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What are the last two digits of 63*35*37*82*71*41?

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What are the last two digits of 63*35*37*82*71*41? [#permalink]

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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 07 Apr 2015, 08:49
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80




Kudos for a correct solution.

63*35 = 05 *37 = 85 * 82 = 70 * 71 = 70*41 = 70
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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First, multiplied 82 and 35, since 2 and 5 will give a 0 in the end. Got last 2 digits 70.
no need to multiply 70 by 41 and 71, since they wont change the last 2 digits (try if you want to, but effective multiplication in case of 41 and 71 is only with 1).
Now, 70 * 3(of 63) gives : last 2 digits as 10.
lastly, 10 * 7(of 37) gives : last 2 digits 70.

Thus D.
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 09 Apr 2015, 13:31
63*35*37*82*71*41=

We have to focus on the last two digits only, so 63*35=05*37=85*82=70
71*41=81 therefore 81*70=70
Hence Answer is D

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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 13 Apr 2015, 07:27
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Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.

Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.

We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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Bunuel wrote:
What are the last two digits of (63)(35)(37)(82)(71)(41)?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80



(63)(35)(37)(82)(71)(41) = (63)(37)(71)(41)(35)(82)
Multiply the units digits in the red product to get: (#####1)(35)(82)
Rewrite blue product as follows: (#####1)(7)(5)(2)(41)
Rearrange: (#####1)(7)(41)(5)(2)
The units digit of (7)(41) is 7, so we get: (#####1)(##7)(10)
= (####7)(10)
= #####70

Answer:
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 05 Feb 2018, 00:10
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80


Kudos for a correct solution.


The equation can be written as \(7*9*7*5*37*41*2*71*41\)
\(49*9*10*37*41*71*41\) = \(401*10*37*41*41*71\)
As you can see, the last digit will be 0, and except for 37 all the digits end with 1.
So, last two Digit will be 70.
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 07 Feb 2018, 11:30
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.

Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.

We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)





Hi I thinks this

"Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70."

same as 1470/100 = 70 right ?

also Remainder 3*-3*-3*1*1*1/10 seem more efficient than 3*7*7*1*1*1/10 ? I hope this way is correct too ?
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 07 Feb 2018, 13:56
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.

Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.

We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)

Hi bunuel ..

Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc..

And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...

Sent from my BND-AL10 using GMAT Club Forum mobile app
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Re: What are the last two digits of 63*35*37*82*71*41? [#permalink]

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New post 09 Feb 2018, 08:37
viv007 wrote:
Bunuel wrote:
Bunuel wrote:
What are the last two digits of 63*35*37*82*71*41?

(A) 10
(B) 30
(C) 40
(D) 70
(E) 80


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

We know that to find the last two digits, we need to find the remainder we get when we divide the product by 100.

Remainder of (63*35*37*82*71*41)/ 100

Note that we can simplify this expression by canceling out the 5 and 2 in the numerator and denominator. But before we do that, here is an important note:

Note: We cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, if we want to cancel off to simplify the question, we can do it, provided we remember to multiply it back again.

So say, we want to find the remainder when 14 is divided by 10 i.e. 14/10 (remainder 4). But we cancel off the common 2 to get 7/5. The remainder here will be 2 which is not the same as the remainder obtained by dividing 14 by 10. But if we multiply 2 back by 2 (the number we canceled off), the remainder will become 2*2 = 4 which is correct.

Take another example to reinforce this – what is the remainder when 85 is divided by 20? It is 5.

We might rephrase it as – what is the remainder when 17 is divided by 4 (cancel off 5 from the numerator and the denominator). The remainder in this case is 1. We multiply the 5 back to 1 to get the remainder as 5 which is correct.

So keeping this very important point in mind, let’s go ahead and cancel the common 5 and 2.

We need the

Remainder of (63*7*37*41*71*41*5*2)/10*5*2

Remainder of (63*7*37*41*71*41)/10

Now using concept 2, let’s write the numbers in form of multiples of 10

Remainder of (60+3)*7*(30+7)*(40+1)*(70+1)*(40+1)/10

Remainder of 3*7*7*1*1*1/10

Remainder of 147/10 = 7

Now remember, we had canceled off 10 so to get the actual remainder so we need to multiply by 10: 7*10 = 70.

When 63*35*37*82*71*41 is divided by 100, the remainder is 70. So the last two digits of 63*35*37*82*71*41 must be 70.

Answer (D)

Hi bunuel ..

Can you enlighten how to approach this kind more other questions like to find last three digit..etc etc..

And in that solution to find last two digit how to approach if nothing is there in divisor means if 10 also divides over there...

Sent from my BND-AL10 using GMAT Club Forum mobile app



Hi Please see this

https://gmatclub.com/forum/compilation- ... ml#p650870
Re: What are the last two digits of 63*35*37*82*71*41?   [#permalink] 09 Feb 2018, 08:37
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