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What is the remainder when 333^222 is divided by 7? [#permalink]
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Question Stats: Hide Show timer StatisticsWhat is the remainder when 333^222 is divided by 7?A. 3B. 2C. 5D. 7 E. 1
Originally posted by jonyg on 21 Jul 2013, 01:16. Renamed the topic, edited the question and the tags.
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What is the remainder when 333^222 is divided by 7? [#permalink]
What is the remainder when 333^222 is divided by 7?A. 3B. 2C. 5D. 7E. 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.
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Re: What is the remainder when 333^222 is divided by 7? [#permalink]
Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? official answer=>e source- random internet What is the remainder when 333^222 is divided by 7? A. 3B. 2C. 5D. 7E. 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.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^n 333^222 / 7Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that.(333)^222 = (336-3)^222Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222So now essentially our task is to find the remainder when 3^222 is divided by 7Again the same routine3^222 = (3^3)^74 = 27^12or, (28-1)^12All terms are divisible but the last i.e (-1)^1212 being a positive power this is equal to 1Answer EAm I missing something ?
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] (333)^222Rem(333/7) = 4=> (4) ^222=> (16) ^111 Rem(16/7) = 2(2)^111 = 2^100 * 2^11Now let us observe the Rem(2^10)/7 2^10 = 1024 => Rem(1024/7) = 2REM(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 reminder when 333^222 is divided by 7? [#permalink]
Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? official answer=>e source- random internet What is the remainder when 333^222 is divided by 7? A. 3B. 2C. 5D. 7E. 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?
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Re: what is the reminder when 333^222 is divided by 7? [#permalink]
iNumbv wrote: Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? 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 : 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 \(0\leq{r}<x.\)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 non-negative 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? [#permalink] A simple one line solution to this problem can be this:
Math Expert Joined: 02 Sep 2009 Posts: 87080
Re: What is the remainder when 333^222 is divided by 7? [#permalink] Hi, I am confused between 2 approaches for these kinds of problemsApproach 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? [#permalink] Hi All,I used the following approach.(333^222)/7(333/7) = Remainder is 44^222 can be written as 2^444 which can be written as (2^3)^148now what we have to do find is((2^3)^148)/7we can write the above expression as((7+1)^148)/7now apply remainder theorem.Hence Remainder is 1. Option E is correct
Math Expert Joined: 02 Sep 2009 Posts: 87080
Re: What is the remainder when 333^222 is divided by 7? [#permalink]
gmatcracker2407 wrote: Hi, I am confused between 2 approaches for these kinds of problemsApproach 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? [#permalink] Hi VeritasPrepKarishma : Can you please help.
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Re: What is the remainder when 333^222 is divided by 7? [#permalink]
Shree9975 wrote: Hi VeritasPrepKarishma : 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=11 is the remainder..hope it helps _________________
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] 333 = (3*111) ^222111/7 = (-1)^222 =1now 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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What is the remainder when 333^222 is divided by 7? [#permalink]
KarishmaParmar wrote: Bunuel wrote: jonyg wrote: what is the reminder when 333^222 is divided by 7? official answer=>e source- random internet What is the remainder when 333^222 is divided by 7? A. 3B. 2C. 5D. 7E. 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.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^n 333^222 / 7Try and bring the dividend to a form ( multiple of divisor +_ something). And usually such Qs are formed like that.(333)^222 = (336-3)^222Each term in the above expression is divisible by 7 ( since 336 is divisible by 7) but the last term which is 3^222So now essentially our task is to find the remainder when 3^222 is divided by 7Again the same routine3^222 = (3^3)^74 = 27^12or, (28-1)^12All terms are divisible but the last i.e (-1)^1212 being a positive power this is equal to 1Answer EAm I missing something ? 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. _________________
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What is the remainder when 333^222 is divided by 7? [#permalink]
jonyg wrote: What is the remainder when 333^222 is divided by 7?A. 3B. 2C. 5D. 7 E. 1 The Euler's of 7 = 7 ( 1-1/7) = 6333 mod 7 = 4222 mod 6 = 04^0 = 1333^222 mod 7 = 1EPS - These type of questions are very popular in the Indian Management Exam for the IIM's i.e CAT and this is the simplest way to solve such sums _________________
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Re: What is the remainder when 333^222 is divided by 7? [#permalink]
jonyg wrote: What is the remainder when 333^222 is divided by 7?A. 3B. 2C. 5D. 7 E. 1 Another approach for similar kinds of sums is using Fermet's theorem. Refer attached photo
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What is the remainder when 333^222 is divided by 7? [#permalink] 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 1Finally, the remainder for the whole number is 1x1 =1 Hope it'll help
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Re: What is the remainder when 333^222 is divided by 7? [#permalink] Given To Find
Approach and Working Out
Correct Answer: Option E _________________
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What is the remainder when 333^222 is divided by 7? [#permalink] (333)^222 /7 ----> Rem of = ?343 = 7 * (49) ----- thus 343 is a Multiple of 7Re-write the Base of the Exponential Term:(333)^222 = (343 - 10)^222 Now Divide this Re-written Expression by 7:(343 - 10)^222 /7 -----> Rem of = ?Using the Binomial Theorem and Formula:(343 - 10)^222 = (343^222) * (-10^0) + (343^221) * (-10^1) + (343^220) * (-10^2) ......... (343^0) * (-10)^222Every Term when we Multiply and Expand out the Binomial (343 - 10) will be Divisible by 7 EXCEPT the LAST TERM = (-10)^222We can continue to Divide (-10)^222 by the Divisor of 7 by SPLITTING the Dividend and Multiplying the "PART" Remainders:(-10)^222 /7 -------> Rof = (-10 /7)Rem of *(-10 / 7)Rem of*(-10 / 7)Rem of.............Remainders continually Multiplied 222 TimesRemove the (-)Negative Remainder in Each Part:-10 + 7 = -3-3 + 7 = +4thus, a (-)Neg. Rem of -10 --------> corresponds to a (+)Pos. Rem of +4 when the Divisor is 7Excess Remainder = (+4) Multiplied 222 Times or ----> (4)^222Again, Divide by the Divisor of 7 to Remove the Excess Remainder:(4)^222 /7 ------> Rof = ?SPOT: (4)^3 = 64 -----> which is +1 More than a Multiple of 7so we Can Re-write the Excess Remainder as:(4^3)^74 = (64)^74 = (63 + 1)^74and we can again use the Binomial Formula ----- only the LAST TERM = (+1)^74 will NOT be Divisible by 7(1)^74 /7 ----> = (1) / 7 ------> yields a Remainder of +1-E- Remainder = +1
What is the remainder when 333^222 is divided by 7? [#permalink] 25 Oct 2020, 19:43 |
How do you find the remainder when dividing by 7?
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