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A new approach to an old question

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A new approach to an old question

by foobarnull » Thu Sep 04, 2008 11:05 am
This RE: If each term in the sum a1 + a2 + … + an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A.38
B.39
C.40
D.41
E.42

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The way I approached this problem was: the units digit of both numbers is 7. 7 only multiplies into a X0 number when multiplied by a 10. So the number of 7s in the set have to be a multiple of 10. Done.

This is also easily confirmed by thinking; a maximum of 50 7s go into 350. That's not an option. 10 down, we have 40.... 11 7s in 77, 39 more free-floaters, 39+1 = 40.

Everyone seems to distill this into two equations, which I get. But I looked far and wide on the web without seeing anyone use the intuitive method. Is my reasoning sound? I'm trying to internalize as many number properties and as much pattern-recognition as I possibly can.

I posted this question here since I'm a MGMAT customer, and a professional opinion would be most helpful. Could use a quick response... testing soon. Thanks!



-----------------

Just read that we're not supposed to discuss MGMAT problems here... not sure if that applies to GMATPrep materials as well. If it does, could a mod please move this thread? Also, I'd appreciate a response via PM in the meanwhile if any MGMAT staff are reading this.
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by Stacey Koprince » Fri Sep 05, 2008 7:35 am
Hi! Yes, we're not supposed to discuss test problems here - the BTG folks want to keep everything organized properly or it would be utter chaos. Here's what you can do: go post this in the relevant folder (in this case GMAT Math), then send me a PM with a link to the post. I'll follow the link to answer your question and voila - you get your answer and we still keep the forums organized properly. :)
(I can't move it myself - sorry!)
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by beatthegmat » Sat Sep 06, 2008 10:44 pm
Moving forward, please post to the correct forum area. Thanks.
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by Stacey Koprince » Sun Sep 07, 2008 8:10 am
Neat approach! I like it! There's one little area where you may have run into trouble but it happened to work out in this problem.

7*x = {a number ending in zero} only if x is a multiple of 10.

Each term is either 7 or 77, so we can rewrite the set {a-sub-1 + a-sub-2 + ... _ a-sub-n} as: 7(a series of 1s and 11s).

So x is the sum of some unknown number of 1s and 11s added together. And then x gets multipled by 7. Note that x is not the same as n (from the a-sub-n in the problem). We do need x to be a multiple of 10, but that doesn't necessarily mean that n also has to be a multiple of 10, so you can't stop here and just pick 40. You have to check one more thing.

Based on our first line, above, the unknown number of 1s and 11s added together should add up to some multiple of 10 in order for the total thing to be 350. In order to get from a mix of 11s and 1s to a sum that is a multiple of 10, we could use ten 11s, ten 1s, nine 1s for one 11 (which is ten terms), eight 1s for two 11s (which is ten terms), seven 1s for three 11s (which is ten terms... hmm, I'm starting to see a pattern here...).

We also know that we need x to equal 50 (because 7*50 = 350). So the 1s and 11s have to add up to 50. And here's where we make the answer choices work for us, as you did. The only one that's a multiple of 10 is (C) 40, so check that one first to see if the little pattern we noticed in the past paragraph works.

And, as you showed, 40 does indeed work:

one 77 = one 11
thirty-nine 7s = thirty nine 1s

one 11 + thirty nine 1s = 50. The problem asks for what "could" equal n, so we're done!
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by cramya » Sun Sep 07, 2008 11:16 am
To find the maximum we have to get maximun number of 7's.

7*40 = 280 but 77 would not fit in teh sequence

7*39 + 77 = 350

So the maximun number possible is 39+1 = 40
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Re: A new approach to an old question

by Stuart@KaplanGMAT » Mon Sep 08, 2008 12:46 am
foobarnull wrote:This RE: If each term in the sum a1 + a2 + … + an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A.38
B.39
C.40
D.41
E.42
Seems to me that nothing is going to be quicker than common sense and backsolving.

We know that the sum is 350 and that the units digit of both possibilities is "7". The only way we're going to get a sum ending in "0" is if the total number of terms is a multiple of 10: choose (C) 40, the only multiple of 10 among the choices.
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by foobarnull » Tue Sep 09, 2008 8:57 am
Thanks Stuart & Stacey. I always find myself surprised by how much easier the Q section is if you've internalized the tricks and shun the pressure. Got a Q48 (down from Q49-50 on prep tests) on the real thing, but that's while being fairly relaxed throughout the test, hardly writing anything down, and finishing both sections ahead of time (5 mins on Q, ~18 mins on V). I consider finishing early on Q bad time management, but far less detrimental than taking too long.

All in all, I might have overdone the "don't sweat details; just see the big picture" bit on Q (worked well on V), but it is so much less taxing an approach as evidenced by this problem.

Cheers.
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