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Intern  Joined: 05 Mar 2014
Posts: 4
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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Hi Bunel,

In a hurry I tried to solve it by Arithmetic progression method.

(7+7) x n/2 = 350 i got n = 50 here..

(also i did tried some few like: (7 +77) x n/2 = 350 and (77+77) x n/2 = 350. )

Am I totally wrong as I approached it by Arithmetic progression method
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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fastest way to solve this IMHO:

350 = 7x50 that mean the sum of term should be 50 if 7 is factored out == 7 (a1+a2+...+an) this can happen if and only if its a factors are multiple of 10 and only C works.

LM wrote:
If each term in the sum a1+a2+a3+.....+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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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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Started with Algebraic way:
7x + 77(n-x)=350
x + 11(n-x)=50 --> this can give multiple solutions like x=6, n=10 or x=17, n=20 ...

Observe all solutions leading to n = multiples of 10. Only Option C (40) is a multiple of 10.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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LM wrote:
If each term in the sum a1+a2+a3+.....+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

You can look at it this way also -
cyclicity of 7:
$$7^1 = 7$$
$$7^2 = 9(last digit)$$
$$7^3 = 3(last digit)$$
$$7^4 = 1(last digit)$$

Sum of the last digits ends with = ..0

so the answer choices must match the $$4*n + 4$$ figure, only 40 matches, hence C.
Intern  Joined: 07 Jul 2015
Posts: 1
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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I took a while to figure this out on my own but here is how I approached this problem:

Since the answer choices were all over 35, there is a very slight possibility that there will be more than 1 multiple of 77.,
Therefore I subtracted 77 (multiple of 7) from 350 (multiple of 7) to get 273 (multiple of 7). Divided 273 by 7 to get 39. Getting 40 as the answer. 39*7 + 1*77 = 350.
Intern  Joined: 17 Aug 2015
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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paranoidvik
The minimum N would be A) 10.

4*77+ 6*7= 350
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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LM wrote:
If each term in the sum a1+a2+a3+.....+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

My Solution:

Consider if we have only 7's then 350 must have 50 number of 7's i.e 350/7=50

But we know we have some 77's too.

Say we have only one "77".

Then one 77 has 7+7+7+7+7+7+7+7+7+7+7 i.e 11 numbers of "7's".

So to accommodate one 77 we must subtract 11 number of 7's.

So we have 50-11=39 number of 7's and 01 number of 77 i.e 39+1=40 Answer C

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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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350=40*7+10*7=39*7+77.
Because a1 to an is either 7 or 77. Therefore, n can be40.

C
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Intern  Joined: 09 Oct 2015
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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Hi All,

Trying to solve the problem by using just one more variable in addition to 'n' which is already a part of the question stem.

Total number of terms: n
Now lets assume the number of terms of 7's and 77's:
7's : x
77's : (n-x)

using the information in the question, we get the following statement:

7(x) + 77( n-x) = 350
7x + 77n - 77x = 350
77n - 70x= 350
7(11n-10x)= 350

11n-10x = 50

Even though plugging in the values will not take too much time, however I suggest we try to understand the statement we just derived:

After subtracting the two values, we should have a '0' in the units place ( 50). To get a '0' we need the units place values of both the terms ( 11n and 10x) to be the same.

10x : this term will always have a '0' in the units place irrespective of the value of x --> this means 11n should have a '0' as well and which is possible is 'n' is a multiple of 10.

40--> multiple of 10.

Hope this helps! Regards,
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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LM wrote:
If each term in the sum $$a_1+a_2+a_3+...+a_n$$ 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

Check the solution in attachment
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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I came up with solution using PoE:

let , number of occurrences of 7 is x ,
number of occurrences of 77 is y ,
total number of terms is n

7x+77y=350 >> x=50-11y -----(1)

we have , x+y=n

substituting value of x from (1)
50-11y+y=n
50-10y=n
As we are subtracting a multiple of 10 from 50
i.e. a number ending with 0 is being subtracted from two digit number ending with 0
Hence , the result should also end with 0

We only have answer choice (C) with such value i.e. 40

Ans : C
Intern  B
Joined: 19 Sep 2012
Posts: 12
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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I have found the following approach more direct and less time-comsuming:
7(n-x) + 77x = 350
7n-7x+77x = 350
7(n+10x)=350
n+10x=50
n=50-10x
If x=1, then n=50-10=40 (The right option)
If x=2, then n=50-20=30
If x=3, then n=50-30=20
If x=4, then n=50-40=10

Am I missing something?
Intern  B
Joined: 27 Dec 2016
Posts: 13
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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I solved it a bit differently.
total number of 7's and 77s = 350
Looking at answer choices we can ensure that total number of 7's must be greater than 77's to achieve the range from 38 to 42.
Here is next math part:
350 - 77=273.
273/7 = 39
39 (7s)+1 (77) = 40
Manager  B
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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From the question stem we know that
a) x+y=n
b) 77x+7y=350 --> 11x+y=50

So, combining these two equations (system of equations) we will get 10x=50-n.

Only C satisfies the equation, since C (e.g.40) is the only answer choice, which has the units digit of 0.
Intern  B
Joined: 19 May 2018
Posts: 16
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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Similar to some of the other common sense approaches outlined above but had a slight variation in my thought process...

Max # of terms = 7 + 7 + 7 + 7 ..... = 350 --> 350/7 = 50 terms (too big for answer choices!)

What happens when we swap out one 7 in the sequence for a 77?

Hmmm well we are adding 70 to the sequence (on a net basis since we added 77 and subtracted 7); thus, we need to subtract or take away 70 from somewhere else in the sequence so that it still equals 350. We only can remove "7s" so we need to remove 10 "7s" (70/7 = 10) in order to deduct 70 from the sequence. In other words, we need to remove 10 terms in the sequence to counteract swapping one "7s" term for a "77s" term.

50 terms - 10 terms = 40 terms.

Intern  B
Joined: 30 Jan 2018
Posts: 1
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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I didn't the get the language, it says either 7 or 77, so we have to use either of one right in the series? Can you explain this?
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Joined: 02 Sep 2009
Posts: 55618
Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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anjalivats wrote:
I didn't the get the language, it says either 7 or 77, so we have to use either of one right in the series? Can you explain this?

It means that some numbers are equal to 7 and others equal to 77.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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There are a few different approaches.

A video explanation of two of them can be found here:

One approach is to think about the maximum and minimum number of 7s we would need in order to reach a sum of 350. The maximum is 350/7 = 50, but 50 isn't an answer choice. We need a number that's lower than 50, so lets find the second-highest number by subtracting a 77 from 350

350 - 77 = 273

273/7 = 39, i.e. our list of numbers could have a 77 and thirty-nine "7s" and the sum would be 350

The video covers this approach and a second one, as well.
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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LM wrote:
If each term in the sum $$a_1+a_2+a_3+...+a_n$$ 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

My approach was simple. From answer choices, comsidering the sum is 350, one could see that mostly there would be MUCH more "7" than "77".

first I tried 7*40+77 which gave me 357 then I tried 39*7+77 this gave me 350

BUT before i reached the solution, i was tryong to use AP i thought formula for fiinding number of terms would help here its quite challenging to figure out quickly which method will work and which one will not!
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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Let us assume x terms of 7 and thus, (n-x) terms of 77
7x + 77(n-x) = 350
x= (11n-50)/10
Now , x is an integer, which is only possible if (11n-50) is a multiple of 10 i.e 'n' should be 40 in the given options (the only multiple of 10). Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum   [#permalink] 09 Jan 2019, 07:00

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