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If each term in the sum a1 + a2 + a3.....+ an is either 7 or [#permalink]
 22 Mar 2006, 23:07
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77% (02:13) wrong based on 3 sessions
If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42
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Re: 7,77....Problem Solving [#permalink]
22 Mar 2006, 23:27
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lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42
350/7=50
77/7=11
so 50-11=39
or 50-2(11)=28
or 50-3(11)=17
or 50-4(11)=6
hence except 39 rest are to small than the answer choices. so go with B.
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I'm getting 40
350 = 50 * 7
& 77 = 11 * 7
50*7 has 50 terms.
we can write it as,
39*7 + 77 = 350 => 40 terms.
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Re: 7,77....Problem Solving [#permalink]
23 Mar 2006, 03:58
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lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42
Let x and y be the number of term 7 and the number of term 77 respectively. WE have :
7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y
we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0
Among the answer choices provided, only C matches.
I go for C.
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Re: 7,77....Problem Solving [#permalink]
23 Mar 2006, 05:24
HIMALAYA wrote: lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42 350/7=50 77/7=11 so 50-11=39 or 50-2(11)=28 or 50-3(11)=17 or 50-4(11)=6 hence except 39 rest are to small than the answer choices. so go with B. You just missed the 77 to count 
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Re: 7,77....Problem Solving [#permalink]
23 Mar 2006, 07:33
vivek123 wrote: HIMALAYA wrote: lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42 350/7=50 77/7=11 so 50-11=39 or 50-2(11)=28 or 50-3(11)=17 or 50-4(11)=6 hence except 39 rest are to small than the answer choices. so go with B. You just missed the 77 to count thats right. yah, i forget to count that. 
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vivek123 wrote: I'm getting 40
350 = 50 * 7
& 77 = 11 * 7
50*7 has 50 terms.
we can write it as,
39*7 + 77 = 350 => 40 terms.
Worked it the same way as Vivek. Laxie`s method would be more efficient with much bigger numbers...
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if each term in the sum of a1+a2+...................+an is either 7 or 77 and the sum equals to 350, which of the following could be equal to n?
38
39
40
41
42
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Re: 7,77....Problem Solving [#permalink]
19 Oct 2010, 00:50
Merging topics ... Please try using the forum search feature ... You should be able to find answers to a lot of questions which have been asked before
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satishreddy wrote: if each term in the sum of a1+a2+...................+an is either 7 or 77 and the sum equals to 350, which of the following could be equal to n?
38
39
40
41
42 Alternate solutions from: tough-ps-problem-please-help-100812.html?hilit=term%20equal%20either#p779175Number 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.
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Re: 7,77....Problem Solving [#permalink]
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: 7,77....Problem Solving [#permalink]
28 Jun 2011, 09:51
laxieqv wrote: lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42 Let x and y be the number of term 7 and the number of term 77 respectively. WE have : 7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0 Among the answer choices provided, only C matches. I go for C. +1 Kudos to you laxieqv !!
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Re: 7,77....Problem Solving [#permalink]
29 Jun 2011, 17:15
x+11y = 50
x+y = n = verifying this with answer choices and checking for integer value for y, we can find out x+y = 40.
Answer is C.
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Re: 7,77....Problem Solving [#permalink]
29 Jun 2011, 17:29
laxieqv wrote: lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42 Let x and y be the number of term 7 and the number of term 77 respectively. WE have : 7x+ 77y= 350 ---> x+11y = 50 ---> x+y= 50- 10y we have : x+y= n = 50- 10y ---> n must have the unit digit of 0 because 50 - 10y must be a number which has unit digit of 0 Among the answer choices provided, only C matches. I go for C. I like this method because it attacks the the question dead on by representing each term as x and y and then deductively finding the answer through POE. Hope I can think this quickly on the exam!
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Re: 7,77....Problem Solving [#permalink]
28 Jul 2011, 12:57
I used a different method altogether. I observed that since all the numbers are 7s, the total number of sevens has to be a multiple of 5 in order to get to 350. out of the answer choices, only 40 is a multiple of 5, So eliminate everything but choice C. I guess, if there were other multiples of 5 then you could use one of the methods above. But as a quick elimination strategy it helped to do the least amount of work.
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Re: 7,77....Problem Solving [#permalink]
28 Jul 2011, 13:03
Substitution ..
it will take 50 7s to get 350 .. options are less than 50 .. so there should be atleast one 77.. 77/7 = 11 ..
so 39 7s + one 77 = 40
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Re: 7,77....Problem Solving [#permalink]
29 Jul 2011, 13:43
C it is.. 350 = 7x+77y x+11y=50 x y 39 1 =40 28 2 17 3 6 4 40 fits into our answer choice. lhotseface wrote: If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350, What is n ?
Note: a1,a2 represent the first term,second...and NOT a*1, a*2 !
A. 38
B. 39
C. 40
D. 41
E. 42
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Re: 7,77....Problem Solving [#permalink]
30 Jul 2011, 21:58
+1 C 7(50)=350 Since 50 is not there in ANSWERS so 77 needs to be ther 77=7(11) So : number of terms if one 77 is there 50-11+1=50-10=40 --> Check the Ans. C matches..
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Re: 7,77....Problem Solving [#permalink]
30 Jul 2011, 22:06
A beautiful variant of this problem - If each term in the sum a1 + a2 + a3.....+ an is either 7 or 77 and the sum is 350 Note: a1,a2 represent the first term,second...and NOT a*1, a*2 ! n has a value of : I. 30 II. 45 III. 10 IV. 50 V. 25 Which of the following can be True ? A - I,II,III B - II,III,V C - I,III,IV D - All C - None Apply the same concepts and you can figure out the solution
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Re: 7,77....Problem Solving
[#permalink]
30 Jul 2011, 22:06
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