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If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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23 Jun 2010, 12:35
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If n is a positive integer, what is the remainder when \(3^{8n+3}+2\) is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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19 Nov 2010, 09:49
azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A 0 B 1 C 2 D 3 E 4 It should be \(3^{8n+3}+2\). The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 > the units digit is 3; 3^2 > the units digit is 9; 3^3 > the units digit is 7; 3^4 > the units digit is 1; 3^5 > the units digit is 3 AGAIN; ... So, the units digit repeats the following pattern {3971}{3971}.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4. Answer: E. Check Number theory chapter of Math Book for more: mathnumbertheory88376.htmlHope it helps.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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06 Jun 2013, 15:00
you can split it up to make it look simpler. if x = 1 (3^8)(3^3) + 2 3^8 = 9^4 = 9 x 9 x 9 x 9 = 81 x 81 (units digit is 1) 3^3 = 27 (units digit is 7) 1 x 7 + 2 = 9 9/5 = 1 r4 Answer is E




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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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23 Jun 2010, 14:40
udaymathapati wrote: Some difficult PS Questions from SET1. Please help me with resolving them...OAs have mentioned at the end.
Q13: If n is a positive integer, what is the remainder when \(3^{8n+3}+2\) is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 3 in power has cyclicity of 4: 1. 3^1=3 (last digit is 3) 2. 3^2=9 (last digit is 9) 3. 3^3=27 (last digit is 7) 4. 3^4=81 (last digit is 1)5. 3^5=243 (last digit is 3 again!) ... To find the last digit of \(3^{8n+3}\), divide the power (which is \(8n+3\)) by cyclicity # ( which is \(4\)) and look at the remainder > \(\frac{8n+3}{4}\) > \(remainder=3\), which means that the last digit of \(3^{8n+3}\) will be the same as the last digit of \(3^3=27\) (last digit is 7). (Side note: If the remainder were 0, then last digit would be the same as the las digit of \(3^4\)). Now, last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any integer with last digit 9 upon division by 5 yields remainder of 4. Answer: E. Hope it's clear.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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23 Jun 2010, 21:35
Bunuel wrote: udaymathapati wrote: Some difficult PS Questions from SET1. Please help me with resolving them...OAs have mentioned at the end.
Q13: If n is a positive integer, what is the remainder when \(3^{8n+3}+2\) is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 3 in power has cyclicity of 4: 1. 3^1=3 (last digit is 3) 2. 3^2=9 (last digit is 9) 3. 3^3=27 (last digit is 7) 4. 3^4=81 (last digit is 1)5. 3^5=243 (last digit is 3 again!) ... To find the last digit of \(3^{8n+3}\), divide the power (which is \(8n+3\)) by cyclicity # ( which is \(4\)) and look at the remainder > \(\frac{8n+3}{4}\) > \(remainder=3\), which means that the last digit of \(3^{8n+3}\) will be the same as the last digit of \(3^3=27\) (last digit is 7). (Side note: If the remainder were 0, then last digit would be the same as the las digit of \(3^4\)). Now, last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any integer with last digit 9 upon division by 5 yields remainder of 4. Answer: E. Hope it's clear. Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1. shouldn't it be 3^0=1 instead?



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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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23 Jun 2010, 22:01
udaymathapati wrote: Bunuel, : If the remainder were 0, then last digit would be the same as the las digit of 3^4=81 means 1. shouldn't it be 3^0=1 instead?
No, as in this case all numbers in power which is multiple of cyclicity would have the last digit of 1. When power is divisible by cyclicity # (remainder 0), then the last digit is the same as the last digit of number in the power of cyclicity #. For example: Cyclicity of 2 is 4, so \(2^{4n}\) (where \(n\) is a positive integer) will have the same last digit as \(2^4\), which is 6. OR: Cyclicity of 4 is 2, so \(4^{2n}\) (where \(n\) is a positive integer) will have the same last digit as \(4^2\), which is 6. Hope it's clear.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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20 Nov 2010, 03:17
put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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20 Nov 2010, 13:20
anandthiru wrote: put n=0, then we have (3^3)+2=29. now 29/5, we have r=4. is my short cut ok ???
counter arguments welcome instead of kudos !!! n is given to be a positive integer so it cannot be 0. That said, whatever will be the last digit of \(3^{(8n+3)} + 2\) for n = positive integer, will be the last digit when n = 0. (We know cyclicity works for n = 0 too) Hence your short cut works.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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22 Nov 2010, 14:38
Answer should be E i.e. 4 remember 3 has a cyclicity of 4 =>3,9,*7,*1 3^(8n+3) +2 => for n=1 => 8*1+3 =>11 =>3(on dividing 11/4) for n=2 =>19=>3(on dividing 19/4) for n=3 =>27=>3(on dividing 27/4) for n=4 =>35=>3(on dividing 35/4) for n=5 =>43=>3(on dividing 43/4) for n=6 =>51=>3(on dividing 51/4) therefore 3^3 gives 7 in units place => 7+2 => 9 in all the cases hence on dividing 9/5 remainder is 4
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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02 Apr 2011, 07:30
gmat1220 wrote: If n is a positive integer, what is the remainder when [3^(8n+3) + 2] is divided by 5? A. 0 B. 1 C. 2 D. 3 E. 4 3^8 +2 = x1 + 2 = x3. In all the posts I have read, the reasoning is the same  the remainder seems to be as simple as dividing the "Unit's digit" than dividing the "whole number" by divisor. I feel the conclusion is unwarranted. Can someone illustrate this? The answer is not my concern  the logic is. 5 is a special case because every number with a unit's digit of 5 or 0 is divisible by 5. So whenever you have a number that ends in 9, the number that is 4 less than it which ends in 5 will be a multiple of 5. Say you get 33759, then definitely 33755 will be a multiple of 5 so your remainder will be 4. Similarly a number ending with 7 will always give 2 as remainder when divided by 5; a number ending in 3 will give 3 as remainder when divided by 5 and so on. You can say the same thing only about one more number i.e. 2. Every time the unit's digits is 1/3/5/7/9 remainder will be 1. Every time unit's digits is 0/2/4/6/8 the remainder will be 0. If you really want to get to the concept, think about this: Is it just coincidence that unit's digits have a cyclicity of 10 (i.e. after 10 numbers you get the same unit's digits) and you can find the remainder using just the unit's digit for only 2 and 5, the factors of 10? Note: Getting into concepts and linking one with the other to get the full picture is what leads to a 5051 in Quant.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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06 Jun 2013, 15:20
Bunuel wrote: azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A 0 B 1 C 2 D 3 E 4 It should be \(3^{8n+3}+2\). The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 > the units digit is 3; 3^2 > the units digit is 9; 3^3 > the units digit is 7; 3^4 > the units digit is 1; 3^5 > the units digit is 3 AGAIN; ... So, the units digit repeats the following pattern {3971}{3971}.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4. Answer: E. Check Number theory chapter of Math Book for more: mathnumbertheory88376.htmlHope it helps. I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question



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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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06 Jun 2013, 16:10
pavan2185 wrote: Bunuel wrote: azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A 0 B 1 C 2 D 3 E 4 It should be \(3^{8n+3}+2\). The units digit of 3 in positive integer power has cyclicity of 4 for the unis digit: 3^1 > the units digit is 3; 3^2 > the units digit is 9; 3^3 > the units digit is 7; 3^4 > the units digit is 1; 3^5 > the units digit is 3 AGAIN; ... So, the units digit repeats the following pattern {3971}{3971}.... \(3^{8n+3}\) will have the same units digit as \(3^3\), which is 7 (remainder when 8n+3 divided by cyclicity 4 is 3). Thus the last digit of \(3^{8n+3}+2\) will be \(7+2=9\). Any positive integer with the unis digit of 9 divided by 5 gives the remainder of 4. Answer: E. Check Number theory chapter of Math Book for more: mathnumbertheory88376.htmlHope it helps. I just came to testing the cycle and felt a dead end. I am not getting why are we zeroing in on 3^3 for the first part of the addition in the question The reason we are focusing on \(3^3\) is because the repeating cycle of \(3^{8n}\) only matters because it is a sequence that loops in a pattern of 4 (which is consequently a divisor of 8). You said that you tested the cycle. So you should understand that, as Bunuel said, 39713971 is the cycle. No matter what multiple of 8 you use (8, 16, 24, 32, etc.), 1 will always be the units digit of the final value. If that is the case we can actually forget the 8n part of the equation, because you know that \(3^{8n}\) will always have a units digit of 1. This means that you have to focus on the \(3^3\) instead. You can look at this is two ways. Either go three more steps in the cycle and land on 7 (397), or because \(3^{8n+3} = 3^{8n}*3^3\) = (number with the units digit of 1) x 27. The units digit will be 7 because 1 x 7 is 7. 7+2 = 9, 9/5 = 1 r4. Answer is E. You can also look at my post above for a method that doesn't involve all this sequencing stuff On another note, be careful not to make this the sequencing a rule for all questions that look like this because it definitely is not a rule. If you see \(3^{7m+3}\), you can't just "cancel out" the 7m. If m=1 the cycle is 3971397, but if m=2 it continues the sequence and does not restart back at 3 but instead starts where it left off after 7 (1397139). The reason we focus on the \(3^3\) in this equation is only because the final value of the "8n sequence" always ends with the same units digit. Kudos if that helps.



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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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07 Jun 2013, 02:11
azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A. 0 B. 1 C. 2 D. 3 E. 4 We know that Remainder(\(\frac{a*b}{c}\)) = Remainder(\(\frac{a}{c}\))*Remainder(\(\frac{b}{c}\)) for integral values of a,b and c. The given expression : \(\frac{(3^{8n}*3^3 + 2)}{5}\) = \(\frac{27*(3^4)^{2n}}{5}\)+\(\frac{2}{5}\) = Remainder of \(\frac{(27)*(81)^{2n}}{5}\) + Remainder of \(\frac{2}{5}\) = Remainder of\(\frac{(2)*(1)^{2n}}{5}\)+Remainder of \(\frac{2}{5}\)= Remainder of \(\frac{4}{5}\) = 4.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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07 Jun 2014, 02:35
eresh wrote: If n is a positive integer, what is the remainder when\(3^{8n+3}\)+ 2 is divided by 5?
A. 0 B. 1 C. 2 D. 3 E. 4 Let us plug in: n = 1; 3^(11) + 2 Use cylicity: Units digit: 3^1 = 3 3^2 = 9 3^3 = 7 3^4 = 1 and so on. 3^11 will have 7 at the end hence the remainder will be 7 + 2 = 9/5 = +4
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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28 Jul 2014, 21:54
suhaschan wrote: If n is a positive integer, what is the remainder when 3^"8n+3" + 2 is divided by 5?
A. 0 B. 1 C. 2 D. 3 E. 4 You should know the concept of cyclicity. Powers of 3 have a cyclicity of 4 i.e. 3^1 = 33^2 = 93^3 = 2 73^4 = 8 13^5 = 24 33^6 = 72 9Look at the last digits of powers: 3, 9, 7, 1, 3, 9, 7, 1, 3, 9... So whenever we have a power of 3, we remove groups of 4 out of it i.e. divide it by 4 and whatever is the remainder tells us the last digit. So say we have 3^13, we divide 13 by 4 and are left with 1 remainder. So this is the first term of the next cycle and will end with 3. Say we have 3^39, we divide 39 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7. If you have \(3^{8n+3}\), we divide 8n+3 by 4 and are left with 3 remainder. So this is the third term of the next cycle and will end with 7. Last term of \(3^{8n+3}+2\) will be 7+2 = 9. When you divide a number that ends with 9 by 5, you get 4 remainder. If you are not sure why this is so, check out the first part of the question discussed in this post: http://www.veritasprep.com/blog/2014/03 ... emainders/Answer (E) Or since you know that cyclicity of 3 is 4, just put n = 0 to get 3^3 + 2 = 29. When you divide 29 by 5, you get remainder 4.
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Re: If n is a positive integer, what is the remainder when 3^(8n + 3) + 2
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09 Aug 2017, 12:51
azule45 wrote: If n is a positive integer, what is the remainder when 3^(8n+3) + 2 is divided by 5?
A. 0 B. 1 C. 2 D. 3 E. 4 We need to determine the remainder of: 3^8n x 3^3 + 2 when it is divided by 5. We only need to know the units digit of the above expression to determine the remainder. Let’s look at the pattern of units digits of powers of 3. Note that we are only concerned with the units digits, so, for example, for 3^3, we concern ourselves only with the units digit of 27, which is 7. Here is the pattern: 3^1 = 3 3^2 = 9 3^3 = 7 3^4 = 1 3^5 = 3 The repeating pattern is 3971. Since the units digit pattern for a base of 3 is 3971, we see that whenever 3 is raised to an exponent that is a multiple of 4, the units digit will be 1. Thus: 3^8n x 3^3 = units digit of 1 x units digit of 7 = units digit of 7 So, units digit of 7 + 2 = units digit of 9, and thus 9/5 has a remainder of 4. Answer: E
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