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# (3508)^2 - (3510*3508) =

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Bunuel wrote:
(3508)² - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

This calls for some factoring

(3508)² - (3510*3508) = (3508)(3508) - (3510)(3508)
= 3508(3508 - 3510)
= 3508(-2)
= -7016

Originally posted by BrentGMATPrepNow on 23 Feb 2018, 09:37.
Last edited by BrentGMATPrepNow on 01 Jun 2020, 09:32, edited 1 time in total.
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Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

Another approach is to keep track of the units digits only

(3508)² - (3510*3508) = (---6) - (---0)
= -(----4)

Since answer choice E is the only answer with units digit 4, E must be correct.

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 23 Feb 2018, 09:42.
Last edited by BrentGMATPrepNow on 10 Apr 2021, 06:26, edited 2 times in total.
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Re: (3508)^2 - (3510*3508) = [#permalink]
Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

(3508)^2 - (3510*3508) = ?

3508 * (3508 - 3510)

3508 * (-2)

-7016

(E)
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Re: (3508)^2 - (3510*3508) = [#permalink]
Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

Let a = 3508 & b = 3510 Or, (a + 2)

So, (3508)^2 - (3510*3508) = a^2 - ab

Or, a^2 - ab = a ( a - b)

Or, a ( a - b) = a (a - a - 2)

Or, a ( a - b) = -2a

THus, Answer must be -2*3508 = -7016 , Answer must be (E)
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Re: (3508)^2 - (3510*3508) = [#permalink]
Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

Factor 3508 from both terms, obtaining:

3508(3508 - 3510)

3508(-2) = -7016

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Re: (3508)^2 - (3510*3508) = [#permalink]
BrentGMATPrepNow wrote:
Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

Another approach is to keep track of the units digits only

(3508)² - (3510*3508) = (---6) - (---0)
= ----6

Since answer choice E is the only answer with units digit 6, E must be correct.

Cheers,
Brent

Please could you help me understand how to calculate units place here? I tried solving through this method I am getting units place as 4
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Re: (3508)^2 - (3510*3508) = [#permalink]
Top Contributor
PoojaKGandhi wrote:
BrentGMATPrepNow wrote:
Bunuel wrote:
(3508)^2 - (3510*3508) =

(A) 7020

(B) 0

(C) -2

(D) -3508

(E) -7016

Another approach is to keep track of the units digits only

(3508)² - (3510*3508) = (---4) - (---0)
= -(----4)

Since answer choice E is the only answer with units digit 6, E must be correct.

Cheers,
Brent

Please could you help me understand how to calculate units place here? I tried solving through this method I am getting units place as 4

Oops, I made a mistake earlier (which I've corrected in the highlighted area above )

When multiplying two integers, the units digit of the entire product will be the same as the units digit of the product of the two units digits.
For example, we know that (7)(2) = 14
This means the product of something like (327)(122) will be -------4

Likewise, the units digit of 3508*3508 will be the unit's digit of (8)(8), which is 4
And the units digit of 3510*3508 will be the unit's digit of (0)(8), which is 0

So, we have (3508)² - (3510*3508) = (---4) - (---0)

Finally we need to recognize that (3508)² < (3510*3508)
So, when we subtract (3510*3508) from (3508)², we will get a NEGATIVE result.
To determine the units digit of this negative result, you can test two values, such that the first number has units digit 4, and the second number has units digit 0 (where the second number is bigger)

For example you could try 4 - 10 = -6 (units digit 6)
Or 4 - 30 = -26 (units digit 6)
Or 24 - 100 = -76 (units digit 6)
Etc
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Re: (3508)^2 - (3510*3508) = [#permalink]
BrentGMATPrepNow wrote:
PoojaKGandhi wrote:
BrentGMATPrepNow wrote:

Another approach is to keep track of the units digits only

(3508)² - (3510*3508) = (---4) - (---0)
= -(----4)

Since answer choice E is the only answer with units digit 6, E must be correct.

Cheers,
Brent

Please could you help me understand how to calculate units place here? I tried solving through this method I am getting units place as 4

Oops, I made a mistake earlier (which I've corrected in the highlighted area above )

When multiplying two integers, the units digit of the entire product will be the same as the units digit of the product of the two units digits.
For example, we know that (7)(2) = 14
This means the product of something like (327)(122) will be -------4

Likewise, the units digit of 3508*3508 will be the unit's digit of (8)(8), which is 4
And the units digit of 3510*3508 will be the unit's digit of (0)(8), which is 0

So, we have (3508)² - (3510*3508) = (---4) - (---0)

Finally we need to recognize that (3508)² < (3510*3508)
So, when we subtract (3510*3508) from (3508)², we will get a NEGATIVE result.
To determine the units digit of this negative result, you can test two values, such that the first number has units digit 4, and the second number has units digit 0 (where the second number is bigger)

For example you could try 4 - 10 = -6 (units digit 6)
Or 4 - 30 = -26 (units digit 6)
Or 24 - 100 = -76 (units digit 6)
Etc

Thank you so much. Now I am understanding better
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Re: (3508)^2 - (3510*3508) = [#permalink]
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Re: (3508)^2 - (3510*3508) = [#permalink]
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