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Joanna bought only $0.15 stamps and $0.29 stamps. How many

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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 19 Dec 2012, 15:14
"Now, two of the numbers (15 and 440) are multiples of 5. That guarantees that the third number, 29y, is also a multiple of 5, and so y must be a multiple of 5 (if it is not immediately clear that 29y needs to be a multiple of 5 here, you can rewrite the equation as 29y = 440 - 15x = 5(88 - 3x), from which we can see that 29y is equal to a multiple of 5). Doing this you greatly cut down on the number of values you need to test; you now only need to check y= 5, 10 and 15 (since if y = 20, the sum is too large)."

Thank you for this explanation. Wow...
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 09 Feb 2013, 19:33
for statement 1, we get 15x + 29y = 440
as we need the value of x,so 15x = 440 -29y
x = (440 - 29y)/15
It's clear from the equation that the value of y should be 5 or 10 to be divided by 15 since x is an integer. if y = 5,you cannot divide it with 15. Now try y =10, so x = 10. Statement 1 is sufficient.

Also you can see from the 2nd statement that x=y .Statement 2 itself is not sufficient as you don't know the total price.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 22 Mar 2013, 06:37
I'd also like to add the following -

(1) She bought $4.40 worth of stamps. - We notice that the ending digit is 0, and one type of stamps costs 0.29. The only way we can have a "0" at the end with $0.29 is to multiply 0.29 with 10,20,30, etc. So if we have 10x0.29, its OK, meaning we have 2.9 worth of 0.29 stamps, and the rest worth $0.15. If we are to have 20, then we would surpass the total value of 4.40$ ( 20*0.29 = 5.8 ). So clearly she bought 10 stamps worth $0.29, and subsequently we can calculate the rest. That's why A is sufficient.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps. - Clearly insufficient.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 22 Mar 2013, 23:47
Expert's post
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.


Just a quick tool for Diophantine equations.

Lets assume the equation 4x+5y = 36. Rearrange it as x =\frac{(36-5y)}{4} or y = \frac{(36-4x)}{5}. Now, lets take the first re-arrangement. We can see that for y=0, we get an integral value for x as x=9 .Now all we have to do is keep adding 4 to the initial value of y=0. Thus, the next value of y which will give an integral solution for x is y=4,8,12 etc. One could also subtract 4 and get subsequent values for y = -4,-8,12 etc.

Taking the second rearrangement, we can see that x=-1, we have y=8. Thus, following the same logic, the next value of x, which will lead to an integral value for y is x= 4,9,14 etc.

Now back to the given problem:

We know that 15x+29y = 440.

Now, 29y = 440-15x

or y =\frac{440-15x}{29} .Now as because the statements on GMAT don't contradict each other, we know that one of the value of x and y for the above equation is x=y=10. Thus, for the above equation, keep adding 29 to x=10 for all the successive values for getting an integral solution for y. Thus, in accordance with the given sum, we can neglect negative integral values and state that x=y=10 is the only possible solution for both x,y>0.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 18 Jun 2013, 17:45
I'd like to provide a demonstration for the rule above.

Let's ax+by=c...................................(1)
an equation with a, b, c positive integers and GCD(a, b)=1.
If (x0, y0) satisfies (1) then
a(x0) + b(y0) = c
Any other solution can be arrived at by varying in opposite directions x0 and y0 so the ax + by keep adding to c.
a(x0 + k) + b(y0 - m) = c
a(x0) + ak + b(y0) - bm = c
a(x0) + b(y0) + ak - bm = c
c + ak - bm = c
ak = bm
it follows that ak is multiple of b (and of course of a) so it is multiple of LCM(a, b)=ab since GCD(a, b) = 1.
In other words ak can be written as ab*z for some integer z.
The same applies to bm. So,
ak = bm =ab*z
So k = b*z and m = a*z.
Returning back to our equation and reemplazing k and m by their expressions in terms of b and a we get:
a(x0 + b*z) + b(y0 - a*z) = c
Which means that the generating formula for all the solutions for (1) is:
[(x0 + b*z), (y0 - a*z)] z is an integer.
the first solutions generated are:
etc
z=-2-->(x0-2*b, y0+2*a)
z=-1-->(x0-b, y0+a)
z=0-->(x0, y0)
z=1-->(x0+b, y0-a)
z=2-->(x0+2*b, y0-2*a)
etc
So the minimum variation from (x0, y0) is:
(x0+b, y0-a) and (x0-b, y0+a)
In the case that solutions must be non-negative, then if both of the above 2 solutions is not non-negative then (x0, y0) is the only non-nergative one.

In the case that solutions must be positive, then if both of the above 2 solutions is not positive then (x0, y0) is the only positive one.

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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 10 Jan 2014, 08:46
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.



We are given: 0.15*x + 0.29*y = TC

1) Gives us TC..... And we assume this is insufficient, but that's not true. Look closely: the least common multiple between these two numbers has to be a multiple of 5, so we basically have a pretty heavy restriction for the possible combinations of the two given TC = 4.40... Also, if we take one of each, we are given: 0.15 + 0.29 = 0.44... Multiply this by 10 (which is a multiple of 5), and thus we have 4.40.. If we multiply by 5, we only get 2.20, if we multiply by 15, we get 6.60, by 20 we get 8.80.. Etc, etc... So the ONLY combination of x and y that works is x = y = 10. We have 10 of each.

2) Clearly insufficient, this only tells us that x = y, but without TC that could be any multiple of 5.

So we go with A.
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Re: Stamps [#permalink] New post 31 Mar 2014, 07:15
Bunuel wrote:
udaymathapati wrote:
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps.


Let x be the # of $0.15 stamps and y the # of $0.29 stamps. Note that x and y must be an integers. Q: x=?

(1) She bought $4.40 worth of stamps --> 15x+29y=440. Only one integer combination of x and y is possible to satisfy 15x+29y=440: x=10 and y=10. Sufficient.

(2) She bought an equal number of $0.15 stamps and $0.29 stamps --> x=y. Not sufficient.

Answer: A.

So when we have equation of a type ax+by=c and we know that x and y are non-negative integers, there can be multiple solutions possible for x and y (eg 5x+6y=60) OR just one combination (eg 15x+29y=440). Hence in some cases ax+by=c is NOT sufficient and in some cases it is sufficient.



Hope it helps.

Hi Bunel,

How to find the equation has more number of solutions or has single solution?

Thanks in advance,
Rrsnathan.
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Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 13 Aug 2014, 19:15
Hi, It looks like we can use the property that a (multiple of x) + (multiple of x) = multiple of x. And from that, we can see that since 4.40 and .15 are multiples of 5, .29 has to be a multiplied by a multiple of 5 as well.

Can someone comment as to whether this property holds true for subtraction/division/multiplication?
Re: Joanna bought only $0.15 stamps and $0.29 stamps. How many   [#permalink] 13 Aug 2014, 19:15
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