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Bunuel
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

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Solution: \(5^2[1 + 2(5^2) + 5^4 - 5^2] = 5^2[(1 + 5^2)^2 - 5^2] = (5(1 + 5^2))^2 - 5^4 = a^2 - b^2 ==> b^2 = 5^4 ==> b= 25\)

Option E
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(a+b) (a-b) = a²-b²
5²+5^4+5^6 = 5² (1+5²+5^4) = 5² (1+5²+5^4+5²-5²) =5² [(1+5²)²-5²] = (5(1+5))² - (5²)²
→ a²-b² = (5(1+5))² - (5²)² → b² = (5²)² → b = 25
Answer: E
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5^2 + 5^4 + 5^6 = (a-b) (a+ b)
5^2( 1 + 25 + 625) = a^2 - b^2
5^2 ( 641 ) + b^2 = a^2
Note = it is given that a is integer so sqareroot of left hand side will also be integer.
Note 2= to make calc easy 26^ 2 is 676 and we are less of 25 ( 676-641) . So the b^2 should be such that 5^2 ( 641 + 25)

Please do note that we took 5^2 common .

b^2 = 25*25
b=25

Hence E is the answer

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Bunuel
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

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5^2 + 5^4 + 5^6
= 5^2 (1+5^2+5^4)
= 5^2 (1+2.5^2+5^4 - 5^2)
= 5^2 ((1+5^2)^2 - 5^2)
= [(5.(1+5^2))^2 - 25^2]

Value of b = 25
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Using Prime Factorization we need to convert this expression into products of two integers as indicated in the question.
5^2+5^4+5^6 = 5^2(1+5^2+5^4) = 25(1+25+625) = 25*651 = (5*5)*(3*217) = 5*5*3*7*31 = 105*155 = (130+25) * (130-25)
Hence b=25 (Answer choice E).
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This is how I did it -

Problem: 5^2 + 5^4 + 5^6

Step1: 5^6 + 5^2 + 5^4 (re-arranged the terms)

Step2: (5^3)^2 + (5)^2 + (5).(5^3) You can see here (a+b)^2 beginning to form with a = 5^3 & b= 5^2 ; Missing 2ab as I only have ab

Step3: (5^3)^2 + (5)^2 + (5).(5^3) + 1 - 1

Step4: (5^3)^2 + (5)^2 + (5).(5^3) + (5).(5^3) - (5).(5^3)

Step5: (5^3 + 5)^2 - (5^2)^2

This is of the form a^2 - b^2 where a = 130 and b = 25
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I cant understand after 26^2+5^2 part the manipulations are beyond me
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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?

5^2 + 5^4 + 5^6 = 25 (1 + 25 + 625) = 16,275 = (a+b)(a-b) = a^2 - b^2 = 130^2 - 25^2

IMO E
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vaibhavkataria97
This is how I did it -

Problem: 5^2 + 5^4 + 5^6

Step1: 5^6 + 5^2 + 5^4 (re-arranged the terms)

Step2: (5^3)^2 + (5)^2 + (5).(5^3) You can see here (a+b)^2 beginning to form with a = 5^3 & b= 5^2 ; Missing 2ab as I only have ab

Step3: (5^3)^2 + (5)^2 + (5).(5^3) + 1 - 1

Step4: (5^3)^2 + (5)^2 + (5).(5^3) + (5).(5^3) - (5).(5^3)

Step5: (5^3 + 5)^2 - (5^2)^2

This is of the form a^2 - b^2 where a = 130 and b = 25
This is amazing! How does one arrive at this? What was your thought process like?
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Hey, This is what I did. I want to know if this is a fine method.
5^2(1+5^2+5^4)
25(1+25+625)
25(651)
5*5*31*7*3 (Prime Factorisation)

- Given that maximum difference in the options is 25. We need to make the products really close to each other
- (31*5) (7*5*3)
- 155*105
- 130+25 130-25
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