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If each term in the sum a1+a2+a3+...+an is either 7 or 77

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

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New post 21 Jun 2014, 13:17
russ9 wrote:
Bunuel wrote:
nonameee wrote:
Can I ask someone to take a look at it as I don't understand the solutions provided (or rather I don't understand how they came up with the solutions)? Thanks.


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

Number of approaches are possible.

For example: as units digit of 350 is zero then # of terms must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

To illustrate consider adding:

*7
*7
...
77
77
----
=350

So, several 7's and several 77's, note that the # of rows equals to the # of terms. Now, to get 0 for the units digit of the sum the # of rows (# of terms) must be multiple of 10. Only answer choice which is multiple of 10 is C (40).

Answer: C.

Or:
\(7x+77y=350\), where \(x\) is # of 7's and \(y\) is # of 77's, so # of terms \(n\) equals to \(x+y\);

\(7(x+11y)=350\) --> \(x+11y=50\) --> now, if \(x=39\) and \(y=1\) then \(n=x+y=40\) and we have this number in answer choices.

Answer: C.

Hope it helps.


Hi Bunuel,

I noticed that you tried to clarify method 1 below but still not connecting.

How are you drawing the conclusion that it has to be a multiple of 10? Why can't it be 2*175 which is NOT a multiple of 10 or 1*350 etc?

Thanks!


Do we have 175's or 350's to sum? We have 7's and 77's. Try to sum those to get the sum with units digit of 0, and you'll see that the number of terms must be 10, 20, 30, ....
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]

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New post 28 Sep 2014, 09:22
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 [#permalink]

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New post 29 Nov 2014, 14:44
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.

Answer: C

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 [#permalink]

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New post 19 Dec 2014, 10:25
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 [#permalink]

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New post 21 Mar 2015, 05:43
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.

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

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New post 09 Sep 2015, 04:39
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.

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

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New post 27 Sep 2015, 02:23
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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 [#permalink]

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New post 10 Oct 2015, 00:31
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 [#permalink]

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New post 19 Oct 2015, 22:47
350=50*7 (no answer)
350=40*7+10*7=39*7+77.
Because a1 to an is either 7 or 77. Therefore, n can be40.

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

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New post 04 Aug 2016, 22:16
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.

Answer C is the correct answer.

40--> multiple of 10.

Hope this helps! :)

Regards,
Shradha

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

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New post 23 Aug 2016, 21:27
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



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

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New post 08 Jan 2017, 08:52
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

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

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New post 12 Feb 2017, 08:12
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?

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

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New post 02 Jul 2017, 22:53
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
Hence, Answer choice C.

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

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New post 02 Sep 2017, 11:43
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.

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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77   [#permalink] 02 Sep 2017, 11:43

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