Bunuel wrote:
How many integers from 1 to 1000, inclusive, have the same remainder when divided by 2, 3, 5, 7?
A. 6
B. 7
C. 8
D. 9
E. 10
Note: 2 is the smallest number in the series. Hence the remainder has to be either 0 or 1.
Why ? remainder cannot exceed the divisor. Case 1: Remainder = 0 If the remainder =0, the number should be a multiple of 2, 3, 5 and 7. The lowest multiple of 2, 3, 5 and 7 is the LCM of 2, 3, 5 and 7 = 210
Now every multiple of 210 between 1 and 1000, inclusive will be a multiple of 2, 3, 5 and 7.
Number of integers between 1 and 1000, inclusive which is multiple of 210 = 4
Case 2: Remainder = 1 Let's assume that the number is of the form -
N = 2x + 1; x is the quotient when N is divided by 2
N = 3y + 1; y is the quotient when N is divided by 3
N = 5z + 1; z is the quotient when N is divided by 5
N = 7p + 1; p is the quotient when N is divided by 7
The common representation of N, which is divisible by 2,3,5 and 7 and leaves a remainder = 1 is
N = LCM(2,3,5,7)q + First common term
First common term = 1
How ?
N = 2*0 + 1 = 1
N = 3*0 + 1 = 1
N = 5*0 + 1 = 1
N = 7*0 + 1 = 1
LCM(2, 3, 5 and 7) = 210
Hence, N = 210q + 1
The first term that satisfies this equation is when q = 0; i.e. N = 1
The second term that satisfies this equation (at q = 1) = 211
The third term that satisfies this equation (at q = 2) = 421
So we see that the numbers are in A.P with common difference 210.
Number of terms below 1000 that satisfies N = 210q + 1 is 5
Total number of terms = 4 + 5 = 9
Option D