Supermaverick wrote:
What is the largest four digit number when it is divided by 4, 6 and 8 leaves remainders 3, 5 and 7 respectively.
a) 9872
b) 9997
c) 9936
d) 9983
e) 9883
Dear
Supermaverick,
I'm happy to respond.
This is a tough question. I'm not sure that this would be on the GMAT, because of the nature of the calculations it requires.
Here's how I would think about it. Let p & q be some positive integers.
We know that the number is of the form (8p + 7). If that's true, then the remainder when we divide by 4 will be 3; since that requirement is automatically included, we can ignore it.
We know that the number is of the form (6q + 5).
First, we can find the smallest positive integer that satisfies these conditions. Go through the numbers that satisfy the first requirement and see whether it satisfies the second.
p = 1 so (8p + 7) = 15. Subtract 5, and we get 10, which is not a multiple of 6.
p = 2 so (8p + 7) = 23. Subtract 5, and we get 18, which is a multiple of 6! That's our number!
The number 23 is of this form.
All the properties of 6 and 8 would repeat at every 24, which is the LCM of 6 and 8. Any number of the form (23 + 24n) would satisfy these conditions. (Let n be some positive integer.)
Here it helps to know about the rules of
multiples.
24*4 = 96 is a multiple of 24
24*400 = 9600 is a multiple of 24
9600 + 240 = 9840 is a multiple of 24
9840 + 96 = 9936 is a multiple of 24
9936 + 24 = 9960 is a multiple of 24
That's close enough to 10,000
Now, add 23 to this multiple of 24.
9960 + 23 =
9983 OA =
(D) Great question. Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test PrepEducation is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)