Bunuel wrote:
seabhi wrote:
Hi,
IMO C. Please correct me if I am wrong.
from stmt 1: x = 7q+1 and x=5q+1 which leads to .. x=35Q+36 as the general formula.
hence multiple values of x will satisfy this. 36, 71, 106 .. you cannot say if the remainder will be 1 or 6.
from stmt2 x < 50 .. not sufficient.
from stmt1 and stmt 2 .. x can be only 36. Hence C.
No, the correct answer is A.
When positive integer x is divided by 7 the quotient is q and the remainder is 1. What is the remainder when x is divided by 10?When positive integer x is divided by 7 the quotient is q and the remainder is 1 --> x = 7q + 1 --> x could be 1, 8, 15, 22, ...
(1) When x is divided by 5 the quotient is q and the remainder is 1 --> x = 5q + 1 --> subtract this from the equation from the stem: 0 = 2q --> q = 0 --> x = 1. 1 divided by 10 gives the remainder of 1. Sufficient.
(2) x is less than 50 --> if x = 1, then the remainder upon division 1 by 10 is 1 but if x = 8, then the remainder upon division 8 by 10 is 8. Not sufficient.
Answer: A.
Hope it helps.
Bunuel the method you mentioned in response to
seabhi has been discussed above. i'm curious -- if he's wrong, how were his calculations incorrect? i built out the #s given from the formulas and got the same thing as him (see below):
- x=5q+1, so: 6, 11, 16, 21, 26, 31,
36, 41, 46, 51, 56, 61, 66,
71- x = 7q+1, so: 8, 15, 22, 29,
36, 43, 50, 57, 64,
71