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# What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y

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What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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Updated on: 23 Apr 2017, 11:35
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What is the last non-zero digit of this expression if x and y are positive integers?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

Spoiler: :: Edit - 4-23-17
Edited to fix a loophole as mentioned by chetan2u. Sorry about it because I did not anticipate the additional possibilities.

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Originally posted by umg on 23 Apr 2017, 08:14.
Last edited by umg on 23 Apr 2017, 11:35, edited 2 times in total.
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What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 09:01
5
5
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

Hi..

Q should have mentioned that x and and y are positive integers

We are looking for last non zero digit...
It will depend on the last non-zero digit of each term
Example...
$$260*3*40$$ would depend on $$6*3*4...$$
$$260^4*27^6*800^{567}$$ would depend on $$6^4*7^6*8^{567}$$
Let's see the statements..

(1) $$x = 10, y = 18$$
X and y are given, so we will be able to find exact value of equation.
Sufficient

(2) $$y^2 - 36y + 324= 0$$
This equation comes down to $$(y-18)^2=0$$ or y=18..
But what about x? This is the tricky part.

We are looking for LAST non-zero digit of each term.
Rest all terms are known except x as y is 18
Here the term is 860^x and it will be same as 6^x.
Since the last non-zero digit is 6 here and 6 to any power will have last digit as 6 only, the value of x will not affect the last non-zero digit of entire term..
6^1=6.....
6^2=36.... Last digit 6
6^3=216.... Last digit again 6
Sufficient

D
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 08:24
imo A

You will need to know the value of both x and y.

how is it OA=d?

what if x is 1/2 and 2? won't we get different answers?
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 08:56
This is tricky. If we're just looking at the last digits, $$860^x$$ should give us a last digit of 0 for all x except x=0, since anything to the zeroth power equals 1. So for all x except x=0. the last digit of the whole expression should be zero. I don't see how the second option ($$y^2−36y+324=0$$) tells us that x isn't zero. But then I guess the last non-zero digit would be the tens digit, so clearly I'm still missing something...
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 09:57
chetan2u wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

Hi..

Q should have mentioned that x and and y are positive integers

We are looking for last non zero digit...
It will depend on the last non-zero digit of each term
Example...
260*3*40 would depend on 6*3*4...
260^4*27^6*800^567 would depend on 6^4*7^6*8^567
Let's see the statements..

(1) $$x = 10, y = 18$$
X and y are given, so we will be able to find exact value of equation.
Sufficient

(2) $$y^2 - 36y + 324= 0$$
This equation comes down to (y-18)^2=0 or y=18..
But what about x? This is the tricky part.

We are looking for LAST non-zero digit of each term.
Rest all terms are known except x as y is 18
Here the term is 860^x and it will be same as 6^x.
Since the last non-zero digit is 6 here and 6 to any power will have last digit as 6 only, the value of x will not affect the last non-zero digit of entire term..
6^1=6.....
6^2=36.... Last digit 6
6^3=216.... Last digit again 6
Sufficient

D

what if x=1/2...then LAST non-zero digit 860^x wont be 6
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Posts: 7956
Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 10:35
sandaki wrote:
chetan2u wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$
Ki
(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

Hi..

Q should have mentioned that x and and y are positive integers

We are looking for last non zero digit...
It will depend on the last non-zero digit of each term
Example...
260*3*40 would depend on 6*3*4...
260^4*27^6*800^567 would depend on 6^4*7^6*8^567
Let's see the statements..

(1) $$x = 10, y = 18$$
X and y are given, so we will be able to find exact value of equation.
Sufficient

(2) $$y^2 - 36y + 324= 0$$
This equation comes down to (y-18)^2=0 or y=18..
But what about x? This is the tricky part.

We are looking for LAST non-zero digit of each term.
Rest all terms are known except x as y is 18
Here the term is 860^x and it will be same as 6^x.
Since the last non-zero digit is 6 here and 6 to any power will have last digit as 6 only, the value of x will not affect the last non-zero digit of entire term..
6^1=6.....
6^2=36.... Last digit 6
6^3=216.... Last digit again 6
Sufficient

D

what if x=1/2...then LAST non-zero digit 860^x wont be 6

Hi..
Yes, you are right.
Pl look at the coloured portion above.
Not only fraction but also negative integer.

It has to be ONLY POSITIVE integer
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 11:02
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

We don't even need choices in this question ...... if it's given in the question that x and y are positive number.
Since 860's last non zero digit will always be 6 only; likewise, 1525's last digit will always be 5 irrespective of the values of x and y.

Easy D.
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What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 12:01
2
gmatexam439 wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

We don't even need choices in this question ...... if it's given in the question that x and y are positive number.
Since 860's last non zero digit will always be 6 only; likewise, 1525's last digit will always be 5 irrespective of the values of x and y.

Easy D.

Nopes.
Not so easy.
Firstly, i too thought the same when i first saw this question.
Easy D.

But there is a trap here.
Let us examine it closely.
Let us consider that both x and y are greater then zero and positive integers.
Let us remove the zero from the 860^x to make it 86^x (as the last digit would be zero.)

Now you see that the last term of the product i.e ((1525)^y will always end with 5.
Hence the last digit of the entire product would still be zero as 5*even ends with zero.
So we need the digit previous to that.
For that we need the value of x.

So D is sufficient.
But your logic wasn't all right.

Off course i might be wrong.

chetan2u is my approach correct ?

P.S => I dont really think this is a GMAT-type question.

Regards
Stone Cold
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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23 Apr 2017, 17:25
1
stonecold wrote:
gmatexam439 wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

We don't even need choices in this question ...... if it's given in the question that x and y are positive number.
Since 860's last non zero digit will always be 6 only; likewise, 1525's last digit will always be 5 irrespective of the values of x and y.

Easy D.

Nopes.
Not so easy.
Firstly, i too thought the same when i first saw this question.
Easy D.

But there is a trap here.
Let us examine it closely.
Let us consider that both x and y are greater then zero and positive integers.
Let us remove the zero from the 860^x to make it 86^x (as the last digit would be zero.)

Now you see that the last term of the product i.e ((1525)^y will always end with 5.
Hence the last digit of the entire product would still be zero as 5*even ends with zero.
So we need the digit previous to that.
For that we need the value of x.

So D is sufficient.
But your logic wasn't all right.

Off course i might be wrong.

chetan2u is my approach correct ?

P.S => I dont really think this is a GMAT-type question.

Regards
Stone Cold

Hi..
You are correct on THAT..
We have some number of 5s in the term...Say 20
If you have less number of 2s in the remaining terms..Say 19 last digit will always be 5..
But if it's more than 20, then ofcourse all the 5s would be converted into 0s and we would require to know digit prior to 0 and the condition we are talking of..

So two cases..
1) number of 5s is less than number of 2s
We will require to know the exact value of y.
2) the number of 5s is MORE than number of 2s..
We don't require exact value as Nonzero digit will be 5 irrespective of value of y.
But yes you have to know that y is less than the power of 2s.
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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24 Apr 2017, 00:36
chetan2u wrote:
chetan2u wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$
Ki
(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

Hi..

Q should have mentioned that x and and y are positive integers

We are looking for last non zero digit...
It will depend on the last non-zero digit of each term
Example...
260*3*40 would depend on 6*3*4...
260^4*27^6*800^567 would depend on 6^4*7^6*8^567
Let's see the statements..

(1) $$x = 10, y = 18$$
X and y are given, so we will be able to find exact value of equation.
Sufficient

(2) $$y^2 - 36y + 324= 0$$
This equation comes down to (y-18)^2=0 or y=18..
But what about x? This is the tricky part.

We are looking for LAST non-zero digit of each term.
Rest all terms are known except x as y is 18
Here the term is 860^x and it will be same as 6^x.
Since the last non-zero digit is 6 here and 6 to any power will have last digit as 6 only, the value of x will not affect the last non-zero digit of entire term..
6^1=6.....
6^2=36.... Last digit 6
6^3=216.... Last digit again 6
Sufficient

D

Hi..
Yes, you are right.
Pl look at the coloured portion above.
Not only fraction but also negative integer.

It has to be ONLY POSITIVE integer

True,
if question had mentioned that x and and y are positive integers,then answer would have been D.

Since it's not mentioned,the answer is A.
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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24 Apr 2017, 03:03
sandaki wrote:

True,
if question had mentioned that x and and y are positive integers,then answer would have been D.

Since it's not mentioned,the answer is A.

Because I made this question, I did not anticipate other possible values of x & y, it seems that I have gotten very rusty these past few months. However, I realized the mistake and have fixed the error. The question now mentions that x and y are positive integers.
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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24 Apr 2017, 13:59
stonecold wrote:
gmatexam439 wrote:
umg wrote:
What is the last non-zero digit of this expression?

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y$$

(1) $$x = 10, y = 18$$

(2) $$y^2 - 36y + 324= 0$$

We don't even need choices in this question ...... if it's given in the question that x and y are positive number.
Since 860's last non zero digit will always be 6 only; likewise, 1525's last digit will always be 5 irrespective of the values of x and y.

Easy D.

Nopes.
Not so easy.
Firstly, i too thought the same when i first saw this question.
Easy D.

But there is a trap here.
Let us examine it closely.
Let us consider that both x and y are greater then zero and positive integers.
Let us remove the zero from the 860^x to make it 86^x (as the last digit would be zero.)

Now you see that the last term of the product i.e ((1525)^y will always end with 5.
Hence the last digit of the entire product would still be zero as 5*even ends with zero.
So we need the digit previous to that.
For that we need the value of x.

So D is sufficient.
But your logic wasn't all right.

Off course i might be wrong.

chetan2u is my approach correct ?

P.S => I dont really think this is a GMAT-type question.

Regards
Stone Cold

Thank you stonecold for pointing that out. and thank you chetan2u for the explanation !!
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What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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28 Apr 2017, 00:14
This question is doing my brain in. Please let me know if my thinking is wrong.

3^327: the last digit of this is 7
4^36: The last digit of this is 6
6^x: The last digit of this is 6
5^y: The last digit of this is 5

Since 7 x 6 x 6 x 5 ends in zero, we need the second last digit of 1525^y. Hence we only need the value of Y to solve the question. Both statements give the value of Y hence D.

Is this correct?
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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28 Apr 2017, 13:43
1
texas wrote:
This question is doing my brain in. Please let me know if my thinking is wrong.

3^327: the last digit of this is 7
4^36: The last digit of this is 6
6^x: The last digit of this is 6
5^y: The last digit of this is 5

Since 7 x 6 x 6 x 5 ends in zero, we need the second last digit of 1525^y. Hence we only need the value of Y to solve the question. Both statements give the value of Y hence D.

Is this correct?

Here is the safest way to approach such questions

$$(573)^3^2^7*(274)^3^6*(860)^x*(1525)^y= (573)^3^2^7*(137*2)^3^6*(86*10)^x*(61*25)^y$$

Notice that 2^(36) will contribute zeros depending upon the value of y.

Hence, if 2y = 36

the last non-zero digit will be decided by

3^(327) * 7^(36) * 6^(x) * 1^(y)
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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y  [#permalink]

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Re: What is the last non-zero digit of (573)^(327)*(274)^(36)*860^x*1525^y   [#permalink] 19 Feb 2019, 10:00
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