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If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]
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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 [#permalink]
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10 May 2010, 19:41
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7x + 77(nx) = 350 > where x is the number of times 7 repeats 7x + 7(11)(nx) = 350 dividing both sides by 7 x + 11 (nx) = 50
trying different options
now (nx) has to be 1 because if its more than 1 (ex. 2) then 11(nx) = 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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If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]
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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 [#permalink]
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26 Dec 2010, 03:25
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Thank you, Bunuel. Your first solution is a good one (the second one is rather nondeterministic for me).



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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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13 Jan 2012, 23:35
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 [#permalink]
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28 Sep 2012, 05:51
I used the answer choices to figure this one out ... Given that we have 77 and 7 , therefore out of all the options we can get exactly 40 options that will yield 350 if we use 7 , 39 times and 77 ones ... 7 x 39 = 273 , adding 77 we get 350.. It took me about two and half minutes to do this by testing each answer out ..
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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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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(5y) 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 [#permalink]
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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 [#permalink]
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06 Nov 2012, 03:03



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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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04 Feb 2014, 12:35
7(1+1 + ... 11+11+.. n terms)=350 1+1 + ... 11+11+.. n terms=50 try out various combinations. Like 11*4+6 (10 terms) The options are bigger numbers. So you decrease 11's and increase 1's. 11*1+39 (40 terms) which is our answer. Hope it helped..
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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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06 Feb 2014, 04:19
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 let us suppose there are all 7's then there must be 50 7's for the sum to be 350 . but it is not one of the answer choices . so let us suppose there is one 77 so there must be 39 7's => total terms 40 which is 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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07 Feb 2014, 17:33
77 is the equivalent of eleven 7s. Look at answer choices. Clearly, the number of 7s should be much more than 77s. Take choice 1: 38. If there are 37 7s and 1 77, we get 37*7 + 77 = 336. To get 350, we need 2 more 7s. So, 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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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 = CIf 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 [#permalink]
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26 Mar 2014, 20:00
Thank you so much Brunel. This is such a simple question but my mind went blank when I was trying to attempt in the practice test! After the test got over, I used the same approach as approach1. It was so simple! Damn



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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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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 [#permalink]
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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
5011 = 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 [#permalink]
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12 Jun 2014, 07:43
I think we can frame another question out of this one.. 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 the minimum value of n? A)10 b) 20 c) 30 d)40 e) 50
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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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13 Jun 2014, 04:21
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Well, I tried it the following way:
a series of 7's and 77's would add up to 350. To start with I maximize 7's and minimize 77's. So only 1 77 should be enough and rest 7's. 350 77  273 273/9 = 39
So 39 * 7 + 1 * 77 = 350
Correct answer therefore in 40.



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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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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 [#permalink]
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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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