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Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 amFollowed by: 5 members; Thanked: 14 times

by Mo2men » Thu Jul 07, 2016 10:16 am

What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

What is the best way to solve? any shortcut??

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Source: Beat The GMAT — Problem Solving | Thu Jul 07, 2016 10:16 am

800_or_bust: Master | Next Rank: 500 Posts; Posts: 199; Joined: Sat Apr 26, 2014 10:53 amFollowed by: 4 members; Thanked: 16 times; GMAT Score:780

by 800_or_bust » Thu Jul 07, 2016 10:30 am

Mo2men wrote:What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

What is the best way to solve? any shortcut??

Hmmm, I attacked this as follows.

First, notice (D) is obviously wrong. If you subtract 7 from a number and the resulting number is a multiple of 7 - well the initial number must also be a multiple of 7. Likewise (E) can be eliminated because it is larger than 7. At most, the number being subtracted could possibly be is 6...

Now, I wasn't 100% sure, but I decided to see if I could find a pattern.

2^0 = 1 Subtract 1 to get a multiple of 7
2^1 = 2 Subtract 2
2^2 = 4 Subtract 4
2^3 = 8 Subtract 1
2^4 = 16 Subtract 2
2^5 = 32 Subtract 4

A pattern should be emerging. In order to get a multiple of 7 from a power of 2, you will either subtract 1, 2 or 4. Only answer choice (A) contains one of these integers. Hence, (A) should be the answer.

What is the OA?

800 or bust!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Jul 07, 2016 10:31 am

Mo2men wrote:What is the least possible value that can be subtracted from 2^586 so that the result is a multiple of 7?

A. 2
B. 3
C. 5
D. 7
E. 11

TEST EASY CASES and LOOK FOR A PATTERN.

2Â³ - 1 = 7.
2â�´ - 2 = 14.
2â�µ - 4 = 28.

2â�¶ - 1 = 63.
2â�· - 2 = 126.
2â�¸ - 4 = 252.

The subtracted numbers exhibit the following pattern:
1, 2, 4.
Implication:
To yield a multiple of 7, we must subtract 1, 2, or 4.

The correct answer is A.

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Thu Jul 07, 2016 2:50 pm

Here's a very straightforward approach:

We want the remainder of 2â�µâ�¸â�¶ when divided by 7. Notice that 2â�µâ�¸â�µ = (2Â³)Â¹â�¹â�µ = (remainder 1)Â¹â�¹â�µ.

Since 2â�µâ�¸â�¶ = 2 * 2â�µâ�¸â�µ = (remainder 2) * (remainder 1), our number has a remainder of 2 when divided by 7. Subtracting 2 will do it, so we're set.

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Mo2men: Legendary Member; Posts: 712; Joined: Fri Sep 25, 2015 4:39 amFollowed by: 5 members; Thanked: 14 times

by Mo2men » Fri Jul 08, 2016 2:06 am

Matt@VeritasPrep wrote:Here's a very straightforward approach:

We want the remainder of 2â�µâ�¸â�¶ when divided by 7. Notice that 2â�µâ�¸â�µ = (2Â³)Â¹â�¹â�µ = (remainder 1)Â¹â�¹â�µ.

Since 2â�µâ�¸â�¶ = 2 * 2â�µâ�¸â�µ = (remainder 2) * (remainder 1), our number has a remainder of 2 when divided by 7. Subtracting 2 will do it, so we're set.

Thanks for your answer.

However, after solving this question, I have another question. what if the same number is divided by 5? what would the reminder? why your approach did not work well here with me???

2â�µâ�¸â�¶ = (2Â³)Â¹â�¹â�µ * 2 = (5+3)Â¹â�¹â�µ * 2

(5+3)Â¹â�¹â�µ will leave reminder 3 when divided by 5

so final reminder = 3 * 2= 6

While I have solved with Mitch's approach

2^1) / 5 will have remainder 2
(2^2) / 5 will have remainder 4
(2^3) / 5 will have remainder 3
(2^4) / 5 will have remainder 1
(2^5) / 5 will have remainder 2

Thus cyclicity of 2^n / 5 is 4

2^286 = 2^2 * 2^284

2^2/5 will have reminder 4

2^284/5 will have reminder 1

so reminder 4*1=4

Thanks [/list]

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Jul 08, 2016 1:37 pm

Mo2men wrote:
Matt@VeritasPrep wrote:Here's a very straightforward approach:

We want the remainder of 2â�µâ�¸â�¶ when divided by 7. Notice that 2â�µâ�¸â�µ = (2Â³)Â¹â�¹â�µ = (remainder 1)Â¹â�¹â�µ.

Since 2â�µâ�¸â�¶ = 2 * 2â�µâ�¸â�µ = (remainder 2) * (remainder 1), our number has a remainder of 2 when divided by 7. Subtracting 2 will do it, so we're set.
Thanks for your answer.

However, after solving this question, I have another question. what if the same number is divided by 5? what would the reminder? why your approach did not work well here with me???

The key idea is to find power of 2 that we know has a remainder of 1 when divided by 5, then work from there.

For instance, we know that 2â�´ has a remainder of 1 when divided by 5. From there

2â�µâ�¸â�¶ =
2â�µâ�¸â�´ * 2Â² =
(2â�´)Â¹â�´â�¶ * 2Â² =
(remainder 1)Â¹â�´â�¶ * (remainder 4) =
(remainder 1) * (remainder 4) =
(remainder 4)

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Fri Jul 08, 2016 1:46 pm

And here's a harder example: suppose we want to know the remainder 2â�µâ�¸â�¶ when divided by 14.

We start with the same approach: looking for a power of 2 that has a remainder of 1 when divided by 14. But we won't be able to find one, since 14 is even and all our powers of 2 will be even. With that in mind, we'll look for something with a fairly small remainder, such as 2.

Since 2â�´ / 14 has remainder 2, let's work with that.

2â�µâ�¸â�¶ =
2â�µâ�¸â�´ * 2Â² =
(2â�´)Â¹â�´â�¶ * 2Â² =
(remainder 2)Â¹â�´â�¶ * (remainder 4)

Now we can work with our remainder 2, treating it as a base. We know that 2Â¹â�´â�¶ breaks down in the same fashion:

2Â¹â�´â�¶ * 2Â² =
(2â�´)Â³â�¶ * 2Â² * 2Â² =
(2â�´)Â³â�· =
(remainder 2)Â³â�·

Then we repeat:

2Â³â�· =
(2â�´)â�¹ * 2 =
(remainder 2)â�¹ * 2

And one more time:

2Â¹â�° =
(2â�´)Â² * 2Â² =
(remainder 2)Â² * 2Â² =
2 * 2 * 2 * 2 = 16

16 / 14 has remainder 2, so we're set: the remainder is 2.

Once you get the hang of this, you can skip a lot of the steps above, but it's a powerful process that generalizes well to any integer base and integer exponent you're likely to see.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jul 08, 2016 2:03 pm

suppose we want to know the remainder 2â�µâ�¸â�¶ when divided by 14.

An alternate approach is to determine the cycle of remainders.

2Â¹/14 --> remainder of 2
2Â²/14 --> remainder of 4
2Â³/14 --> remainder of 8
2â�´/14 --> remainder of 2

Since 2â�´ yields the same remainder as 2Â¹ -- implying the beginning of a new cycle -- the cycle of remainders is as follows:
2, 4, 8...2, 4, 8...2, 4, 8...

Implication:
Every exponent that is a multiple of 3 will yield a remainder of 8, with the next biggest exponent beginning a new cycle and yielding a remainder of 2.

285 is a multiple of 3 since the sum of its digits is a multiple of 3.
Thus:
2Â²â�¸â�µ --> remainder of 8
2Â²â�¸â�¶ --> remainder of 2 (the start of a new cycle).

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