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While making rows of students a teacher found that on making 6 equal r
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11 Jul 2023, 00:14
11 Jul 2023, 00:14
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Question Stats:
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While making rows of students a teacher found that on making 6 equal rows or 5 equal rows, one student is left in both the cases whereas all the students could easily be arranged if 7 equal rows are made. What could be the minimum number of students?
A. 31
B. 91
C. 126
D. 241
E. 300
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
A. 31
B. 91
C. 126
D. 241
E. 300
Post A Detailed Correct Solution For The Above Questions And Get Kudos.
Question From Our Project: Free GRE Prep Club Tests in Exchange for 20 Kudos
\(\Longrightarrow\) GRE - Quant Daily Topic-wise Challenge
\(\Longrightarrow\) GRE NUMBER PROPERTIES
Re: While making rows of students a teacher found that on making 6 equal r
[#permalink]
07 Aug 2023, 22:29
07 Aug 2023, 22:29
Expert Reply
OE
As we can see that on making 6 equal rows or 5 equal rows of students, one student
is left.
Hence, using the dividend divisor relationship the number of students can be written
as,
๐ = 6๐ผ1 + 1 โ โโโ 1, 7, 13 โฆ โฆ as well as;
๐ = 5๐ผ2 + 1 โ โโโ 1, 6, 11 โฆ โฆ
That means, if we subtract 1 from the number ๐, ๐-1 will be divisible by 6 and 5, both.
If we are looking for a number which is divisible by 5 and 6, that must be divisible by
30 {๐ฟ๐ถ๐ ๐๐ (6,5)}.
Hence, merging these two arithmetic progressions we get that ๐ โ 1 will be a
number which is divisible by 30 {๐ฟ๐ถ๐ ๐๐ (6,5)}will leave a remainder 1
We get that ๐ will be a number when divided by 30 {๐ฟ๐ถ๐ ๐๐ (6,5)}will leave a
remainder.
Thus, the number of students can be written as,
๐ = 30๐ผ1 + 1 โ โโโ 1, 31, 61, 91 โฆ โฆ
Now, since the total number of students could be easily arranging in 7 equal rows,
thus ๐ will be divisible by 7
Therefore, the minimum possible number of students which must satisfy ๐ = 30๐ผ1 +
1 and also will be divisible by 7 will be 91
Ans. (B)
As we can see that on making 6 equal rows or 5 equal rows of students, one student
is left.
Hence, using the dividend divisor relationship the number of students can be written
as,
๐ = 6๐ผ1 + 1 โ โโโ 1, 7, 13 โฆ โฆ as well as;
๐ = 5๐ผ2 + 1 โ โโโ 1, 6, 11 โฆ โฆ
That means, if we subtract 1 from the number ๐, ๐-1 will be divisible by 6 and 5, both.
If we are looking for a number which is divisible by 5 and 6, that must be divisible by
30 {๐ฟ๐ถ๐ ๐๐ (6,5)}.
Hence, merging these two arithmetic progressions we get that ๐ โ 1 will be a
number which is divisible by 30 {๐ฟ๐ถ๐ ๐๐ (6,5)}will leave a remainder 1
We get that ๐ will be a number when divided by 30 {๐ฟ๐ถ๐ ๐๐ (6,5)}will leave a
remainder.
Thus, the number of students can be written as,
๐ = 30๐ผ1 + 1 โ โโโ 1, 31, 61, 91 โฆ โฆ
Now, since the total number of students could be easily arranging in 7 equal rows,
thus ๐ will be divisible by 7
Therefore, the minimum possible number of students which must satisfy ๐ = 30๐ผ1 +
1 and also will be divisible by 7 will be 91
Ans. (B)
gmatclubot
Re: While making rows of students a teacher found that on making 6 equal r [#permalink]
07 Aug 2023, 22:29
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