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M26-02

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M26-02  [#permalink]

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New post 16 Sep 2014, 01:24
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Re M26-02  [#permalink]

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New post 16 Sep 2014, 01:24
6
11
Official Solution:

What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


Must know for the GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\), which means that the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).

So:

\({(a^m)}^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\).

Thus, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. \(7^1=7\) (last digit is 7)

2. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...

1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Thus, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of \(3^9\) will be 3 (first in pattern, as \(9=4*2+1\)).

So, we have that the units digit of \({(17^3)}^4=17^{12}\) is 1 and the units digit of \(1973^{3^2}=1973^9\) is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like \(11-13=-2\), which gives the final answer that the units digit of \({(17^3)}^4-1973^{3^2}\) is 2.


Answer: B
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Re: M26-02  [#permalink]

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New post 01 Nov 2014, 06:59
1
Good Question, with too many important concepts. I am wondering what will be the Units digit if the second number was not smaller than 1st

Ex: (117^12)-(73^9)

Will the units digit be 8 ?
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New post 01 Nov 2014, 07:22
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Re: M26-02  [#permalink]

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New post 10 Jan 2015, 05:28
Bunuel wrote:
awal_786@hotmail.com wrote:
Good Question, with too many important concepts. I am wondering what will be the Units digit if the second number was not smaller than 1st

Ex: (117^12)-(73^9)

Will the units digit be 8 ?


Yes, then the units digit would be 8. For example, 7^12 - 3^9 = 13,841,267,518.



But how would we know whether the 2nd no is smaller than the 1st unless we actually compute it.

Wouldn't this actually be very time consuming ?
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New post 10 Jan 2015, 05:55
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buddyisraelgmat wrote:
Bunuel wrote:
awal_786@hotmail.com wrote:
Good Question, with too many important concepts. I am wondering what will be the Units digit if the second number was not smaller than 1st

Ex: (117^12)-(73^9)

Will the units digit be 8 ?


Yes, then the units digit would be 8. For example, 7^12 - 3^9 = 13,841,267,518.



But how would we know whether the 2nd no is smaller than the 1st unless we actually compute it.

Wouldn't this actually be very time consuming ?


Why is second number much larger then the first one? Consider this, even if we had (100^3)^4 (instead of (17^3)^4) and 1000^(3^2) (instead of 1973^(3^2)) --> (100^3)^4=100^12=10^24 and 1000^(3^2)=1,000^9=10^27.

Hope it's clear.
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Re: M26-02  [#permalink]

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New post 04 Jul 2015, 05:26
Bunuel wrote:
Official Solution:

What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


Must know for the GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\), which means that the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).

So:

\({(a^m)}^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\).

Thus, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. \(7^1=7\) (last digit is 7)

2. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...

1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Thus, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of \(3^9\) will be 3 (first in pattern, as \(9=4*2+1\)).

So, we have that the units digit of \({(17^3)}^4=17^{12}\) is 1 and the units digit of \(1973^{3^2}=1973^9\) is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like \(11-13=-2\), which gives the final answer that the units digit of \({(17^3)}^4-1973^{3^2}\) is 2.


Answer: B


Thanks Bunuel for the question.

While, I can understand that the unit's digit will be 1 and 3 , but I don't get this that the difference will be 2.

Since the second number is larger, the difference in my case is coming out as 1-3 = -2, but the answer is 2.

Please if you can help.

Cheers
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Re: M26-02  [#permalink]

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New post 05 Jul 2015, 08:36
anewbeginning wrote:
Bunuel wrote:
Official Solution:

What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


Must know for the GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\), which means that the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).

So:

\({(a^m)}^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\).

Thus, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. \(7^1=7\) (last digit is 7)

2. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...

1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Thus, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of \(3^9\) will be 3 (first in pattern, as \(9=4*2+1\)).

So, we have that the units digit of \({(17^3)}^4=17^{12}\) is 1 and the units digit of \(1973^{3^2}=1973^9\) is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like \(11-13=-2\), which gives the final answer that the units digit of \({(17^3)}^4-1973^{3^2}\) is 2.


Answer: B


Thanks Bunuel for the question.

While, I can understand that the unit's digit will be 1 and 3 , but I don't get this that the difference will be 2.

Since the second number is larger, the difference in my case is coming out as 1-3 = -2, but the answer is 2.

Please if you can help.

Cheers


The question asks about the units digit of the difference, which is 2 not -2, which cannot be.
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Re: M26-02  [#permalink]

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New post 05 Jul 2015, 13:02
1
Bunuel wrote:
anewbeginning wrote:
Bunuel wrote:
Official Solution:

What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


Must know for the GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\), which means that the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).

So:

\({(a^m)}^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\).

Thus, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. \(7^1=7\) (last digit is 7)

2. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...

1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Thus, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of \(3^9\) will be 3 (first in pattern, as \(9=4*2+1\)).

So, we have that the units digit of \({(17^3)}^4=17^{12}\) is 1 and the units digit of \(1973^{3^2}=1973^9\) is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like \(11-13=-2\), which gives the final answer that the units digit of \({(17^3)}^4-1973^{3^2}\) is 2.


Answer: B


Thanks Bunuel for the question.

While, I can understand that the unit's digit will be 1 and 3 , but I don't get this that the difference will be 2.

Since the second number is larger, the difference in my case is coming out as 1-3 = -2, but the answer is 2.

Please if you can help.

Cheers


The question asks about the units digit of the difference, which is 2 not -2, which cannot be.



What if we consider it as 1 and 3 so the carry over will yield 11-3 =8

why is that not possible
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New post 05 Jul 2015, 13:10
anewbeginning wrote:
Bunuel wrote:
Bunuel wrote:
Official Solution:

What is the units digit of \((17^3)^4-1973^{3^2}\)?

A. 0
B. 2
C. 4
D. 6
E. 8


Must know for the GMAT:

I. The units digit of \((abc)^n\) is the same as that of \(c^n\), which means that the units digit of \((17^3)^4\) is the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the same as that of \(3^{3^2}\).

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus: \(a^{m^n}=a^{(m^n)}\) and not \({(a^m)}^n\), which on the other hand equals to \(a^{mn}\).

So:

\({(a^m)}^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\).

Thus, \({(7^3)}^4=7^{(3*4)}=7^{12}\) and \(3^{3^2}=3^{(3^2)}=3^9\).

III. The units digit of integers in positive integer power repeats in specific pattern (cyclisity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. \(7^1=7\) (last digit is 7)

2. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (last digit is 7 again!)

...

1. \(3^1=3\) (last digit is 3)

2. \(3^2=9\) (last digit is 9)

3. \(3^3=27\) (last digit is 7)

4. \(3^4=81\) (last digit is 1)

5. \(3^5=243\) (last digit is 3 again!)

...

Thus, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty number 4) and the units digit of \(3^9\) will be 3 (first in pattern, as \(9=4*2+1\)).

So, we have that the units digit of \({(17^3)}^4=17^{12}\) is 1 and the units digit of \(1973^{3^2}=1973^9\) is 3. Also notice that the second number is much larger than the first one, thus their difference will be negative, something like \(11-13=-2\), which gives the final answer that the units digit of \({(17^3)}^4-1973^{3^2}\) is 2.


Answer: B



The question asks about the units digit of the difference, which is 2 not -2, which cannot be.



What if we consider it as 1 and 3 so the carry over will yield 11-3 =8

why is that not possible


Please read the highlighted part and this post: m26-184441.html#p1467411
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Re: M26-02  [#permalink]

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New post 04 Feb 2016, 05:05
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when a question trhows at u a trick like the one above with the second number larger than the first one try to deconstruct the ambiguity with simple numbers. If 111-1973 then the results ends in 2. If 11111 - 1973 ---> 8.
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New post 27 May 2016, 21:24
Thanks Bunuel, this is an excellent question!
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New post 15 Jan 2017, 02:21
I think this is a high-quality question and I don't agree with the explanation. Answer is 8, since unit digit for 1st number is 1 and for second number is 3, so while subtracting 1 carry over will come, which results in 11 -3 = 8.
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New post 15 Jan 2017, 02:25
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New post 28 Aug 2017, 10:23
Hi

I want to know why not 8. for ex if we have 21-3== then units digit is 8
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New post 28 Aug 2017, 10:25
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New post 17 Dec 2017, 02:44
We can write the the equation as
17^12-1973^6
or
7^12-3^6
now 7 has the cycle of 4. therefore the unit digit is given by 12/4 => reminder is 0 or the 4th term of 7's cycle i.e 1
similarly 3 has the cycle of 4. therefore the unit digit is given by 6/4 => reminder is 2 or the 2nd term of 3's cycle i.e 9
therefore unit digit is given by (11-9 =2)
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New post 20 Jul 2018, 07:52
Hi, this is my appoach 17^3 will have the same units digit as 7.7.7 which is 3 then 3^4 will have the same units digit as 3.3.3.3 which is 1 ( keep this in mind) , now the second part 1973^3^2 will have the same units digit as 3^3^2 , 3^3 will have a units digit 7 then 7^2 will have a units digit of 9.
So 1-9 in subtraction we borrow 1 to the first one will be 11-9 is 2, so answer is 2.

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New post 20 Jul 2018, 15:47
Is this a 750 question?

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New post 21 Jul 2018, 00:28
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Re: M26-02   [#permalink] 21 Jul 2018, 00:28

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