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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 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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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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19 Dec 2014, 10:25
Started with Algebraic way: 7x + 77(nx)=350 x + 11(nx)=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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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: Sum of the last digits ends with = .. 0so 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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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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27 Sep 2015, 02:23
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 [#permalink]
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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 5011=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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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.
C



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If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]
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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 : (nx) using the information in the question, we get the following statement: 7(x) + 77( nx) = 350 7x + 77n  77x = 350 77n  70x= 350 7(11n10x)= 350 11n10x = 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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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 Check the solution in attachment
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If each term in the sum a1+a2+a3+...+an is either 7 or 77 [#permalink]
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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=5011y (1)
we have , x+y=n
substituting value of x from (1) 5011y+y=n 5010y=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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12 Feb 2017, 08:12
I have found the following approach more direct and less timecomsuming: 7(nx) + 77x = 350 7n7x+77x = 350 7(n+10x)=350 n+10x=50 n=5010x If x=1, then n=5010=40 (The right option) If x=2, then n=5020=30 If x=3, then n=5030=20 If x=4, then n=5040=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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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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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=50n.
Only C satisfies the equation, since C (e.g.40) is the only answer choice, which has the units digit of 0.




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