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Algebra :: M26-02

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Algebra :: M26-02  [#permalink]

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New post 02 Jan 2013, 06:53
2
25
What is the units digit of \((173)^4-1973^{3^2}\)

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.
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Re: Algebra :: M26-02  [#permalink]

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New post 02 Jan 2013, 07:07
4
LM wrote:
What is the units digit of \((173)^4-1973^{3^2}\)

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.

Because 1973^9 is greater than 173^4 so the difference will be negative... Final answer will be a negative number with units digit (3-1)=2... So answer will be B

Hope it helps!
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Re: Algebra :: M26-02  [#permalink]

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New post 22 Jun 2013, 19:48
LM wrote:
What is the units digit of \((173)^4-1973^{3^2}\)

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

I think the answer should be 8 (1-3 or 11-3=8). Unable to figure out why the official answer is not 8.


I made the same mistake, but the answer must be 2. One of the traps here is that second term is larger than first term, so the second term will "go on top" in standard subtraction. Think of this way too: if you multiply \(-1\) through the expression, then the result will also be 2.

I hope this helps
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Re: Algebra :: M26-02  [#permalink]

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New post 07 Mar 2014, 20:54
This is good question.

if the second number is smaller, then would it be 11-3 = 7?
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Re: Algebra :: M26-02  [#permalink]

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New post 08 Mar 2014, 07:20
2
8
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?


It would be 11-3=8.

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 that same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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: Algebra :: M26-02  [#permalink]

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New post 21 Apr 2014, 01:15
Bunuel wrote:
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?


It would be 11-3=8.

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 that same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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.


Bunnel,

Could you please explain following statement?

11-13=-2, which gives the final answer that the units digit of (17^3)^4-1973^{3^2} is 2

Thanks
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Re: Algebra :: M26-02  [#permalink]

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New post 21 Apr 2014, 01:32
2
PathFinder007 wrote:
Bunuel wrote:
Mountain14 wrote:
This is good question.

if the second number is smaller, then would it be 11-3 = 7?


It would be 11-3=8.

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 that same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is that 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 (cyclicity): 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=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (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 th units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclicty 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 then 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.


Bunnel,

Could you please explain following statement?

11-13=-2, which gives the final answer that the units digit of (17^3)^4-1973^{3^2} is 2

Thanks


(positive number ending with 1) - (greater number ending with 3) = (negative number ending with 2)

11-23=-12
31-133=-102
41-123=-82
....

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

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New post 21 Apr 2014, 01:47
Thanks Bunnel for your quick response.
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Re: Algebra :: M26-02   [#permalink] 21 Apr 2014, 01:47
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