Set S cosists of five consecutive integers,

[jnelson0612]

Would really wanna know how to generate a set that suits our requirement.

"the sum of these two sets(sets of consecutive integers) will be equal only when the sum is zero. " is incorrect.
Ex: Set S could be 5,6,7,8 & 9 while Set T could be 2,3,4,5.6.7 & 8 and these sets have equal sums.
There are umpteen other examples.

Well, I have Set S which has 5 consecutive integers, and Set T which has seven consecutive integers, but the SUM is equal. Hmm. Even though Set S only has 5 numbers, it is equal to a Set with 7 numbers. Well, I need to make S have higher numbers so it can counter-balance the extra numbers added in Set T. Then start trying possibilities. I might try to match up some of the numbers in both sets so I only have to worry about the ends. You can see that in the set you mentioned; both sets have 5, 6, 7, 8. S has a 9 and T has 2, 3, 4. 2+3+4=9 so these numbers work.

[sachin.w]

Thanks Jamie,

Is there any algebraic way to do this? because once we the solution made up of variables, we could just substitute the values and make umpteen sets.

[RonPurewal]

Thanks Jamie,

Is there any algebraic way to do this? because once we the solution made up of variables, we could just substitute the values and make umpteen sets.

sure, you could use the general representation of consecutive integers in algebraic form.

e.g.
you could write the seven consecutive integers as n, n + 1, ..., n + 6, so that their sum is 7n + 21.
you could write the five consecutive integers as m, m + 1, ..., m + 4, so that their sum is 5m + 10.
then you would be looking for integers such that 7n + 21 = 5m + 10.
you would still be guessing the values of those integers (since there's no algebraic way to generate them), but that's probably easier than guessing (and adding) entire sequences of numbers.
still, given the amount of work required to set this up in the first place, you probably aren't saving any time.

--

the other thing you can do -- and this is something that will be helpful for any "consecutive integers" situation -- is define a variable for the MIDDLE number of each set, rather than the first (smallest) one. if you do this, then the sums will work out to be very "nice" numbers.
so, the 7 integers in this problem can be written as
n - 3, n - 2, n - 1, n, n + 1, n + 2, n + 3
with a sum of 7n.
the 5 integers can be written as
m - 2, m - 1, m, m + 1, m + 2
with a sum of 5m.
then you just need to find integers such that 7m = 5n, which is a heck of a lot easier.
for instance, one such pair is m = 5 and n = 7, in which case you get the two sets that jamie listed above.

[NinaP494]

Why is it so easy to fall for a tricky second statement? Clearly it is not a very tough question at all. Is it because there is some residue memory from st1? Is it a better strategy to always start with st2 on DS questions? thanks

[RonPurewal]

if you personally have the issue of "remembering" statement 1 when you use statement 2, then you just have to be AWARE of that problem. then, whenever you use statement 2, just double-check that you're NOT "carrying over" anything from statement 1.

you can try using statement 2 first, if you want. if that helps you NOT make this kind of error, then, go ahead and keep doing it!
there's no gain or loss of efficiency either way -- you have to use the 2 statements individually no matter what, so the order is immaterial -- so, you may as well try. if it helps, it helps.

[RonPurewal]

also—
this test has no "trick questions", EVER. you'll have to think carefully and precisely, but you will never have to think of some unusual tricky "gotcha!" interpretation of a statement.
in other words, none of the statements should be "tricky".

basically, i'm just picking at your choice of words, there, in the last post. it's ok if a problem is challenging, or detail-oriented, or complicated... but it should NEVER be "tricky".

[NinaP494]

if you personally have the issue of "remembering" statement 1 when you use statement 2, then you just have to be AWARE of that problem. then, whenever you use statement 2, just double-check that you're NOT "carrying over" anything from statement 1.

Thanks Ron! I think this is a simple yet really powerful advice. It will definitely work.

[RonPurewal]

yup.

preventing procedural errors is really just a matter of self-awareness.