Please show me a precise calculation where 3 divided by 7 gives a remainder of 3. Because i'm always arriving at the remainder of 2 and why 33 is not considered as the smallest value of n? do we have to factorize it or just pick the smallest single digit number in 33 which happens to be 3?
NehaKalani
When positive integer n is divided by 5, the remainder is 3, and when n is divided by 6, the remainder is 3. What is the remainder when the smallest possible integer value of n is divided by 7?
A. 0
B. 1
C. 2
D. 3
E. 4
Why is it wrong to think this way?
n = 5
x+3 but also n = 6
x+3
So n can be 8, 18, 23, 28, 33, 38, 43, 48 and so on...
But n also can be 9, 15, 21, 27, 33, 39 and so on...
The smallest value of n that satisfies both the conditions is 33.
And 33 when divided by 7 leaves a remainder of 5.
Bunuel, can you please explain why my approach is incorrect?
When positive integer n is divided by 5, the remainder is 3: n = 5q + 3. n can be 3, 8, 13, 18, 23, 28, 33, ...
When positive integer n is divided by 6, the remainder is 3: n = 6p + 3. n can be 3, 9, 15, 21, 27, 33, ...
As you can see the smallest possible value of n is 3, not 33. 3 divided by 7 gives the reminder of 3.
Notice that q and p are quotients and each of them can be 0.