If one integer is greater than another integer by 3, and the : Problem Solving (PS)

LM

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If one integer is greater than another integer by 3, and the [#permalink]

26 Jan 2012, 07:35

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B

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If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A. 15
B. 7
C. 3
D. 2
E. 1

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abby17

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Re: Integers [#permalink]

26 Jan 2012, 08:55

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Let the two integers be 'a' and 'b'
Then, according to the question
a-b=3 and $a^3 - b^3$=117
Now,
$(a-b)^3 = a^3 - b^3 -3ab(a-b)$
Substituting the values of (a-b) and $a^3 - b^3$ we get
$3^3$ = 117 - 3ab*3
27 = 117 - 3ab*3
9ab=90
ab=10

Now to find the value of (a+b) use
$(a+b) = \sqrt{(a-b)^2 +4ab}$
Again substitute the values
(a+b)=$\sqrt{9+40}$
(a+b)=$\sqrt{49}$
(a+b)=7

So the answer is B

L

Bunuel

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Re: Integers [#permalink]

26 Jan 2012, 09:14

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LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A) 15
B) 7
C) 3
D) 2
E) 1

Actually there are two values of the sum possible.

Let one integer be $x$ and another $y=x-3$. Basically we are asked to find the value $x+y=2x-3$

Given: $x^3-y^3=117$ --> $(x-y)(x^2+xy+y^2)=117$ --> substitute $y$: $(x-x+3)(x^2+x^2-3x+x^2-6x+9)=117$ --> $x^2-3x-10=0$ --> $x=-2$ or $x=5$ --> $2x-3=-7$ or $2x-3=7$. Only 7 is among the answer choices.

Answer: B.
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Re: Integers [#permalink]

26 Jan 2012, 09:45

yes Bunuel is right. I forgot to consider the negative value

(a+b)=$\sqrt{49}$
(a+b)=-7 or +7

jns

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Re: If one integer is greater than another integer by 3, and the [#permalink]

25 Sep 2013, 03:16

LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A. 15
B. 7
C. 3
D. 2
E. 1

I did not remember the formula so did it this way - Given a and b are 2 integers. a = b+3.
C, D and E are too small to produce a difference of 117. for eg. in choice c: a + b = 3. Assuming a to be the largest possible number, i.e, 3, a^3 = 27. 27-anything cannot result in 117. You can try with b= -3 or -2 also.
We are left with option B and A. Tried B first as that is smaller. Unit's digit of a^3 - b^3 should be 7. Let a= 6 and b=3. Considering only unit digits for cubes: 6-7 is not equal to 7.
Let a = 5 and b = 2. 5-8 = 7. Or 125 - 8 = 117. Got it. B is the answer

This method will not be useful when the options have big numbers. So it is better to memorize the formula

B

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Re: If one integer is greater than another integer by 3, and the [#permalink]

21 Jan 2014, 22:41

Is memorizing the formula the most efficient way to solve this problem?

jlgdr

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Re: If one integer is greater than another integer by 3, and the [#permalink]

11 Feb 2014, 07:31

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LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A. 15
B. 7
C. 3
D. 2
E. 1

Whoaa! This was a long one!

Let's see a = b+3

So we have (b+3)^3 - (b)^3 = 117

Now remember this a^3-b^3 = (a-b)(a^2+ab+b^2)

So we have (3)((b+3)^2+(b)(b+3)+b^2) = 117

After solving b^2+3b-10 = 0

So b=2 or b=-5

We are trying to find the value of a+b = 2b+3

Therefore if b=2, then 2b+3 = 7 which is actually listed as one of the answer choices.

B

PareshGmat

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Re: If one integer is greater than another integer by 3, and the [#permalink]

05 Mar 2014, 01:57

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Let one integer be x, other is (x+3)

We require to find the value of 2x+3

(x+3)^3 - x^3 = 117

Solving it, x^3 gets cancelled

9x^2 + 27x + 27 = 117

Solving the above, x= 2 & x+3 = 5

Answer = 2x+3 = 7

B

vponde

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Re: If one integer is greater than another integer by 3, and the [#permalink]

15 Oct 2016, 05:04

Bunuel wrote:

LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A) 15
B) 7
C) 3
D) 2
E) 1

Actually there are two values of the sum possible.

Let one integer be $x$ and another $y=x-3$. Basically we are asked to find the value $x+y=2x-3$

Given: $x^3-y^3=117$ --> $(x-y)(x^2+xy+y^2)=117$ --> substitute $y$: $(x-x+3)(x^2+x^2-3x+x^2-6x+9)=117$ --> $x^2-3x-10=0$ --> $x=-2$ or $x=5$ --> $2x-3=-7$ or $2x-3=7$. Only 7 is among the answer choices.

Answer: B.

Sorry, can someone explain me why $x + y = 2x - 3$ ?

L

Bunuel

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Posts: 92747

Re: If one integer is greater than another integer by 3, and the [#permalink]

15 Oct 2016, 05:10

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Expert Reply

vponde wrote:

Bunuel wrote:

LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A) 15
B) 7
C) 3
D) 2
E) 1

Actually there are two values of the sum possible.

Let one integer be $x$ and another $y=x-3$. Basically we are asked to find the value $x+y=2x-3$

Given: $x^3-y^3=117$ --> $(x-y)(x^2+xy+y^2)=117$ --> substitute $y$: $(x-x+3)(x^2+x^2-3x+x^2-6x+9)=117$ --> $x^2-3x-10=0$ --> $x=-2$ or $x=5$ --> $2x-3=-7$ or $2x-3=7$. Only 7 is among the answer choices.

Answer: B.

Sorry, can someone explain me why $x + y = 2x - 3$ ?

Since $y=x-3$, then $x+y=x+(x-3)=2x-3$
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Alexey1989x

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Re: If one integer is greater than another integer by 3, and the [#permalink]

16 Dec 2016, 02:40

Genereal approach using the formula is too time consuming. Can anyone advise if there is another way to solve that issue without substituting the numbers?
thanks

falakdoshi

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Re: If one integer is greater than another integer by 3, and the [#permalink]

01 Jan 2017, 11:49

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Hi Alexey,

We could solve this with the help of the answer choices as well, since the numbers are small.

1) X + X + 3 = SUM

2X + 3 = SUM

Now,

Choose of the answer choices: Say 7 for this sum.

2X + 3 = 7

Therefore 2X = 4

X = 2

Now the numbers are;

2 and 5.

2^3 = 8 and 5^ 3 = 125

Difference in cubes = 117.

Answer (B)

Each option should not take more than 20 - 30 seconds this way.

B

mcwags

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Re: If one integer is greater than another integer by 3, and the [#permalink]

25 Mar 2017, 19:44

Can someone please explain to me why I can't use the following technique?

x-y=3 and x^3-y^3=117

Using the difference of cubes formula:

(x-y)(x^2+xy+y^2)=117

3(x^2+xy+y^2)=117

(x^2+xy+y^2)=39

(x+y)(x+y)=39

Therefore, (x+y)= 13 or 3

P

gracie

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Joined: 07 Dec 2014

Posts: 1074

Re: If one integer is greater than another integer by 3, and the [#permalink]

26 Mar 2017, 14:48

LM wrote:

If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

A. 15
B. 7
C. 3
D. 2
E. 1

x^3-(x-3)^3=117➡
x^3-(x^3-9x^2+27x-27)=117➡
x^2-3x-10=0
(x-5)(x+2)=0
x=5
x-3=2
5+2=7
B

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Re: If one integer is greater than another integer by 3, and the [#permalink]

02 May 2017, 11:14

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mcwags wrote:

Can someone please explain to me why I can't use the following technique?

x-y=3 and x^3-y^3=117

Using the difference of cubes formula:

(x-y)(x^2+xy+y^2)=117

3(x^2+xy+y^2)=117

(x^2+xy+y^2)=39

(x+y)(x+y)=39

Therefore, (x+y)= 13 or 3

x^2 + xy + y^2 =! (x+y)^2

x^2 + 2xy + y^2 = (x+y)^2

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