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3. 'a' and 'b' are two given constant positive integers. A student is asked to find two positive integers 'q' and 'r' are such that
a = bq + r , 0 < r < b. The findings of the student are q = 3 and 5 ; r = 2 and 4.
Is the student correct in his finding ? Justify your answer.
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Answered by
17
Your answer is --
No, student is not correct in his finding
Because , it is given that 'a' and 'b' are the two CONSTANT number .
It is possible only when q = 1 and r = 0.
as ,
➡ 6 = 6×1 + 0
➡ 4 = 4×1 + 0
➡ 10 = 10×1 + 0
➡ 3 = 3×1 + 0
➡ 5 = 5×1 + 0
In the above example a and b are constant .
So, q = 1 and r = 0
According to question
student take q = 3 and 5 ; r = 2 and 4.
✔✔ hence, his finding is not correct .
-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-
【 Hope it helps you 】
No, student is not correct in his finding
Because , it is given that 'a' and 'b' are the two CONSTANT number .
It is possible only when q = 1 and r = 0.
as ,
➡ 6 = 6×1 + 0
➡ 4 = 4×1 + 0
➡ 10 = 10×1 + 0
➡ 3 = 3×1 + 0
➡ 5 = 5×1 + 0
In the above example a and b are constant .
So, q = 1 and r = 0
According to question
student take q = 3 and 5 ; r = 2 and 4.
✔✔ hence, his finding is not correct .
-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-
【 Hope it helps you 】
QGP:
Similarly a=5q+4 is also similarly valid.
But now, the question asks whether the two relations: a=3q+2 and a=5q+4 can be simultaneously valid?
In other words, if a number gives a remainder of 2 when divided by 3, will the number give a remainder of 4 when divided by 5? We have to find whether any case exists for which the two previous statements are valid simultaneously
Try the question now.
i think u dont understand question
a and b are constant
Yes "a" and "b" are constant. But we don't know their values. From what I understand of the question, We have to check if there can be any constant values of a and b which can satisfy both a=3b+2 and a=5b+4 simultaneously
sir , give me example
Well, we have to simply solve the two equations a=3b+2 and a=5b+4, and check whether the values of a and b are positive or not. If not, then there cannot be any possible values of a and b under the given conditions, and the student has found wrong value of q and r
dhnyawad
Answered by
5
Here is your solution :
When, q = 3 and r = 2.
=> a = bq + r
Substitute the value of q and r,
•°• a = 3b + 2 ---------- ( 1 )
When, q = 5 and r = 4.
=> a = bq + r
Substitute the value of q and r,
=> a = 5b + 4 ---------- ( 2 )
From , ( 1 ) and ( 2 ) we get ,
=> 3b + 2 = 5b + 4
Subtracting 3b to both sides,
=> 3b + 2 - 3b = 5b + 4 - 3b
=> 2 = 2b + 4
Subtracting 2 to both sides,
=> 2 - 2 = 2b + 4 - 2
=> 0 = 2b + 2
Subtracting 2 to both sides,
=> 0 - 2 = 2b + 2 - 2
=> -2 = 2b
Dividing both sides by 2,
=> ( -2 ) / 2 = 2b / 2
=> -1 = b
•°• b = -1
Substitute the value of b in ( 1 ),
=> a = 3b + 2
=> a = 3( -1 ) + 2
=> a = -3 + 2
=> a = -1.
We have the values of a and b as negative numbers but in the question it is given that a and b are positive numbers.
Hence, the student is wrong.
Hope it helps !!
When, q = 3 and r = 2.
=> a = bq + r
Substitute the value of q and r,
•°• a = 3b + 2 ---------- ( 1 )
When, q = 5 and r = 4.
=> a = bq + r
Substitute the value of q and r,
=> a = 5b + 4 ---------- ( 2 )
From , ( 1 ) and ( 2 ) we get ,
=> 3b + 2 = 5b + 4
Subtracting 3b to both sides,
=> 3b + 2 - 3b = 5b + 4 - 3b
=> 2 = 2b + 4
Subtracting 2 to both sides,
=> 2 - 2 = 2b + 4 - 2
=> 0 = 2b + 2
Subtracting 2 to both sides,
=> 0 - 2 = 2b + 2 - 2
=> -2 = 2b
Dividing both sides by 2,
=> ( -2 ) / 2 = 2b / 2
=> -1 = b
•°• b = -1
Substitute the value of b in ( 1 ),
=> a = 3b + 2
=> a = 3( -1 ) + 2
=> a = -3 + 2
=> a = -1.
We have the values of a and b as negative numbers but in the question it is given that a and b are positive numbers.
Hence, the student is wrong.
Hope it helps !!
:-)
Good Answer!
Thanks !!
its so long
I didn't get what do you want to say ?
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