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gauravsoni


3. (1555 x 1557 x 1559) / 13 and \((102^2 - 101^2) /17\) , what is the sum of the remainders of these divisions ?

    1. 17
    2. 27
    3. 25
    4. 15


Remainder of (1555 x 1557 x 1559) / 13
Try dividing 1555 by 13 and you get the remainder 8. So the remainder when you divide 1557 by 13 will be 10 (2 more than last time) and when you divide 1559 by 13, the remainder will be 12 (another 2 more).
Using negative remainder concept will make the calculations easier here. The negative remainders are -5, -3 and -1 respectively.
(13a - 5) * (13b - 3) * (13c - 1) when divided by 13 will give a remainder -5*-3*-1 = -15 which is -2 (since 13 less will again give a multiple of 13) which when converted to positive remainder gives 11.


Remainder of \((102^2 - 101^2) /17\)
Remainder of (102 + 101)(102 - 101)/17
Remainder of 203/17 is 16

Sum of remainders = 11 + 16 = 27

Answer (B)

Check out these posts for a discussion on these concepts:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/04 ... unraveled/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... emainders/ (negative remainders)
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/05 ... ek-in-you/
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4. Two different numbers when divided by the same divisor leaves 11 and 12 as remainders respectively and when their sum was divided by the same divisor , remainder was 4. what is the divisor?

    1.36
    2. 28
    3. 14
    4. 19

If you go through the posts mentioned above, this question should be very easy for you.

When you add the numbers together, you will have 11 + 12 = 23 balls extra and since the actual remainder is only 4, it means you were able to make another group of 19 balls. This means the divisor must be 19.
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2. What is the smallest number by which 31250 should be divided to make it a perfect square ?

    1. 4
    2. 2
    3. 1
    4. 5


A couple of posts on factorization you might find helpful:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... ly-number/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... t-squares/