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If (61^2-1)/h is an integer, then h could be divisible by each...

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If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 07 Jun 2018, 03:43
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If \(\frac{(61^2-1)}{h}\) is an integer, then \(h\) could be divisible by each of the following EXCEPT:
A) 8
B) 12
C) 15
D) 18
E) 31

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If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 07 Jun 2018, 03:56
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CAMANISHPARMAR wrote:
If \(\frac{(61^2-1)}{h}\) is an integer, then \(h\) could be divisible by each of the following EXCEPT:
A) 8
B) 12
C) 15
D) 18
E) 31


\(\frac{(61^2-1)}{h}\) can be written as

\(\frac{(61^2-1^2)}{h}\)

=\(\frac{(61+1)(61-1)}{h}\)
= \(\frac{(62*60)}{h}\)
=\(\frac{31*2^3*3*5}{h}\)

\(2^3\),\(2^2\)*3, 3*5 & 31 are the factors of \(\frac{31*2^3*3*5}{h}\)

Hence, 18 is not divisible by the given expression.

Ans. D
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Re: If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 07 Jun 2018, 04:05

Solution



Given:
    • The expression \(\frac{(61^2 – 1)}{h}\) is an integer

To find:
    • From the given options, which one may not be able to divide h

Approach and Working:
We can simplify the given expression as follows:
    • \(\frac{(61^2 – 1)}{h} = \frac{(61 + 1) (61 – 1)}{h} = \frac{62 * 60}{h}\)
    • If this expression is an integer, then h must a factor of the numerator
    • \(62 * 60 = 2 * 31 *2^2 * 3 * 5 * = 2^3 * 3 * 5 * 31\)

Now, verifying with the options,
    • \(2^3 * 3 * 5 * 31 = 8 * 3 * 5 * 31\) (satisfies Option A)
    • \(2^3 * 3 * 5 * 31 = 12 * 2 * 5 * 31\)(satisfies Option B)
    • \(2^3 * 3 * 5 * 31 = 2^3 * 15 * 31\) (satisfies Option C)
    • \(2^3 * 3 * 5 * 31 = 2^3 * 3 * 5 * 31\) (satisfies Option E)

Hence, the correct answer is option D.

Answer: D
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If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 07 Jun 2018, 15:08
CAMANISHPARMAR wrote:
If \(\frac{(61^2-1)}{h}\) is an integer, then \(h\) could be divisible by each of the following EXCEPT:
A) 8
B) 12
C) 15
D) 18
E) 31



It's given that = \((61^2\)-1)/h = integer.

we understand that h is basically a divisor of 61^2-1. it means that h is a factor of 61^2-1. so , we can trace out the value of h as follows:

using (\(a^2\)-\(b^2\)) =(a+b) (a-b)

\(61^2\)-1 = (61+2) (61-1)
= 62*60

So, h = 62*60

now scan answer choices by which h is not divisible. Only option C fits as 15 = 3*5. (60) (62) is not divisible by 3.

Thus option C is the correct answer.
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Re: If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 07 Jun 2018, 17:39
(61^2-1^2)/h = I
(61-1)(61+1)/h = I
60*62/h = I

A) 8 - Possible because 15*4 = 60 and 31*2 = 62 so we have 4*2 = 8 as a factor
b) 12 - possible because 12 is a factor of 60
c) 15 -> possible because 15 is factor of 50
d) 18 -> not possible because 18 has two 3s and the number has only one
e) 31 -> possible because 31 is a factor of 62

Ans D
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Re: If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 08 Jun 2018, 11:20
CAMANISHPARMAR wrote:
If \(\frac{(61^2-1)}{h}\) is an integer, then \(h\) could be divisible by each of the following EXCEPT:
A) 8
B) 12
C) 15
D) 18
E) 31


Note that (61^2 - 1) is a difference of squares. Since 61^2 - 1 = (61 + 1)(61 - 1) = 62 x 60 = 2 x 31 x 2^2 x 3 x 15, we see that h can’t be divisible by 18 since there most h can only have one factor of 3, but 18 has two factors of 3.

Alternate Solution:

Note that (61^2 - 1) is a difference of squares, which is factored as (61 + 1)(61 - 1), or 62 x 60.

Let’s now factor 62 x 60 into primes: 2 x 31 x 2 x 2 x 3 x 5 = 2^3 x 3 x 5 x 31.

We see that the original expression can thus be expressed as 2^3 x 3 x 5 x 31/h. Now, we know that this expression yields an integer answer, so h must have factors that cancel out with the factors in the numerator. Let’s consider each answer choice:

Choice A: Since 8 = 2^3, which also is present in the numerator, h could be divisible by 8.

Choice B: Since 12 = 2^2 x 3, which also is present in the numerator, h could be divisible by 12.

Choice C: Since 15 = 3 x 5, which also is present in the numerator, h could be divisible by 15.

Choice D: Since 18 = 2 x 3^2, we see that there are not enough 3’s in the numerator to cancel out 3^2 in the factorization of h. Thus, Choice D is the correct answer.

Choice E: Since 31 is present in the numerator, we see that h could be divisible by 31.

Answer: D
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If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 08 Jun 2018, 11:37
CAMANISHPARMAR wrote:
If \(\frac{(61^2-1)}{h}\) is an integer, then \(h\) could be divisible by each of the following EXCEPT:
A) 8
B) 12
C) 15
D) 18
E) 31

\((61^2-1) = (61^2-1^2)\)

Or, \((61^2-1) = (61 + 1 )(61 - 1)\)

Or, \((61^2-1) = 62*60\)

\(62 = 2*31\)

\(60 = 2^2*3*5\)

So, We have \((61^2-1) = (61 + 1 )(61 - 1) = 2^3*3*5*31\)

(A) 8 = 2^3 (Divisible)
(B) 12 = 2^2*3 (Divisible)
(C) 15 = 3*5 (Divisible)
(D) 18 = 2*3^2
(E) 31 = 31 (Divisible)

Thus, Answer must be (D)
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If (61^2-1)/h is an integer, then h could be divisible by each... [#permalink]

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New post 08 Jun 2018, 16:09
If you don't see the shortcut immediately, you can calculate the expression and set it equal to a variable that represents an intenger (I).

(61^2 -1)/h = I

61^2-1 = h*I
3721-1 = h*I
3720= h*I

Prime factor 3720 = 2^3* 3*5*31

Now you can start crossing off answers that you are able to see

3*5=15
2^3= 8
31
2^2*3 = 12

The only # that can't be calculated is 18
Answer D
If (61^2-1)/h is an integer, then h could be divisible by each...   [#permalink] 08 Jun 2018, 16:09
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