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?

(1) When x is divided by 5 the quotient is q and the remainder is 1
(2) x is less than 50

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I got the correct answer but took quite a while. I listed out all the possible value of x and realize remainder when x is divided by 10 has to be 1. Any other faster method?

8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85
6, 11, 16, 21, 26, 31, 36, 41, 46, 56, 61, 71

from the ques stem we know-
x= 7q + 1
from 1) x = 5q + 1
subtracting we get x =1. hence when we divide x i.e. 1 by 10 we get remainder 1. sufficient.
stmt 2 is clearly insufficient. lots of possible values for x.

You don't get x=1 when you subract, you'd get 0=2q....also when you divide 1 by 10 the remainder is 10..

Precisely what i thought but from Question stem and St 1 we have
but note that it is for the same value of q that we get the remainder 1

x= 7q+1 and x=5q+1 -------> For the same value of q

Also 7q+1=5q+1 --------> 2Q=0 or q=0

Thus the number is 1 and when divided by 10 the remainder is 1 only.

Ans A is correct

if the Question would have stated x= 7q+1 and x=5a+1 then x= 35C+1 and possible value of x = 1,36,71, 106,141.....

then even taking statement 2 the answer would be E. The Question stem needs to mention that x is a 2 digit number.

Hope it helps
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From the question stem, x =7q+1.

F.S 1 states that x = 5q+1. Thus, we know that 7q+1=5q+1 --> q=0.Note that q is also an integer. Thus, x=1.Remainder when 1 is divided by 10 is 1. Sufficient.

F.S 2 for x=8, remainder when divided by 10 is 8.However, when x=15,the remainder when x is divided by 10 is 5. 2 different numerical values, hence Insufficient.

A.



Nope . This is incorrect.
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Hi,
Full solution to avoid any confusion-
x= 7q + 1
from 1) x = 5q + 1
Now subtracting these two equations yields q =0. Now put q=0 in any equation and you will get x=1. hence the no is 1. Now if we divide x (which is nothing but 1) we will remainder as 1.

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.
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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

Official Explanation from Veritasprep:

In quotient remainder problems, number picking is often a valid technique to show patterns and answer the question. For this problem, however, it is not very helpful as the question is really testing a tricky property of division with integers.

From statement one, you learn that when x is divided by 5 the quotient is q and the remainder is 1. In the question stem, you learned that when x
is divided by 7 the quotient is also q and the remainder is also 1. Most students understand that if you divide a number by two different divisors and get the same remainder, you must be at the LCM (lowest common multiple) of those two numbers plus the remainder (and then at the same location down the number line to infinity). For instance if you divide a number by 5 and 7 and get a remainder of 1, then that can happen at 35+1, 70+1, 105+1, etc. However, if you read carefully you see that the quotient is the same in both operations, a puzzling result. Clearly if you divide 36 by 7 it will give you a different quotient then when you divide 36 by 5. The only way this can happen is if x is equal to 1. Remember that when a smaller integer is divided by a larger integer, the quotient is always 0 and the remainder is the dividend itself. Here when 1 is divided by 7 the quotient is 0 and the remainder is 1 and when 1 is divided by 5 the quotient is also 0 and the remainder is 1. Statement 1 is sufficient as x must be 1 and the remainder when x is divided by 10 is also 1.

Statement 2 is clearly not sufficient by itself so the answer is A. This statement is here for people who miss that the quotient is the same and think that x could be any multiple of 35+1. By knowing that x is less than 50 it would lock in the value at 36. However, from the discussion above it is clear that this would be incorrect.
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