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Re: 218. If x and y are positive integers, what is the remainder [#permalink]
18 Feb 2011, 11:30

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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.

Re: 218. If x and y are positive integers, what is the remainder [#permalink]
19 Feb 2011, 02:11

Expert's post

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).

Re: If x and y are positive integers, what is the remainder when [#permalink]
04 Feb 2014, 05:49

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 _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Re: If x and y are positive integers, what is the remainder when [#permalink]
20 Oct 2014, 01:31

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.

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. _________________

Re: If x and y are positive integers, what is the remainder when [#permalink]
20 Oct 2014, 03:22

Expert's post

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.

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. _________________

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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