Sequence ... Each term is either 7 or 77

[dslewis]

If each term in the sum Asubscript1 + Asubscipt2 + ... + AsubscriptN is either 7 or 77 and the sume equals 350 which of the following could be the value of N

38
39
40
41
42

[RonPurewal]

well, first, think about the qualitative aspects of the sequence: if the sequence consisted entirely of 7's, then there would be fifty terms in the sequence. these answer choices are reasonably close to fifty, so it stands to reason that by far the majority of the terms will be 7's. therefore, try as few 77's as possible.

try only one 77:
remaining terms = 350 - 77 = 273
this would be 273 / 7 = 39 sevens
so ... you'd have one '77' and thirty-nine '7's

this works!

answer = c

[mdh3000]

Ron,

Could you also approach the question this way?

Since the units digit of 350 is zero, you know that the number of terms in the equation must be such that:

n*7 = number with units digit of zero

The only time this is true is if n is 10 or a multiple thereof, and 40 is the only answer that satisfies that.

mdh

[RonPurewal]

Ron,

Could you also approach the question this way?

Since the units digit of 350 is zero, you know that the number of terms in the equation must be such that:

n*7 = number with units digit of zero

The only time this is true is if n is 10 or a multiple thereof, and 40 is the only answer that satisfies that.

mdh

wow, yeah, that's brilliant.
i feel so lame all of a sudden!

just goes to show - one of the hallmarks of gmat problems is that many (if not most) of them can be approached in a great variety of ways.
so, don't hesitate; pursue the first approach that seems legitimate, and you'll by likely to find success.

[LP1]

but how are we accounting for 77 in the soln posted by 'mdh3000' since the sequence cud be formed by either 7 or 77?

[RonPurewal]

but how are we accounting for 77 in the soln posted by 'mdh3000' since the sequence cud be formed by either 7 or 77?

* we'll thank you to please take the extra 0.5 second to write "could" instead of "cud".

* the genius of mdh3000's solution is that it depends only on the UNITS DIGIT (i.e., "ones place" / "ones digit") of the numbers in question. since that digit is the same for either 7 or 77, there is no need to distinguish between those two numbers.

[sprparvathy]

Can we also approach this as follows:

Let x be the no. of 7s and y be the no. of 77s

Therefore, x+y = n

Also it follows then that 7x + 77y = 350

=> x + 11y = 50

trying different combinations for y

y = 1 => x = 39 ( x+y = 40)
y = 2 => x = 28 (x +y = 30) and so on...

But the only 40 is available in the answer choices. Therefore, ans is C.

[vivekcall81]

can we do it this way:
lets say 7 be T times in the sequence then 77 will be N-T times in the sequence.
(N-T)*77 + T*7 = 350
(N-T)11 + T = 50
11N-10T = 50
to get 50, N has to be multiple of 10
from the options:
N = 40
440-50=390, 390/10=39 T=39, N=40

proof:
40*7 + (40-39)*7 = 350

[akhp77]

One alternate method
7 * 50 = 350
7 * 11 + 7 * 39 = 350
77 * 1 + 7 * 39 = 350
So one solution could be 1+39 = 40

[michael_shaunn]

I too thought in the way Ron has approached the solution.But it's also great to see the unit digit's approach and the second appraoch where 77 is treated as 11 times 7 i.e basically what we are doing is that we are counting the total number of 7's in 350 and we can orally find the solutions for the question and then seeing how we can take out 7's in multiples of 11(which is basically 77).

like......50-11+1 or 50-2*11+2 or 50-3*11+3 0r 50-4*11+4.

[RonPurewal]

all of the above solutions are valid.

some of them require much more unusual insights than do others -- i'm looking in particular at the solution that requires randomly breaking up 7*50 into 7*11 + 7*39 (i'm not really sure how one would come upon such an insight, especially under the psychological and time pressures of the test) -- but all of them would work if executed properly.