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19 May 2007, 05:45
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X is positive interger less than 500.When x is divided by 7 the remainder is 1; when x is divided by 3, the remainder is 2. How many such numbers are possible?
A 18
B 20
C 22
D 24
E 25.
OA D
Please explain your answere.
Cheers!
Javed.
A 18
B 20
C 22
D 24
E 25.
OA D
Please explain your answere.
Cheers!
Javed.
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19 May 2007, 07:31
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Gotcha... !! 24 is the ans.
My answer is a bit rookie but I got the answer.atleast a line of thinking... Correct me if I am wrong anywhere..
I started with a trial and error on the type of such numbers...
8, 29, 50 , 71..... => Such numbers come at a gab of 21 numbers.
So the total number of such numbers should be between 20 and 25.
Applying A.P formula, d = 21, lets say n = 24
So, a24 = 8 + 23.21 = 483.
A number greater than this will cross 500.
Hence , there are 24 numbers that when divided 7 give remainder as 1 and when by 3 give remainder as 2.
Thanks...
My answer is a bit rookie but I got the answer.atleast a line of thinking... Correct me if I am wrong anywhere..
I started with a trial and error on the type of such numbers...
8, 29, 50 , 71..... => Such numbers come at a gab of 21 numbers.
So the total number of such numbers should be between 20 and 25.
Applying A.P formula, d = 21, lets say n = 24
So, a24 = 8 + 23.21 = 483.
A number greater than this will cross 500.
Hence , there are 24 numbers that when divided 7 give remainder as 1 and when by 3 give remainder as 2.
Thanks...
19 May 2007, 09:35
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Bookmarks
Dividing 7 with 3 gives remainder 1. So, 7xn+1 gives a 3m+2 once in three times i.e. for 8, 29, ...
AP formula gives An = A + (n-1)d
500 (or so) = 8 + (n-1)21
492/21 (or so) = (n-1)
23.42 (or so) = (n-1)
But n should be an integer, so n-1 = 23 => n = 24
AP formula gives An = A + (n-1)d
500 (or so) = 8 + (n-1)21
492/21 (or so) = (n-1)
23.42 (or so) = (n-1)
But n should be an integer, so n-1 = 23 => n = 24
