Hi nahid078,
Did this question come from a GMAT book or from a 'math' book?
I ask because, when posting questions, you really should post the ENTIRE question (including the answer choices). If this is a GMAT question, then this type of prompt could easily be solved by TESTing THE ANSWERS. Without including that information though, we're now limited in how we can answer the question.
GMAT assassins aren't born, they're made,
Rich
Sweets
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Rich is right - the answer choices make this question much easier. Let's have a go anyway...nahid078 wrote:after distributing the sweets equally among 25 children, 8 sweets remain. Had the number of children been 28, 22 sweets would have been left after equal distribution. What was the total number of sweets?
When it comes to remainders, we have a nice rule that says:
If N divided by D, leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
Okay, onto the question..........
Let T = TOTAL # of sweets
After distributing the sweets equally among 25 children, 8 sweets remain.
In other words, when T is divided by 25, the remainder is 8.
By the above rule, the possible values of T are 8, 33, 58, 83, 108, ....
Had the number of children been 28, 22 sweets would have been left after equal distribution.
In other words, when T is divided by 28, the remainder is 22.
By the above rule, the possible values of T are 22, 50, 78, 106, 134, 162, 190, 218,....
Hmmm, we haven't found a number in common with each list yet.
So, let's use some logic.
In the first part, we can see that the possible values of T will have a UNITS DIGIT of either 8 or 3
In the second part, we can see that the possible values of T will have a UNITS DIGIT that's even.
So, we can see that the value of T that satisfies BOTH conditions will have 8 as its UNITS DIGIT
Let's take a look at our second list.
Once we're at 218, we can see that we won't get back to a units digit of 8 until we add FIVE 28's Since 5 x 28 = 140, and adding 140 to 218 will give us a UNITS DIGIT of 8 again.
So, let's keep adding 140 to the second list and see where we get a value that also satisfies the first condition.
218 + 140 = 358...PERFECT.
358 divided by 25 leaves remainder 8.
So, there were 358 sweets.
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Cheers,
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x/25 = a remainder 8 -> 25a + 8 = xnahid078 wrote:after distributing the sweets equally among 25 children, 8 sweets remain. Had the number of children been 28, 22 sweets would have been left after equal distribution. What was the total number of sweets?
x/28 = b remainder 22-> 28b + 22= x
subtracting both equations, we have
25a-28b-14=0
25a = 14 + 28b
25a = 7(2+4b)
a = (7/25)(2+4b)
since a is an integer, 2+4b has to be a mutiple of 25
choices are 25,50,75,100 etc
if we subtact 2 from the choices above we are left with 4b. choices for 4b are
23,48,73,98 etc
4b cannot be odd
4 x any positive integer = even number, therefore 2+4b has to be multiple of 50
2 + 4b = 50 -> b=12
28b + 22= x
28*12 + 22 = 358
