If the quantity \(5^2 + 5^4 + 5^6\) is written as \((a + b)(a – b),\) in which both \(a\) and \(b\) are integers, which

M7MBA wrote: ↑

Sun Jun 14, 2020 1:52 pm

If the quantity \(5^2 + 5^4 + 5^6\) is written as \((a + b)(a – b),\) in which both \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

A. 5
B. 10
C. 15
D. 20
E. 25

[spoiler]OA=E[/spoiler]

Source: Manhattan GMAT

So, we have \(5^2 + 5^4 + 5^6 = (a + b)(a – b),\)

\(5^2 + 5^4 + 5^6 = 5^2(1 + 5^2 + 5^4)\)

\(5^2(1 + 5^2 + 5^4) = 25(1 + 25 + 525)=25*651\)

So, we have \((a + b)(a – b) = a^2-b^2 = 25*651\)

Let's try option values.

A. 5 => \(b^2 = 25 => a^2-b^2 = a^2 - 25 = 25*651 => a^2 = 25*651 + 25 = 25*652 => a = 5*√652\). This cannot be the answer as it does not give a as an integer.

B. 10 => \(b^2 = 10 => a^2-b^2 = a^2 - 100 = 25*651 => a^2 = 25*651 + 100 = 25*655 => a = 5*√655\). This cannot be the answer as it does not give a as an integer.

C. 15 => \(b^2 = 225 => a^2-b^2 = a^2 - 225 = 25*651 => a^2 = 25*651 + 225 = 25*660 => a = 5*√660\). This cannot be the answer as it does not give a as an integer.

D. 20 => \(b^2 = 400 => a^2-b^2 = a^2 - 400 = 25*651 => a^2 = 25*651 + 400 = 25*667 => a = 5*√667\). This cannot be the answer as it does not give a as an integer.

E. 25 => \(b^2 = 625 => a^2-b^2 = a^2 - 625 = 25*651 => a^2 = 25*651 + 625 = 25*676 => a = 5*√676 = 5*26= 130\). This is the answer as gives a as an integer.

Correct answer: E

Hope this helps!

-Jay
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M7MBA wrote: ↑

Sun Jun 14, 2020 1:52 pm

If the quantity \(5^2 + 5^4 + 5^6\) is written as \((a + b)(a – b),\) in which both \(a\) and \(b\) are integers, which of the following could be the value of \(b?\)

A. 5
B. 10
C. 15
D. 20
E. 25

[spoiler]OA=E[/spoiler]

Solution:

We can factor the expression as:

5^2(1 + 5^2 + 5^4)

25(1 + 25 + 625)

25(651)

25(3 x 217)

5^2 x 3 x 7 x 31

We can say that the given expression is equal to, for example, 25 x 651 or 75 x 217 and set a + b = 651 and a - b = 25 or a + b = 217 and a - b = 75. To solve for b, we need to subtract the two equations, however, it will yield a large value for b. For example, 2b = 626 for the former example and 2b = 142 for the latter. However, since the value of b in the given choices is quite small, the two factors must be close to each other. Furthermore, since the units digit is either 5 or 0, the difference between a + b and a - b must have a units digit of 0. Therefore, we must find a pair of factors that satisfy these two conditions.

Upon inspection on 5^2 x 3 x 7 x 31, we see that both factors must have a factor of 5. So, we could have:

(5 x 3 x 7) x (5 x 31)

105 x 155

So by letting a + b = 155 and a - b = 105 and subtracting these two equations, we have:

2b = 50

b = 25

Answer: E