9975 wrote: Hi VeritasPrepKarishma : BunuelCan 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. 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
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Re: What is the remainder when 333^222 is divided by 7?
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25 Sep 2015, 01:51
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



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Re: What is the remainder when 333^222 is divided by 7?
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15 Feb 2016, 10:17
Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1
official answer=>e
source random internet 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: {241}. 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. Hope it helps. 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^n333^222 / 7 Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that. (333)^222 = (3363)^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, (281)^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 ?
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What is the remainder when 333^222 is divided by 7?
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15 Feb 2016, 10:30
KarishmaParmar wrote: Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? a.3 b.2 c.5 d.7 e.1
official answer=>e
source random internet 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: {241}. 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. Hope it helps. 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^n333^222 / 7 Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that. (333)^222 = (3363)^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, (281)^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 ? 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 cyclicityonthegmat213019.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.



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Re: What is the remainder when 333^222 is divided by 7?
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02 Dec 2016, 03:32
jonyg wrote: What is the remainder when 333^222 is divided by 7?
A. 3 B. 2 C. 5 D. 7 E. 1 \(333 = 4 (mod_7)\) \(\frac{4^{222}}{7} = \frac{2^{444}}{7}\) \(2^3 = 1 (mod_7)\) \(\frac{(4^3)^{148}}{7} = \frac{1^{148}}{7} = \frac{1}{7}\) Remainder is \(1\)



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Re: What is the remainder when 333^222 is divided by 7?
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09 Feb 2018, 02:15
Hi Bunuel, why is the remainder not 2? With the questions I've been going through, the method was always the same. When a question asks for the remainder when dividing a number^integer, it is basically asking for the unit digit of the number. And so I did the unit digit of 3 ^222 which would be 9. but then 9/7 gives a remainder of 2. (Example of a similar question: 2^243 divided by 5, what is remainder? then the remainder is 2 because of cyclisity it ends up being 7, which if 7/5 gives a remainder of 2. Or 43^86 divided by 5, the remainder is 4 because 9/5 gives a remainder of 5) Could you please explain why this method didn't work for this example? Thanks in advance!



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Re: What is the remainder when 333^222 is divided by 7?
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09 Feb 2018, 02:36
lisa.cornelia wrote: Hi Bunuel, why is the remainder not 2? With the questions I've been going through, the method was always the same. When a question asks for the remainder when dividing a number^integer, it is basically asking for the unit digit of the number. And so I did the unit digit of 3 ^222 which would be 9. but then 9/7 gives a remainder of 2. (Example of a similar question: 2^243 divided by 5, what is remainder? then the remainder is 2 because of cyclisity it ends up being 7, which if 7/5 gives a remainder of 2. Or 43^86 divided by 5, the remainder is 4 because 9/5 gives a remainder of 5) Could you please explain why this method didn't work for this example? Thanks in advance! Have you checked this post: https://gmatclub.com/forum/whatisthe ... l#p1248585 ?
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