If 5a + 7b = k, where a and b are positive integers, what is the large

\(lcm (5,7)=35\)

Seven 5s can be converted to Five 7s OR Five 7s can be rewritten as Seven 5s.

But question stem tells that 'a' and 'b' are positive integers , so 'a' cannot be zero . if this is the case then we can have
5*7+ 7*5=70

We cannot have more than seven 5 .

Answer D

For those who selected E , note that unit digit is 4
we have 2 answers straightforward. .
60+14
49+25

We 'll start putting the values of with a = 10 and b = 1.
Keep decreasing the value of a by 1 and increasing the value of b by 1.

At a = 10 and b = 1, 5a + 7b = 57
At a = 9 and b = 2, 5a + 7b = 59
At a = 8 and b = 3, 5a + 7b = 61

We see a pattern, for every decrease in a and every decrease in b, the result is increasing by 2.

So,
At a = 4 and b = 7, 5a + 7b = 69.

Which is option C.

If 5a + 7b = k, where a and b are positive integers, what is the largest possible value of k for which exactly one pair of integers (a, b) makes the equation true?

a=5 & b=7 will give 25+49 = 74

Ill go with E


(edit) : just understood the "pair" funda of the question. I'll need to revise my answer after some serious thinking.

A word of advice for newcomers like me. Don't post in excitement. Think!!!

c. 69
5a+7b=k
b=2
and b=7
d. 70
e. 74
b=2
b=7

i will go with d

5*a1 + 7*b1 = k
5*a2 + 7*b2 = k
5*(a1 - a2) = 7*(b2 - b1)
since we are dealing with integers we can assume that a1 - a2 = 7*q and b2 - b1 = 5*q where q is integer, so whenever we get a pair for (a;b) we can find another one by simply adding 7 to "a" and subtracting 5 from "b" or vice versa, subtracting 7 from "a" and adding 5 to "b".
Lets check how it works for our numbers, starting from the largest:

E)74 = 5*12 + 7*2 (a1 = 12, b1 = 2), subtract 7 from "a" and add 5 to "b" respectively, so a2 = 5 and b2 = 7, second pair - bad
D)70 = 5*7 + 7*5 (a1 = 7, b1 = 5), if we add 7 to "a" we will have to subtract 5 from b but b can't be 0, so - no pair, if we subtract 7 from "a", we'll get a = 0 which also isn't allowed - no pair, thus this is the only pair for (a;b) that works, good!, thus
D is the answer

Ans D
a=7 and B=5 = 70
a=0 and B=5 or a=7 B=0 not permitted as it says 5a + 7b = k, where a and b are positive integers

Sorry, Why do you say that a1-a2=7q instead of 7((b2-b1)/5) ? and the same for b2-b1=5q ? I'm missing your assumption. are you saying that (b2-b1)/5 must be an integer and so the only way to have an integer is that that difference is a multiple of 5 ?

Thank you !

Yes, that is what I mean, if the difference between values of B is not a multiple of 5 (or multiple of 7 for values A), you will never find a value for A(or respectively B) to get the same number K, I just made it simplier by taking a variable Q which is an integer

Thanks a lot!

But if a1-a2=7q, and so a2=a1-7q. Why did you say plus and minus 7 ?

sorry to bother you !

coz q can be negative

MANHATTAN GMAT OFFICIAL SOLUTION:

Work Backwards from the Answers

The largest answer choice (and the largest possible value for k) is 74, so the largest possible value for b is 10. (If b were 11, then 7b would equal 77, which is too large.)

What about for a? Use similar reasoning: the largest possible value for a must be 14 because 5(14) = 70, which is less than 74.

The possibilities for b, then, are 1, 2, 3…9. The possibilities for a are 1, 2, 3…14.

List out the possibilities for 5a and 7b:

Because the question asks for the largest possible value of k, start with the largest answer choice. Do any of these pairs add to 74? Something from the 7b column would need to be added to a multiple of 5 (that is, a number that ends in either 5 or 0). Check the units digits. One possibility is that 7b + something with a unit’s digit of 5 = something with a unit’s digit of 4, so 7b would need to have a unit’s digit of 9. For example, 49 + 25 = 74. Is (E) the answer?

Hang on. The question asks for the largest possible number for which exactly one pair of integers (a, b) will work. We’ve just found one pair, but is there another? Keep checking. No other values for 7b end in 9, but we could also have 7b + a unit’s digit of 0 = a unit’s digit of 4, in which case the 7b would have to have a unit’s digit of 4. Are there any values for 7b that end in 4? Yes: 14. 14 + 60 = 74. For 74, then, at least two pairs (a, b) exist. Answer (E) is not correct.

What about answer (D), 70? Here, if the 5a value ends in 5, then 7b would also have to end in 5. Are there any values of 7b that end in 5? Yes, one: 35. 35 + 35 = 70, so one possible (a, b) pair does exist. Is there another? No other values of 7b end in 5. We also have to check what would happen if 5a ends in 0, though—in that case, 7b would also have to end in 0. The only value for 7b that ends in 0 is 70, but that would leave 5a = 0, and that’s not permitted (the problem says that a is a positive integer).

The only possible (a, b) pair, then, that makes k = 70 is (7, 5).

The correct answer is (D).

It is also possible to solve the problem with theory, but the solution is pretty tricky.

Note that the least common multiple of 5 and 7 is 5 x 7 = 35. Therefore, if a solution may be found where a > 7, then there will definitely be at least two solutions to the equation, since it will be possible to “trade” seven 5’s for five 7’s. For instance, if a = 8, then 5(8) + 7b = k, meaning that k would be the sum of eight 5’s and some number of 7’s. In that case, we could trade seven of the eight 5’s for five 7’s, so that k can also be written as 5(1) + 7(b + 5).

For the same reason, if b > 5, there will also be at least two solutions to the equation.

The maximum possible values for a and b (when told that there can’t be more than one solution pair for a and b) are thus 7 and 5, respectively. Can any other pairs of values for a and b also produce k = 70? Check as we did above; the answer is no. The only pair is a = 7 and b = 5, so k = 70 is the answer.

The correct answer is (D).


_________________

5a+7b=k➡5a/7=k/7-b
if 5a/7 is an integer<70, then 5a must equal 35
a=7
b=5
5*7+7*5=70
D

We can go through the given answer choices and check the given values from largest to smallest until we’ve found our answer.

E. 74

If k = 74, a could be 5 and b could be 7 since 5(5) + 7(7) = 25 + 49 = 74. However, we also can reduce b by 5 and increase a by 7 to obtain a sum of 74. That is, a could be 12 and b could be 2, and we also have 5(12) + 7(2) = 60 + 14 = 74. Thus, we have two different pairs of integers that make the equation true andE is not the answer.

D. 70

If k = 70, a could be 7 and b could be 5 since 5(7) + 7(5) = 35 + 35 = 70. Like in choice E, we also can reduce b by 5 and increase a by 7 to obtain a sum of 70. That is, a could be 14 and b could be 0, and we also have 5(14) + 7(0) = 70 + 0 = 70. However, we are told that a and b are positive integers. So a and b can’t be 14 and 0, respectively. There are no other ways to make a sum of 70 if both a and b are positive integers. Thus, we have exactly one pair of integers (7,5) that satisfies the equation and D is the correct answer.

Answer: D
_________________

Below is my approach and its pretty quick:
Make the equation as this: 5a/7 +b = k/7
as both a and b are positive integers, k/7 also needs to be integer.
So,k will be multiple of 7
Narrow down the answers for k as 35 and 70.
If we take k as 70, can we get the value? yes, with a= 7 and b =5
So answer is D.

After looking at the values for k and the given statement, noticed that the value should have 5 and 7 as its factors.

Now with this in mind, if you look at the answer options

Only 2 values have that 35 and 70.

Out of which 70 is the maximum value with a and b as 7,5 respectively.

Giving answer as D
_________________

We know how to solve integer solutions to equations in two variables. Get the first solution by hit and trial and then get subsequent solutions by adjusting the values of a and b with the values of co-efficients.

So when we get a value for (a, b), the next value will be obtained by
(a - 7) and (b + 5) or
(a + 7) and (b - 5)

If we want a single solution, we should ensure that the moment we go to (a - 7), it should become 0 (because we need only positive integer solutions). Also (b - 5) should become 0.
So a = 7, b = 5 are the greatest values that a and b can take.

If a = 8, (a - 7), (b + 5) will be a second acceptable solution.
Similarly, if b = 6, (a + 7), (b - 5) will be a second acceptable solution.

5*7 + 7*5 = 70 so maximum value of k is 70.

Answer (D)
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Thanks a lot , apu. really the best stuff. u make this problem a routine one. enjoying a lot.

I approach this question by listing down all the multiple of either 5 or 7:

5: 5, 10, 15, 20, 25, 30, 35, ..., 70, 75
7: 7, 14, 21, 28, 35, 42, 49, ..., 70, 77

Since the largest number in the answer selections are 74, E is out. Correct answer is D (70) with the assumption of:
(a, b) = (0,10) or (14, 0)

The question is asking for the largest number.

With extremely high probability the answer will not be the largest number among the answer choices, so it makes sense to immediately check the second largest.

With just a few substitutions of potential pairs, 7 and 5 quickly appear as the only pair for D

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