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# M26-02

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Math Expert
Joined: 02 Sep 2009
Posts: 56244

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16 Sep 2014, 01:24
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20
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Difficulty:

75% (hard)

Question Stats:

42% (01:17) correct 58% (01:25) wrong based on 176 sessions

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

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

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Joined: 02 Sep 2009
Posts: 56244

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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.

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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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 56244

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01 Nov 2014, 07:22
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.
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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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10 Jan 2015, 05:55
2
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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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.

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.

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
Math Expert
Joined: 02 Sep 2009
Posts: 56244

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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.

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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04 Feb 2016, 05:05
3
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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27 May 2016, 21:24
Thanks Bunuel, this is an excellent question!
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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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15 Jan 2017, 02:25
ankitsa wrote:
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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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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28 Aug 2017, 10:25
deepudiscover wrote:
Hi

I want to know why not 8. for ex if we have 21-3== then units digit is 8

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

Sent from my SM-G950F using GMAT Club Forum mobile app
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21 Jul 2018, 00:28
SemNoah wrote:
Is this a 750 question?

Sent from my SM-G950F using GMAT Club Forum mobile app

Yes, it's a hard question.
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Re: M26-02   [#permalink] 21 Jul 2018, 00:28

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