A positive number is divided by 100 to get a remainder thrice as the q...
Let the number be N and the quotient when divided by 100 be k. Then remainder is 3k. 3k < 100
N = 100k + 3k = 103k,
Also, N is divisible by 11.
=> k = 11p, where p is an integer.
=> N = 103*11p = 1133p...
As N < 100000, it implies that p can range from 1 to [100000/1133] i.e between 1 and 88
=> So, p can can range from 1 to 88
Also 3k < 100 => 3 * 11p < 100 => p < 4 Hence, p can take values 1,2,3
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A positive number is divided by 100 to get a remainder thrice as the q...
Solution:
Let's suppose the positive number is 'x'.
Given,
x = 100q + 3q
x = 103q
As x is divisible by 11,
103q is divisible by 11
11 divides 103q if and only if 11 divides q.
q can take values from 1 to 9071 (as 103 * 9071 = 935413)
Therefore, the possible values of q are 1, 2, 3, ... , 9071.
We need to find the number of such values of q which give x less than 100000.
As x = 103q, the maximum value of x can be 103 * 9071 = 935413.
Therefore, the maximum value of q for which x is less than 100000 is 970.
We need to find the number of values of q from 1 to 970 which give x less than 100000.
Let's calculate the value of x for q = 1, 2, 3, ... , 970.
We get x = 103, 206, 309, ... , 99989.
The number of values of q for which x is less than 100000 is 3.
Therefore, the required number of such numbers is 3.