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MaxFabianKirchner

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08 Jul 2025, 08:47

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Let T be the total number of students. The information from the math session implies that T = 5m + 4 for some integer m. The information from the biology session implies that T = 6b + 4 for some integer b. Together, these statements establish that T has a remainder of 4 when divided by both 5 and 6.

If T-4 is a multiple of both 5 and 6, it must be a multiple of their least common multiple, which is 30. Therefore, the total number of students must be of the form T = 30k + 4, where k is a positive integer. The possible values for T are 34, 64, 94, and so on.

The history session is divided into h groups of 7 and one remaining group of r students, where r is fewer than 7. This relationship is T = 7h + r. We must test the possible values of T. If T = 34, then 34 = 7h + r. Dividing 34 by 7 gives a quotient of 4 and a remainder of 6. This corresponds to h=4 and r=6. Both values are available in the respective columns of the prompt. If T = 64, dividing by 7 gives a quotient of 9, which is not an available option for the number of groups. Higher values of T also fail to match the available options.

Therefore, the only valid solution is that there are 4 groups of 7 and 6 remaining students.

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08 Jul 2025, 08:54

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M= 4+5x
B= 4+6y
H= <7 +7z

Number of total students is same
4+5x=4+6y ; 5x=6y
(X,Y) possible values= (6,5),(12,10).......

M= 4+5x=4+5*6=34

H= <7+7z=34
Possible if z=4 and remaining value is 6
6 +7(4)=34

Z=4 and remaining value =6

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08 Jul 2025, 08:55

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Math: one group of 4, rest groups of 5 → N=4+5a
Biology: one group of 4, rest groups of 6 → N=4+6b
History: one group less than 7, rest groups of 7 → N=7c+r, where r<7

I would suggest to directly put value combinations in c and r so that a and 6 are integers.
Only possible combination, c = 4 , r = 6

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08 Jul 2025, 08:57

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For math
number of students=4+5x=?

For biology
number of students=4+6y=?

Looking at these both since the number of students needs to be same , 5x=6y ,suppose if x =6,y=5 both will be 30.So this looks okay

The total number of students could be 34.

For history

m+7k=34,if k=4
m+28=34
m=6

And I can see answers for 6 and 4.

So for Groups of 7 the answer is 4,and the remaining answer is 6.

Bunuel

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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.

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08 Jul 2025, 09:06

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At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Let the number of students be n

n = 4 + 5k = 4 + 6m = 7o + p ; where p < 7

4 + 5k = 4 + 6m
5k = 6m
k is a multiple of 6 and m is a multiple of 5

Let k = 6; m = 5
n = 4 + 5*6 = 34 = 4 + 6*5 = 7*4 + 6

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.

Groups of 7	Remaining students
4	6

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08 Jul 2025, 09:11

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Math = 4+5x
Biology = 4+6x
here it shows total students are 4+ (multiple of 30) --> 34,64,94

now these need to be split in 7y + (number less than 7)
34 = 7(4) + 6
64 = 7(9) +1
94 = 7(13)+ 3
here 34 matches the answer with 4 groups of 7 and 6 remaining students

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08 Jul 2025, 09:21

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I will present a quick approach to this question.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.

Read sentence 1 and 2 together to derive at a initial conclusion. As for both math and biology session one group is of 4 students and remaining students can be distributed in either 5 or 6 member groups, the total number of students must = 4+ 30k (where k is a positive integer)
Possible number of students thus can be 34,64,94 ... and so on

• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Now when total number students are 34, then 4 group of 7 students can be made with 6 students in one group. Therefore choice '4' for first column and '6' for second column is our answer.

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08 Jul 2025, 09:25

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Finding total students (N):
Math session: N = 4 + 5k (where k ≥ 0)

So N = 4 (mod 5)

Biology session: N = 4 + 6j (where j ≥ 0)

So N = 4 (mod 6)

Since both give remainder 4:
N ≡ 4 (mod 30)
Possible values: N = 4, 34, 64, 94,

Checking small values:
For N = 4:

Math: 4 = 4 + 5(0)
Biology: 4 = 4 + 6(0)
History: 4 students total → 0 groups of 7, remaining group has 4 students
Since 4 < 7

For N = 34:

Math: 34 = 4 + 5(6)
Biology: 34 = 4 + 6(5)
History: 34 ÷ 7 = 4 remainder 6 → 4 groups of 7, remaining group has 6 students
Since 6 < 7

Verification for N = 34:

Math: 1 group of 4 + 6 groups of 5 = 4 + 30 = 34
Biology: 1 group of 4 + 5 groups of 6 = 4 + 30 = 34
History: 4 groups of 7 + 1 group of 6 = 28 + 6 = 34

Groups of 7: 4
Remaining students: 6

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08 Jul 2025, 09:26

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let total students be N

for math session

the students are divided so that one group has 4 students and all other groups have 5 students each.
possible values are 14,19,24,29,34,39

For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
value of N possible
10,16,22,28,34,40
from 1 &2
we can have common value 34 so N is 34

For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.
7*4+6 = 34
Groups of 7 = 4
Remaining students =6
4,6 is correct option

At a summer camp, a class of students participates in three different sessions: math, biology, and history.

• For the math session, the students are divided so that one group has 4 students and all other groups have 5 students each.
• For the biology session, they are divided so that one group has 4 students and all other groups have 6 students each.
• For the history session, the same students are divided so that one group has fewer than 7 students and all other groups have 7 students each.

Select for Groups of 7 the number of groups that have 7 students each in the history session, and select for Remaining students the number of students in the remaining history group, that would be jointly consistent with the given information. Make only two selections, one in each column.

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08 Jul 2025, 09:27

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n=5p+4
n=6k+4

n=30z+4
z=1, n=34

7*4 + 6

so 4 groups, 6 remaining..

Ans 4 & 6

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08 Jul 2025, 09:33

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Bunuel

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Number of students = n

n = 5x + 4

n = 6y + 4

5x + 4 = 6y + 4

5x = 6y

Therefore x = 6 and y = 5

Number of students = 34

34 = 7*4 + 6

Answer

Groups of 7 = 4
Remaining students = 6

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Bunuel

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Solving the solution we have 34 students in mathematics and biology classes.sustituting in history class we get 7 students for 4 groups amd and remaining 6 students under 7

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08 Jul 2025, 09:37

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Let S be the total number of students in the class.

From the math session information:
One group has 4 students, and all other groups have 5 students each.
This means S=4+5kM for some integer kM ≥0.
(If kM =0, then S=4 and there's only one group.

If kM >0, there are kM +1 groups).
This can be written as S≡4(mod5).
Or S≡−1(mod5).

From the biology session information:
One group has 4 students, and all other groups have 6 students each.
This means S=4+6kB for some integer kB ≥0. (If kB =0, then S=4 and there's only one group.

If kB >0, there are kB +1 groups).
This can be written as S≡4(mod6).
Or S≡−2(mod6).

Now we need to find an S that satisfies both congruences:
S≡4(mod5)
S≡4(mod6)

Since S has the same remainder when divided by 5 and 6, S must be of the form LCM(5,6)⋅q+4.
LCM(5,6)=30.
So, S=30q+4 for some non-negative integer q.

Let's list possible values for S:
If q=0, S=4.
If q=1, S=34.
If q=2, S=64.
If q=3, S=94.
And so on.

Let's check if S=4 is valid for the group structures.

Math: If S=4, it's 1 group of 4. (kM =0). Valid.

Biology: If S=4, it's 1 group of 4. (kB = 0). Valid.

Now, let's consider the history session information:
The same students (S) are divided so that one group has fewer than 7 students and all other groups have 7 students each.
Let N7 be the number of groups with 7 students each, and R be the number of students in the remaining group.
So, S = 7*N7 +R, where 0≤R<7.
The problem states "one group has fewer than 7 students", which means R cannot be 0. So 1≤R<7.

Let's test the possible values of S:

If S=4:
4 = 7*N7 +R

Since R≥1, N7 must be 0.

Then R=4. This satisfies 1≤R<7.

So, for S=4: Number of groups of 7 (N7) = 0. Remaining students (R) = 4.

Now, let's look at the available options. The options for "Groups of 7" are 2, 4, 6, 8, 10.
Since N7 =0 is not among the options for "Groups of 7", S=4 is not the total number of students we are looking for.

If S=34:
34 = 7*N7 +R

We know 1≤R<7.

If N7 = 4, 7×4=28. R=34−28=6. This is a valid remainder (1≤6<7).

So, for S=34: Number of groups of 7 (N7) = 4. Remaining students (R) = 6.

Let's check if these values are in the options:
"Groups of 7": 4 (is in options)
"Remaining students": 6 (is in options)

This combination (Groups of 7 = 4, Remaining students = 6) is jointly consistent with the given information.

Let's check other values of S just to be thorough, but we've likely found our answer.

If S=64:
64 = 7*N7 +R
N7 =9, 7×9=63. R=64−63=1. This is a valid remainder (1≤1<7).

So, for S=64: Number of groups of 7 (N7) = 9. Remaining students (R) = 1.

Is N7 = 9 in the "Groups of 7" options? No.

If S=94:
94 = 7*N7 +R
N7 =13, 7×13=91. R=94−91=3. This is a valid remainder (1≤3<7).

So, for S=94: Number of groups of 7 (N7) = 13. Remaining students (R) = 3.

Is N7 =13 in the "Groups of 7" options? No.

The only S value from our list that produces a valid N7 and R pair present in the options is S=34.

Therefore, the number of groups that have 7 students each in the history session is 4, and the number of students in the remaining history group is 6.

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08 Jul 2025, 09:43

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All three setups force the total class size to be 4 more than a multiple of both 5 and 6, so the smallest non‐trivial solution is 34 students. When you divide 34 by 7, you get four full 7‐student groups (4 × 7 = 28) with 6 students left over. Hence in the history session there are 4 groups of 7 and 6 remaining students.

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08 Jul 2025, 09:45

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The groups of 7 has uniform number of students than other groups.

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08 Jul 2025, 09:50

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Notes:

Math, Biology, History

During math: 1 group has 4 and other groups have 5 students
During Bio: 1 group has 4 and other groups have 6
During History 1 group has less than 7 and the others have 7

Math = 5x+4
Bio = 6y +4

Potential: x=6, y=5 so
Total students could be 34

x=12 and y=10 so total students could be 64 and going up in 30's

Lets try 34
7*4 = 28
34-28 = 6

Both of those work in the grid. we will also check 64
7*9 =63
64-63 = 1 Not on the grid. does not work.

Group of 7 = 4
Remaining Students = 6

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08 Jul 2025, 09:51

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The total number of students must satisfy both the math and biology groupings. For math, the total equals 4 plus a multiple of 5; for biology, 4 plus a multiple of 6. This means the total number is 4 plus a common multiple of 5 and 6, which is 30, so the total students equal 4 plus 30 times some integer. Testing the smallest such total, 34, for the history session where groups have 7 students each except one smaller group, we find that 34 equals 7 times 4 plus 6. Since 6 is less than 7, this fits the history condition perfectly. Therefore, the number of groups with 7 students is 4, and the remaining group has 6 students.

Best way is to test all numbers from starting and do trial-&-error

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