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# A Tricky Question on Negative Remainders

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Re: A Tricky Question on Negative Remainders [#permalink]
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

P.s.
Sorry for the _____ but I did not know how to format properly the table....
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Re: A Tricky Question on Negative Remainders [#permalink]
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

P.s.
Sorry for the _____ but I did not know how to format properly the table....

When getting the pattern of the units digit of x^k you should always start with k=1, not k=0.
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Re: A Tricky Question on Negative Remainders [#permalink]
Yes you are right because k=0 will always give 1 as results...so no meaning.

But apart from that is right the approach?

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Re: A Tricky Question on Negative Remainders [#permalink]
Bunuel wrote:
Hi everyone,

I am new and I was just reading the post but I was somehow lost when following the solution...so I tried to solve myself and I did in this way:

I wrote down the following in order to find a possible pattern (R=remainder):

k | 3^k | R when divided by 5 |
0 | 1___|_______1________|
1 | 3___|_______3________|
2 | 9___|_______4________|
3 | 27__|_______2________|
4 | 81__|_______1________|
5 | 243_|_______3________|

Stop! OK I found a general pattern for the remainder cyclical (1,3,4,2)...then since 3^(7^11) is equal to 3^77 then is sufficient to look at the line where k is 7 (because 77 is eleven times 7)...that is the same as the line with k=3 so R=2! Indeed...

6 | not needed |_______4________|
7 | not needed |_______2________|

So the answer is 2 that is (C).

Am I wrong? It was quite fast in this way...

Thanks to everybody!

P.s.
Sorry for the _____ but I did not know how to format properly the table....

When getting the pattern of the units digit of x^k you should always start with k=1, not k=0.

The red part is not correct: 3^7^11 = 3^77 but 3^(7^11) = 3 to the power of (7^11).

Theory on Exponents: math-number-theory-88376.html
Tips on Exponents: exponents-and-roots-on-the-gmat-tips-and-hints-174993.html

Check Units digits, exponents, remainders problems directory in our Special Questions Directory.
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Re: A Tricky Question on Negative Remainders [#permalink]
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.
3^3 = 27
27 / 5 = remainder of 2
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Re: A Tricky Question on Negative Remainders [#permalink]
Tactic101 wrote:
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.

When you divide 3^k by 5 or 10, and want to know the remainder, then you want to know where you'll be in the units digit pattern of 3^k. Because the units digit pattern of 3^k repeats in 'blocks' (cycles) of 4 terms, you need to know the remainder when you divide k by 4 in order to know where you'll be in that pattern. In this question, k = 7^11, so you'd need to find the remainder when you divide 7^11 by 4. What you found above is the remainder when you divide 7^11 by 10. So you correctly determined that 7^11 has a units digit of 3, but that alone doesn't tell you the remainder when you divide 7^11 by 4 (since some numbers that end in 3 will give a remainder of 1 when you divide by 4, and some will give a remainder of 3 when you divide by 4). It's because it's not straightforward to find the remainder when you divide 7^11 by 4 that the posters above needed some lengthier solutions.
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Re: A Tricky Question on Negative Remainders [#permalink]
Tactic101 wrote:
Is there something wrong in my methodology?

7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3.
3^3 = 27
27 / 5 = remainder of 2

Hi
"7^11 .. 7 has cycle of 7, 9, 3, 1.. so the remainder is 3". Here the units digit would be 3 and not the remainder. Am I correct?
how to solve ahead if we know 3^m3/5? could someone help pls.
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Re: A Tricky Question on Negative Remainders [#permalink]
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Re: A Tricky Question on Negative Remainders [#permalink]
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