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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. QOTD#14 When x is divided by 3, the remainder is 1. When x  Question banks Downloads My Bookmarks Reviews Important topics
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TAGS: GRE Prep Club Legend  Joined: 07 Jun 2014
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GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
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QOTD#14 When x is divided by 3, the remainder is 1. When x [#permalink]
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Expert's post 00:00

Question Stats: 62% (02:14) correct 37% (03:00) wrong based on 16 sessions
When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?

Drill 4
Question: 5
Page: 293

[Reveal] Spoiler: OA
4

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Sandy
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GRE Prep Club Legend  Joined: 07 Jun 2014
Posts: 4810
GRE 1: Q167 V156 WE: Business Development (Energy and Utilities)
Followers: 118

Kudos [?]: 1897 , given: 397

Re: QOTD#14 When x is divided by 3, the remainder is 1. When x [#permalink]
Expert's post
Explanation

To solve this question, write it out.

Since there are fewer numbers that yield a remainder of 2 when divided by 7, start there. The first such number is 2, and thereafter they increase by 7; the rest of the list is thus 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, and 93. Rather than list out all the numbers that yield a remainder of 1 when divided by 3, just select the numbers that meet the requirement from the list you already have: Only 16, 37, 58, and 79 do, so there are 4 values for x.

Hence the correct answer is 4.
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Sandy
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Re: QOTD#14 When x is divided by 3, the remainder is 1. When x [#permalink]
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Expert's post
sandy wrote:
When x is divided by 3, the remainder is 1. When x is divided by 7, the remainder is 2. How many positive integers less than 100 could be values for x?

When it comes to remainders, we have a nice rule that says:

If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Now onto the question....
When x is divided by 7, the remainder is 2
Possible values of x are: 2, 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93

When x is divided by 3, the remainder is 1
Possible values of x are: 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 73, 76, 79, 82, 85, 88, 91, 94, 97

The two sets have 4 numbers in common.
[Reveal] Spoiler:
4

ASIDE: Notice that each of the shared values (16, 37, 58, and 79) are 21 greater than the previous shared value. Also notice that 21 is the least common multiple (LCM) of 3 and 7.
So, once we found 1 value in common, we could have just kept adding 21 to that value to find the subsequent values.

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Brent Hanneson – Creator of greenlighttestprep.com  Intern Joined: 24 Oct 2017
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Re: QOTD#14 When x is divided by 3, the remainder is 1. When x [#permalink]
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in 1st case when N=4/3..remainder=1
2nd case when N=9/7...remainder=2
looking at condition we can conclude that 3,9,27,81.....in short, the power of 3 is increasing there and the limit is till 100 hence last number to consider is 81...making it total 4 numbers. Re: QOTD#14 When x is divided by 3, the remainder is 1. When x   [#permalink] 22 Jan 2018, 08:19
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