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M26-02 Updated on: 01 Jun 2025, 10:57
Originally posted by wanderingpoppy on 31 Jul 2024, 03:10.
Last edited by wanderingpoppy on 01 Jun 2025, 10:57, edited 1 time in total.
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M26-02 14 Nov 2024, 17:42
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Bunuel do you have any more similar problems?
Bunuel
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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M26-02 15 Nov 2024, 00:16
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braddouglas10
Bunuel do you have any more similar problems?
Bunuel
What is the units digit of \((17^3)^4-1973^{3^2}\)?

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

Similar questions to practice:
https://gmatclub.com/forum/what-is-the- ... 51159.html
https://gmatclub.com/forum/what-is-the- ... 16185.html

Hope it helps.
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M26-02 16 Nov 2024, 14:00
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Thank you Bunuel for taking time out of your busy day to help me!

Your answers to questions have been a blessing. This site owes you so much!
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braddouglas10
Bunuel do you have any more similar problems?
Bunuel
What is the units digit of \((17^3)^4-1973^{3^2}\)?

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

Similar questions to practice:
https://gmatclub.com/forum/what-is-the- ... 51159.html
https://gmatclub.com/forum/what-is-the- ... 16185.html

Hope it helps.
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M26-02 20 Jan 2025, 23:03
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I like the solution - it’s helpful.
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M26-02 05 Mar 2025, 09:03
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M26-02 Updated on: 24 Mar 2025, 06:57
Originally posted by Vineet1102 on 24 Mar 2025, 06:12.
Last edited by Vineet1102 on 24 Mar 2025, 06:57, edited 1 time in total.
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I still cannot understand the solution. I cannot understand why it is 2 and not 8?
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M26-02 24 Mar 2025, 06:42
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Vineet1102
I did not quite understand the solution. I cannot understand why it is 2 and not 8?
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We have:

(smaller positive number with the units digit of 1) - (greater positive number with the units digit of 3) =

= (a number with the units digit of 2).

For example, 1 - 3 = -2, 1 - 13 = -12, 21 - 53 = -32, ...
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M26-02 24 Mar 2025, 06:59
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Got it! Thanks, did not know about this detail
Bunuel
Vineet1102
I did not quite understand the solution. I cannot understand why it is 2 and not 8?

We have:

(smaller positive number with the units digit of 1) - (greater positive number with the units digit of 3) =

= (a number with the units digit of 2).

For example, 1 - 3 = -2, 1 - 13 = -12, 21 - 53 = -32, ...

Show more
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M26-02 24 Mar 2025, 07:04
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Vineet1102
Got it! Thanks, did not know about this detail
Bunuel
Vineet1102
I did not quite understand the solution. I cannot understand why it is 2 and not 8?

We have:

(smaller positive number with the units digit of 1) - (greater positive number with the units digit of 3) =

= (a number with the units digit of 2).

For example, 1 - 3 = -2, 1 - 13 = -12, 21 - 53 = -32, ...

Show more


Glad it helped. That point is also explained in detail in the official solution.



Bunuel
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 \(c^n\).

Hence, 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. \({(a^m)}^n=a^{mn}\) and \(a^{m^n}=a^{(m^n)}\).

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

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers 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!)

...

Hence, the units digit of \(17^{12}\) is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of \(1973^9\) is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of \({(17^3)}^4=17^{12}\) is 1, and the units digit of \(1973^{3^2}=1973^9\) is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were \(100^{12}\) instead of \(17^{12}\), and the second number were \(1000^9\) instead of \(1973^9\), the first number, \(100^{12}=10^{24}\), would still be smaller than the second number \(1,000^9=10^{27}\). Therefore, \(17^{12} < 1973^9\), and the units digit of the difference is 2.


Answer: B
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M26-02 03 Jul 2025, 02:27
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Bunuel

Remember the rule for working with stacked exponents, which is to begin with the highest exponent and work our way down.

3^3^2=3^(3^2)=3^9

May i ask what you mean with Working our way down, because it is very vague. The base is 3. The exponent 3 is raised to the exponent 2. When you say Working our way down, there is no Down by the first exponent 3, since it is the first exponent.

This is how interpret " Working our way down": 3^3^2. 2^3 is 8, thus 3^8. I donot get the wording of Working our way down.

How would you solve the following: 3^3^2^6^4, with 3 being the base and the stacked exponents being 3^2^6^4?

Thanks in advance!
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M26-02 03 Jul 2025, 02:38
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Rebaz
Bunuel

Remember the rule for working with stacked exponents, which is to begin with the highest exponent and work our way down.

3^3^2=3^(3^2)=3^9

May i ask what you mean with Working our way down, because it is very vague. The base is 3. The exponent 3 is raised to the exponent 2. When you say Working our way down, there is no Down by the first exponent 3, since it is the first exponent.

This is how interpret " Working our way down": 3^3^2. 2^3 is 8, thus 3^8. I donot get the wording of Working our way down.

How would you solve the following: 3^3^2^6^4, with 3 being the base and the stacked exponents being 3^2^6^4?

Thanks in advance!
Show more

This is very simple:

\(3^{3^2}\)

You start from the top: 3^2 = 9, and then calculate 3^9.

For \(5^{4^{3^2}}\), again start from the top: 3^2 = 9, so we get \(5^{4^9}\). Next, calculate 4^9, which is 262144, so we get \(5^{262144}\).
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M26-02 17 Aug 2025, 03:12
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M26-02 05 Oct 2025, 05:00
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I don’t quite agree with the solution. we arent asked "difference in magnitude", we shouldnt care which of the two numbers is bigger, we are asked ....1 - ....3 and that will be 8 no matter what.
if magnitude was asked then we would have done - If the second number is larger than the first, then we should flip the subtraction in the units place
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M26-02 10 Nov 2025, 08:22
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whollyshiv
I don’t quite agree with the solution. we arent asked "difference in magnitude", we shouldnt care which of the two numbers is bigger, we are asked ....1 - ....3 and that will be 8 no matter what.
if magnitude was asked then we would have done - If the second number is larger than the first, then we should flip the subtraction in the units place
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You are missing a key point. The expression in the question is (17^3)^4 - 1973^(3^2), which means the subtraction order is fixed. Since 1973^(3^2) is larger, the result is negative, so the units digit comes from 1 - 3 = -2, giving 2 as the final units digit. Please go and review the discussion carefully for more context.
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