courtesy of user 'karenmeow'
if each term in sum a1+a2+...+an is either 7 or 77, and the sum equals 350, which can be n?
choices: 38, 39, 40, 41, 42
if each term in the sum a1 + a2 + ... is either 7 or 77
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well, first, think about the qualitative aspects of the sequence: if the sequence consisted entirely of 7's, then there would be fifty terms in the sequence. these answer choices are reasonably close to fifty, so it stands to reason that by far the majority of the terms will be 7's. therefore, try as few 77's as possible.
try only one 77:
remaining terms = 350 - 77 = 273
this would be 273 / 7 = 39 sevens
so ... you'd have one '77' and thirty-nine '7's
this works!
answer = c
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notice that if you're adept at noticing patterns in DIGITS, you can also notice the following:
both 7 and 77 end with '7', so a look at the units digit of the sum will give telltale information about the number of terms.
specifically, the final sum, 350, ends with a zero. this means that, in the units column, you've added together a bunch of 7's and gotten a number ending in 0. the only way this can happen is if the number of 7's is a multiple of ten: there could be 10, 20, ... "7"s to be added together.
the only such number in the answer choices is 40, so that must be the correct answer.
try only one 77:
remaining terms = 350 - 77 = 273
this would be 273 / 7 = 39 sevens
so ... you'd have one '77' and thirty-nine '7's
this works!
answer = c
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notice that if you're adept at noticing patterns in DIGITS, you can also notice the following:
both 7 and 77 end with '7', so a look at the units digit of the sum will give telltale information about the number of terms.
specifically, the final sum, 350, ends with a zero. this means that, in the units column, you've added together a bunch of 7's and gotten a number ending in 0. the only way this can happen is if the number of 7's is a multiple of ten: there could be 10, 20, ... "7"s to be added together.
the only such number in the answer choices is 40, so that must be the correct answer.
Ron has been teaching various standardized tests for 20 years.
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Hi Ron,
I solved it this way:-
Lets say there are x number of 7's and y number of 77's
so 7x+77y=350
i.e. x+11y=50
Normally I take the highest and lowest values when substituting values.In this case, x will be 39 and y is 1. x+y is n which will be 40.
Let me know if this is correct.
I solved it this way:-
Lets say there are x number of 7's and y number of 77's
so 7x+77y=350
i.e. x+11y=50
Normally I take the highest and lowest values when substituting values.In this case, x will be 39 and y is 1. x+y is n which will be 40.
Let me know if this is correct.
Last edited by Sunny22uk on Tue Jul 15, 2008 9:25 am, edited 1 time in total.
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if you're going to go to the trouble of writing equations, then you may as well write a pair of simultaneous equations, so that you can just solve them the way you'd solve any other system.Sunny22uk wrote:Hi Ron,
I solved it this way:-
Lets say there are x number of 7's and y number of 77's
so 7x+77y=750
i.e. x+11y=50
Normally I take the highest and lowest values when substituting values.In this case, x will be 39 and y is 1. x+y is n which will be 40.
Let me know if this is correct.
you can indeed write x + 11y = 50, as above. and then your second equation comes from the fact that there are forty numbers in the sequence: x + y = 40
so:
x + 11y = 50
x + y = 40
subtract
gives 10y = 10
y = 1
there you go.
whatever works!
Ron has been teaching various standardized tests for 20 years.
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Hi sunny,Sunny22uk wrote:Hi Ron,
I solved it this way:-
Lets say there are x number of 7's and y number of 77's
so 7x+77y=750
i.e. x+11y=50
Normally I take the highest and lowest values when substituting values.In this case, x will be 39 and y is 1. x+y is n which will be 40.
Let me know if this is correct.
where from u got 750?
the qustion states the sum is 350
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Here, we are assuming that X+Y = 40 terms and backsolving. The criteria for choosing I assume if when we get integer values for X and Y by backsolving. Can there be a scenario where couple of answer choices can solve this criteria when we are backsolving? If so, how to narrow it down. Appreciate your reponse.
lunarpower wrote:if you're going to go to the trouble of writing equations, then you may as well write a pair of simultaneous equations, so that you can just solve them the way you'd solve any other system.Sunny22uk wrote:Hi Ron,
I solved it this way:-
Lets say there are x number of 7's and y number of 77's
so 7x+77y=750
i.e. x+11y=50
Normally I take the highest and lowest values when substituting values.In this case, x will be 39 and y is 1. x+y is n which will be 40.
Let me know if this is correct.
you can indeed write x + 11y = 50, as above. and then your second equation comes from the fact that there are forty numbers in the sequence: x + y = 40
so:
x + 11y = 50
x + y = 40
subtract
gives 10y = 10
y = 1
there you go.
whatever works!
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well, sure there can, but not on a problem whose setup is like this one.ildude02 wrote:Can there be a scenario where couple of answer choices can solve this criteria when we are backsolving?
the setup of this problem is basically two simultaneous linear equations in two variables, so it's guaranteed to have one unique solution (unless the equations are multiples of each other, which they aren't).
you might get something where the problem asks you to maximize some quantity with some other equation given as a condition. for instance, you might be given that x + 11y = 50 and then asked to maximize the value of, say, -2x + 3y. in that case, it's sufficient just to find the extremes of x + 11y = 50 - namely, (50, 0) and (0, 50/11), or (50, 0) and (6, 4) if you're restricted to integers - and then just plug them into the expression you're trying to maximize.
Ron has been teaching various standardized tests for 20 years.
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I think the "pattern recognition" can be really helpful where drawing equations can be a tedious and time consuming task, To Ron-Thanks for illustrating the approach here.
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Sum of x times 7 and y times 77 gives a number whose unit can be 4, 1, 8, 5, 2, 9, 6, 3, 0.
The unit will be 0 only if x+y = 10 (which is n the same as n).
So to get 350, n must be a multiple of 10, hence the only acceptable solution is 40.
Answer C)
Please correct me if the above reasoning is wrong.
The unit will be 0 only if x+y = 10 (which is n the same as n).
So to get 350, n must be a multiple of 10, hence the only acceptable solution is 40.
Answer C)
Please correct me if the above reasoning is wrong.
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I did the following way.
350 = 7+77+7+77+........
350 = 7(1+11+1+11+....)
7*50 = 7(1+11+1+11+....)
So (1+11+1+11+....) must be equal to 50.
Since the question asks which of the following, we can say that one 11 and thirty nine 1's will give a sum of 50. Hence the number of terms is 1+39 which is 40.
350 = 7+77+7+77+........
350 = 7(1+11+1+11+....)
7*50 = 7(1+11+1+11+....)
So (1+11+1+11+....) must be equal to 50.
Since the question asks which of the following, we can say that one 11 and thirty nine 1's will give a sum of 50. Hence the number of terms is 1+39 which is 40.
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Clever approach!sidceg wrote:I did the following way.
350 = 7+77+7+77+........
350 = 7(1+11+1+11+....)
7*50 = 7(1+11+1+11+....)
So (1+11+1+11+....) must be equal to 50.
Since the question asks which of the following, we can say that one 11 and thirty nine 1's will give a sum of 50. Hence the number of terms is 1+39 which is 40.
This tells us that the number of terms must be divisible by 10, and only answer choice C works.
Cheers,
Brent
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Another possible approach it to look for a pattern:If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?
A)37
B)39
C)40
D)41
E)42
Since both 7 and 77 have 7 as their units digit, we know that if we take any two terms, their sum will have a units digit of 4 (e.g., 7 + 7 = 14, 7 + 77 = 84, 77 + 77 = 154)
Similarly, if we take any three terms, their sum will have a units digit of 1. (e.g., 7 + 7 + 7 = 21, 7 + 7 + 77 = 91, etc.)
Now let's look for a pattern.
The sum of any 1 term will have units digit 7
The sum of any 2 terms will have units digit 4
The sum of any 3 terms will have units digit 1
The sum of any 4 terms will have units digit 8
The sum of any 5 terms will have units digit 5
The sum of any 6 terms will have units digit 2
The sum of any 7 terms will have units digit 9
The sum of any 8 terms will have units digit 6
The sum of any 9 terms will have units digit 3
The sum of any 10 terms will have units digit 0
The sum of any 11 terms will have units digit 7 (at this point, the pattern repeats)
From this, we can conclude that the sum of any 20 terms will have units digit 0
And the sum of any 30 terms will have units digit 0, and so on.
We are told the sum of the terms is 350 (units digit 0), so the number of terms must be 10 or 20 or 30 or . . .
Since C is a multiple of 10, this must be the correct answer.
Cheers,
Brent
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Most of these explanations seem overly complex.
We know that 350 = 7 * 50, so we must have fifty 7s in some fashion.
Every time we replace 7s with a 77, we turn ELEVEN terms into ONE: in other words, we subtract 10 total terms.
So we could have 50 terms (all 7s), 40 terms (39 7s and one 77), 30 terms (28 7s and two 77s), etc. The only choice given here that works is C, so we're done.
We know that 350 = 7 * 50, so we must have fifty 7s in some fashion.
Every time we replace 7s with a 77, we turn ELEVEN terms into ONE: in other words, we subtract 10 total terms.
So we could have 50 terms (all 7s), 40 terms (39 7s and one 77), 30 terms (28 7s and two 77s), etc. The only choice given here that works is C, so we're done.