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

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

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New post 10 May 2010, 09:36
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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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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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New post 26 Dec 2010, 02:36
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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

Number of approaches are possible.

For example, approach #1:

Since the units digit of 350 is zero then the number 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.

Approach #2:

\(7x+77y=350\), 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.
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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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New post 10 May 2010, 19:41
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7x + 77(n-x) = 350 --> where x is the number of times 7 repeats
7x + 7(11)(n-x) = 350
dividing both sides by 7
x + 11 (n-x) = 50

trying different options

now (n-x) has to be 1 because if its more than 1 (ex. 2) then 11(n-x) = 22 and x will be 37 or more which takes the total beyond 50.

Therefore now trying options we get

38 --> 37 + 11(1) = 48
39 --> 38 + 11(1) = 49
40 --> 39 + 11(1) = 50 ... this is the right answer
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Re: Tough PS problem- please help!  [#permalink]

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New post 11 Sep 2010, 08:16
Same concept alternate solution

Let the number of 77's be X, we know \(X>=0\)

\(7*(n-X) + 77*X = 350\)
\(7*(n-X) + 7*X + 70*X = 350\)
\(7*n + 70*X = 350\)

Since \(X>=0\), \(X\) can be \(0,1,2,3,4,5\) .... Correspondingly the value of n would be \(50,40,30,20,10,0\) (always a multiple of 10)

So the answer in this case is 40.
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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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New post 26 Dec 2010, 03:25
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Thank you, Bunuel. Your first solution is a good one (the second one is rather non-deterministic for me).
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Re: Sequence  [#permalink]

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New post 28 Apr 2011, 11:39
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Hi,
One other approach could be this.
350/7 =50.
Implies, it takes 50 7s or 0 77s to get a 350.
Considering 77 is 11 7s put together, try 1 77 and 39 7s (50-11 7s).
There u go. 1+39 = 40 !
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Re: 7,77....Problem Solving  [#permalink]

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New post 27 Jun 2011, 22:26
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Since, there is no 50 in the answer choices (350/7 = 50), we know there is at least one 77.

350 - 77 = 273
273/7 = 39
39+1 = 40.

If 40 wasn't there, I would have subtracted 77 from 273 and continued in a similar way.
Ans. 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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New post 13 Jan 2012, 23:35
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Thanks Bunuel for both the solutions. Very much appreciated.
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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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New post 30 Sep 2012, 06:33
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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


This is as good as saying all the numbers are either 1 or 11 and the sum equals 50
Let 1s be x and 11s be y, thus making n = x+y
x + 11y = 50
(x+y)+10y = 50
x+y = 10(5-y)
Hence x+y must be a multiple of 10 i.e. n must be multiple of 10. Only choice C

In case if the question asks the possible values of n, we can conclude that n could be 10,20,30,40,50
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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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New post 04 Nov 2012, 13:02
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,

Thanks for all your help, as usual. I answered the question using the second approach but would love to understand the first approach better. I must be missing something simple, but could you further explain why the number off terms must be a multiple of 10 to get a "0" in the units digit? That statement is tripping me up.
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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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New post 06 Nov 2012, 03:03
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egiles wrote:
Hi Bunuel,

Thanks for all your help, as usual. I answered the question using the second approach but would love to understand the first approach better. I must be missing something simple, but could you further explain why the number off terms must be a multiple of 10 to get a "0" in the units digit? That statement is tripping me up.


Consider this 7+7+7+7+7+7+7+7+7+7=10*7=70 (the sum of ten 7's equals to 10 times that term).

Hope it helps.
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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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New post 07 Mar 2014, 02:49
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\(\frac{350}{7} = 50\);

As all the options are below 50, there should be atleast one 77 present

To add one 77, we require to remove eleven 7's (7 x 11)

50 - 11 + 1 = 40 = Answer = C

If 40 wasn't there, then again same method:

40 - 11 + 1 = 30
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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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New post 26 Mar 2014, 21:44
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


We will start with the bigger numbers. There can be maximum of 50 Sevens and one 77 will replace 11 Sevens.

For there to be 50 terms: there will be One 77 (11 sevens) and 39 sevens. Hence 39 + 11 = 50 sevens.

Hence the answer is 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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New post 17 Apr 2014, 01:28
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350= 7a + 77b (a= no of 7s & b = no of 77s)

If b=0 then a=50
If a=0 then b=4 (+some remainder)

both these answers are not in the options.And looking at the options and above range we can conclude that the sum contains atleast 1 term as 77

put b=1,

350= 7a + 77(1)
350 - 77 = 7a

50-11 = a

a=39

a+b = 39+1 = 40 which is in the options

hence 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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New post 21 Jun 2014, 13:12
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!
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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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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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If each term in the sum A1 + A2 + ....... + An  [#permalink]

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New post 20 Aug 2016, 23:40
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janety0312 wrote:
Qs: 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 (answer)
d) 41
e) 42


Could someone teach me how to solve it? Thanks a lot~!



Given \(A1 + A2 + ... An = 350\)
Let there be \(x\) 7 and \((n-x)\) 77 in the set.

so \(x * 7 + (n-x) * 77 = 350\)
= \(7 * (x + (n-x)*11) = 7* 50\)
= \((x + 11n - 11x) = 50\)
= \(11n - 10x = 50\)
Since right hand side is a multiple of 10,then n has to be a multiple of 10 for integer \(x\) value and only C)40 . \(( x = (11 n - 50)/10)\)


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

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New post 19 Aug 2018, 04:41
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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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum  [#permalink]

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New post 19 Aug 2018, 08:58
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Re: If each term in the sum a1+a2+a3+...+an is either 7 or 77 and the sum   [#permalink] 19 Aug 2018, 08:58

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