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When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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Updated on: 21 Nov 2018, 04:44
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Interesting Applications of Remainders – Practice question 2 When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)? To solve question 3: Question 3To read the article: Interesting Applications of Remainders
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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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21 Nov 2018, 04:30
Numbers 2, 3, 7 and 8 have a cyclicity of 4. Numbers 4 and 9 have a cyclicity of 2. Numbers 0, 1, 5 and 6 have a cyclicity of 1.
8 has a cyclicity of 4. When 48 is divided by 4, the remainder is 0. When the remainder is 0, then the cyclicity of the number should be considered as the remainder. Hence \(8^{48} = 8^4\).
8, when divided by 3, gives a negative remainder of 1. then \(1^4 = 1^4\) = 1. \(R_1 = 1\)
Also \(8^4 = 64*64\). The remainder when its divided by 5 is 4*4 = 16/5 = 1. \(R_2 = 1\)
\(R_1+R_2 = 2\).
B is the answer.



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When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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Updated on: 31 Mar 2019, 03:01
Solution Given:When \(8^{48}\) is divided by • 3, the remainder is \(R_1\) • 5, the remainder is \(R_2\) To find:• The value of \(R_1 + R_2\) Approach and Working:• \(R_1\) = \((\frac{{8^{48}}}{3})_R = [\frac{{(91)^{48}}}{3}]_R = [\frac{{(1)^{48}}}{3}]_R = (\frac{1}{3})_R = 1\) • \(R_2\) =\([\frac{{(Units\ digit\ of\ 8^{48)}}}{5}]_R\) Now, units digit cycle of 8 is 8, 4, 2, 6. As \(8^{48}\) can be written as \(8^{4k}\), the units digit is 6. • Hence, \(R_2\) = \((\frac{6}{5})_R\) = 1 • Therefore, \(R_1 + R_2\) = 1 + 1 = 2
Hence, the correct answer is option B. Answer: B
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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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10 Dec 2018, 12:58
EgmatQuantExpert wrote: Interesting Applications of Remainders – Practice question 2 When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)? To solve question 3: Question 3To read the article: Interesting Applications of Remainders hi chetan2u, Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1? I understand cyclicity of 8 is 4. I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ?
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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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10 Dec 2018, 18:19
Hey stne, let me give a try. 9 is completely divisible by 3, giving a remainder of 0. When 8 is divisible by 3, it gives a remainder of 2 or 1. And (1)^48 will be 1. Hence the final remainder. Hope it helps Posted from my mobile device



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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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10 Dec 2018, 18:34
stne wrote: EgmatQuantExpert wrote: Interesting Applications of Remainders – Practice question 2 When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)? To solve question 3: Question 3To read the article: Interesting Applications of Remainders hi chetan2u, Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1? I understand cyclicity of 8 is 4. I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ? Hi... We are looking for remainders when \(8^{48}\) is divided by 3 and 5.. 1) Remainder when \(8^{48}\) is divided by 5 As you said 8 has cylicity of 4 and here the remainder will depend on the units digit.. So units digit are 8,4,2,6... As 48 is 4*12, so units digit will be same as that of 8^4.. Hence units digit will be 6 and remainder will be 1 since 6 divided by 5 gives us 1 as remainder. 2) Remainder when \(8^{48}\) is divided by 5 Here we will use the property of divisibility by 3 : If the sum of digits of number is divisible by 3, the number itself is divisible by 3. So 8 will leave 2 as remainder... 8^2 or 64 should leave 2^2 or 4, which is equal to 1, as remainder.. 64=3*21+1 Next 8^3 will give 2^3 or 8, which is same as 2... So the remainders have a cylicity of 2,1,2,1,2,1.... Odd power will give 2 and even power give 1 as remainder. Here 48 is power and hence even. Therefore, remainder will be 1.. Of course other ways are to convert them in binomial expansion.. 8^(48)=(91)^48... When you expand this, all terms will have 9 except last term.. Expansion : 9^48+9^47*(1)^1+9^46*(1)^2....+9^(1)^47+(1)^48... Only (1)^48 is not divisible by 3, remainder = 0+0+0+0...+0+1=1 But you can always find your answer with out use of binomial expansion as GMAT does not test or rely on binomial expansion. Combined remainder = 1+1=2... B
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When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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Updated on: 11 Dec 2018, 03:12
chetan2u wrote: stne wrote: EgmatQuantExpert wrote: Interesting Applications of Remainders – Practice question 2 When \(8^{48}\) is divided by 3 and 5 respectively, the remainders are \(R_1\) and \(R_2\). What is the value of \(R_1\) + \(R_2\)? To solve question 3: Question 3To read the article: Interesting Applications of Remainders hi chetan2u, Can you explain the first part of this question in a better way, how \(\frac{8^{48}}{3}\) gives a remainder of 1? I understand cyclicity of 8 is 4. I understand \({8^{48}}\) gives a unit digit of 6 , but 6 divided by 3 gives no remainder , so what am I missing here ? Hi... We are looking for remainders when \(8^{48}\) is divided by 3 and 5.. 1) Remainder when \(8^{48}\) is divided by 5 As you said 8 has cylicity of 4 and here the remainder will depend on the units digit.. So units digit are 8,4,2,6... As 48 is 4*12, so units digit will be same as that of 8^4.. Hence units digit will be 6 and remainder will be 1 since 6 divided by 5 gives us 1 as remainder. 2) Remainder when \(8^{48}\) is divided by 5 Here we will use the property of divisibility by 3 : If the sum of digits of number is divisible by 3, the number itself is divisible by 3. So 8 will leave 2 as remainder... 8^2 or 64 should leave 2^2 or 4, which is equal to 1, as remainder.. 64=3*21+1 Next 8^3 will give 2^3 or 8, which is same as 2... So the remainders have a cylicity of 2,1,2,1,2,1.... Odd power will give 2 and even power give 1 as remainder. Here 48 is power and hence even. Therefore, remainder will be 1.. Of course other ways are to convert them in binomial expansion.. 8^(48)=(91)^48... When you expand this, all terms will have 9 except last term.. Expansion : 9^48+9^47*(1)^1+9^46*(1)^2....+9^(1)^47+(1)^48... Only (1)^48 is not divisible by 3, remainder = 0+0+0+0...+0+1=1 But you can always find your answer with out use of binomial expansion as GMAT does not test or rely on binomial expansion. Combined remainder = 1+1=2... B Hi chetan2u, Thank you, but for the second point , remainder when divided by 3 , why can't we use the unit digit concept? When finding the remainder while dividing by 5 we use the unit digit concept but don't use this concept when dividing by three, why? How do we decide when to use unit digit concept and when not to use unit digit concept?
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Originally posted by stne on 11 Dec 2018, 03:00.
Last edited by stne on 11 Dec 2018, 03:12, edited 1 time in total.



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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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11 Dec 2018, 03:11
Hi stne, The multiples of 5 are 5,10,15,20... So all numbers ending with 5 or 0 are multiple of 5 and only those that end with 5 and 0 are multiple of 5. Similarly all ending with 0 are multiple of 10. All ending with even number are a multiple of 2. But what about 3... The multiple of 3 are 3,6,9,12,15,18,21,24,27,30.. So 6 is a multiple but 16 is not , 26 is not. That is why the property of divisibility of 3 is SUM of digits of number should be divisible by 3.. Say 23415 : 2+3+4+1+5=15 1+5=6 and 6 is divisible by 3, so 23415 is divisible by 3.. But say 23: 2+3=5, 5 not divisible by 3, so 23 is not divisible by 3
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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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11 Dec 2018, 03:21
chetan2u wrote: Hi stne, The multiples of 5 are 5,10,15,20... So all numbers ending with 5 or 0 are multiple of 5 and only those that end with 5 and 0 are multiple of 5. Similarly all ending with 0 are multiple of 10. All ending with even number are a multiple of 2. But what about 3... The multiple of 3 are 3,6,9,12,15,18,21,24,27,30.. So 6 is a multiple but 16 is not , 26 is not. That is why the property of divisibility of 3 is SUM of digits of number should be divisible by 3.. Say 23415 : 2+3+4+1+5=15 1+5=6 and 6 is divisible by 3, so 23415 is divisible by 3.. But say 23: 2+3=5, 5 not divisible by 3, so 23 is not divisible by 3 Hi chetan2u, Thank you, will need to brush up my remainder concept .BTW there is a small typo in point number 2 in the link below: https://gmatclub.com/forum/when848is ... l#p2188053You have 5 instead of 3, you may want to correct it, so that others are not confused. Thank you.
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Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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16 Dec 2018, 08:46
Approach for people not familiar with binomial and/or negative remainders:
8^1 = 8 8^2 = 64 8^3 = 512 8^4 = 4,096 8^5 = 32,768
(you should have 8^3 memorised imo).
now, take each power and divide by 3 to determine cyclicity of remainders (then do the same with 5).
8^1 / 3 = r2 8^2 / 3 = r1 8^3 / 3 = r2
(at this point I stopped, but we could test 8^4 for completeness sake).
8^4 / 3 = r1
Pattern established. Therefore 8^48 yields r1.
Do the same for 5.
8^1 / 5 = r3 8^2 / 5 = r4 8^3 / 5 = r2 8^4 / 5 = r1
(pattern probably ends here  you should know 8 has cyclicity of 4. Nonetheless, if your arithmetic is good, then find 8^5 and divide by 5 to confirm).
8^5 / 5 = r3
Therefore, 8^48 yields remainder 1 when divided by 5.
Hence, R1 = 1, R2 = 1, and R1+R2 = 2
B




Re: When 8^48 is divided by 3 and 5 respectively, the remainders are R1
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