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If x and y are positive integers, what is the remainder when [#permalink]
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If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1.
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banksy wrote: 218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4). Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1}  {9, 1}  ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0. (1) x = 25. Not sufficient. (2) y = 1 > \(y=odd\) > remainder=0. Sufficient. Answer: B. Similar question: divisiblitywith109075.htmlCheck Number Theory chapter of Math Book for more: mathnumbertheory88376.html
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Re: 218. If x and y are positive integers, what is the remainder [#permalink]
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19 Feb 2011, 01:27
HI Bunuel What does this mean ? remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4) Regards, Subhash
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Re: 218. If x and y are positive integers, what is the remainder [#permalink]
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subhashghosh wrote: HI Bunuel
What does this mean ?
remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4)
Regards, Subhash Theory: First of all note that the last digit of xyz^(positive integer) is the same as that of z^(positive integer); • Integers ending with 0, 1, 5 or 6, in the positive integer power, have the same last digit as the base (cyclicity of 1): xyz0^(positive integer) ends with 0; xyz1^(positive integer) ends with 1; xyz5^(positive integer) ends with 5; xyz6^(positive integer) ends with 6; • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4; For example last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... So cyclicity of the last digit of 3 in power is 4. • Integers ending with 4 and 9 have a cyclicity of 2. Now, last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... means that: 3^1, 3^5, 3^9, 3^13, ..., 3^(4x+1), all will have the same last digit as 3^1 so 1; 3^2, 3^6, 3^10, 3^14, ..., 3^(4x+2), all will have the same last digit as 3^2 so 9; 3^3, 3^7, 3^11, 3^15, ..., 3^(4x+3), all will have the same last digit as 3^3 so 7; 3^4, 3^8, 3^12, 3^16, ..., 3^(4x), all will have the same last digit as 3^4 so 1; So to get the last digit of 3^x, (where x is a positive integer) you should divide x by 4 (cylcility) and look at the remainder: If remainder is 1 then the last digit will be the same as for 3^1 (so the first digit from the pattern {3, 9, 7, 1}); If remainder is 2 then the last digit will be the same as for 3^2 (so the second digit from the pattern {3, 9, 7, 1}); If remainder is 3 then the last digit will be the same as for 3^3 (so the third digit from the pattern {3, 9, 7, 1}); If remainder is 0 then the last digit will be the same as for 3^4 (so the fourth digit from the pattern {3, 9, 7, 1}); Next, as 4+4x (the power of 3^(4+4x)) is clearly divisible by 4 (remainder 0) then the last digit of 3^(4+4x) is the same as the last digit of 3^4 so 1. You can apply this to integers ending with other digits as well (with necessary modification of pattern). Hope it's clear. P.S. Check this for more: mathnumbertheory88376.html
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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04 Feb 2014, 06:49
banksy wrote: If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?
(1) x = 25. (2) y = 1. Its a value question and hence we need a definate value. 3^(4+4x) will be 3^8, 3^12, 3^16....when x = 1, 2, 3..... So we can safely conclude that 3^(4 + 4x) will always have remainder of 1(based on cyclicity) . So value of X does not matter. Hence 1 is not sufficient  or does not help us to respond uniquely. Second statement tells us that Y=1, which is exactly what need to reply. We dont have to go into the details of calculations to find out the remainder. Answer would be B
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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20 Oct 2014, 02:31
Bunuel wrote: banksy wrote: 218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4). Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1}  {9, 1}  ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0. (1) x = 25. Not sufficient. (2) y = 1 > \(y=odd\) > remainder=0. Sufficient. Answer: B. Similar question: divisiblitywith109075.htmlCheck Number Theory chapter of Math Book for more: mathnumbertheory88376.htmlHi Bunuel, I didn't get the second point, y=odd. If we are provided the value of, irrespective of the fact that whether the value is even or odd, we can calculate the remainder, since we know the cyclicity. Please correct me if I am wrong.
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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20 Oct 2014, 04:22
Thoughtosphere wrote: Bunuel wrote: banksy wrote: 218. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10? (1) x = 25. (2) y = 1. If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?Last digit of \(3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... So cyclicity of the last digit of 3 in power is 4. Now, \(3^{4+4x}=3^{4(1+x)}\) will have the same last digit as \(3^4\) which is 1 (remainder upon division the power 4+4x by cyclicity 4 is 0, which means that \(3^{(4+4x)}\) will have the same last digit as 3^4). Now \(3^{(4 + 4x)}+9^y\) will be divisible by 10 if \(y=odd\), in this case the last digit of \(9^{odd}\) will be 9 (9 has a cyclicity of 2: {9, 1}  {9, 1}  ...) so the last digit of \(3^{(4 + 4x)}+9^y\) will be 1+9=0. (1) x = 25. Not sufficient. (2) y = 1 > \(y=odd\) > remainder=0. Sufficient. Answer: B. Similar question: divisiblitywith109075.htmlCheck Number Theory chapter of Math Book for more: mathnumbertheory88376.htmlHi Bunuel, I didn't get the second point, y=odd. If we are provided the value of, irrespective of the fact that whether the value is even or odd, we can calculate the remainder, since we know the cyclicity.Please correct me if I am wrong. Sorry, but I don;t understand what you mean there... Anyway, y=odd, means that the units digit of 9^y is 9. Since we established that the units digit of 3^(4 + 4x) is 1, irrespective of x, then the units digit of the sum is 1+9=0, which ensures the divisibility by 10.
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If x and y are positive integers, what is the remainder when [#permalink]
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20 Oct 2014, 06:04
Dear Bunuel, What I mean to say is that, irrespective of whether y is even or odd, if the value of Y is given, We'll be able to find the remainder. If it is even, the remainder would be 1 + 1 = 2 and if its odd, it would be 1 + 9 = 0.
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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20 Oct 2014, 06:13
banksy wrote: If x and y are positive integers, what is the remainder when 3^(4 + 4x) + 9^y is divided by 10?
(1) x = 25. (2) y = 1. Here's an alternate Solution \(3^(4 + 4x) + 9^y = 9^(2x+2) + 9^y\) \({9^(2x+2) + 9^y} / 10\) \(= (1)^even + (1)^y\) The above equation implies that, the value of x is of no use since we'll get remainder 1 from that part of the equation. The second part of the equation which involves y, will be of value to us. 1  Insufficient 2  y = 1. The remainder will be 1 from here, so the final remainder will be \(1  1 = 0\)
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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20 Oct 2014, 06:25
Thoughtosphere wrote: Dear Bunuel, What I mean to say is that, irrespective of whether y is even or odd, if the value of Y is given, We'll be able to find the remainder. If it is even, the remainder would be 1 + 1 = 2 and if its odd, it would be 1 + 9 = 0. Yes, knowing whether y is odd or even is sufficient to answer the question but still not getting what are you trying to say.
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Re: If x and y are positive integers, what is the remainder when [#permalink]
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Bunuel wrote: subhashghosh wrote: HI Bunuel
What does this mean ?
remainder upon division the power 4+4x by cyclicity 4 is 0, which means that 3^{(4+4x)} will have the same last digit as 3^4)
Regards, Subhash Theory: First of all note that the last digit of xyz^(positive integer) is the same as that of z^(positive integer); • Integers ending with 0, 1, 5 or 6, in the positive integer power, have the same last digit as the base (cyclicity of 1): xyz0^(positive integer) ends with 0; xyz1^(positive integer) ends with 1; xyz5^(positive integer) ends with 5; xyz6^(positive integer) ends with 6; • Integers ending with 2, 3, 7 and 8 have a cyclicity of 4; For example last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... So cyclicity of the last digit of 3 in power is 4. • Integers ending with 4 and 9 have a cyclicity of 2. Now, last digit of \(xyz3^{positive \ integer}\) repeats in blocks of 4: {3, 9, 7, 1}  {3, 9, 7, 1}  ... means that: 3^1, 3^5, 3^9, 3^13, ..., 3^(4x+1), all will have the same last digit as 3^1 so 1;
3^2, 3^6, 3^10, 3^14, ..., 3^(4x+2), all will have the same last digit as 3^2 so 9; 3^3, 3^7, 3^11, 3^15, ..., 3^(4x+3), all will have the same last digit as 3^3 so 7; 3^4, 3^8, 3^12, 3^16, ..., 3^(4x), all will have the same last digit as 3^4 so 1; So to get the last digit of 3^x, (where x is a positive integer) you should divide x by 4 (cylcility) and look at the remainder: If remainder is 1 then the last digit will be the same as for 3^1 (so the first digit from the pattern {3, 9, 7, 1}); If remainder is 2 then the last digit will be the same as for 3^2 (so the second digit from the pattern {3, 9, 7, 1}); If remainder is 3 then the last digit will be the same as for 3^3 (so the third digit from the pattern {3, 9, 7, 1}); If remainder is 0 then the last digit will be the same as for 3^4 (so the fourth digit from the pattern {3, 9, 7, 1}); Next, as 4+4x (the power of 3^(4+4x)) is clearly divisible by 4 (remainder 0) then the last digit of 3^(4+4x) is the same as the last digit of 3^4 so 1. You can apply this to integers ending with other digits as well (with necessary modification of pattern). Hope it's clear. P.S. Check this for more: http://gmatclub.com/forum/mathnumbertheory88376.htmlHi, I think there is typo error. Unit digit should be 3 instead of 1 for 3^1 for highlighted portion.



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Re: If x and y are positive integers, what is the remainder when [#permalink]
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22 Jun 2017, 11:12
Imo B This is tricky question, we need only statement 2 to know whether it is divisible by 10 as power of 3^(4 + 4x) is always going to be multiple of 4 .So we will get 1 as the units digit according to the cyclicity of number 3. If we know the value of y we can give answer whether it is divisible by 10




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