M26-02

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 \(c^n\).

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

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:

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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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.

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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Thanks Bunuel, this is an excellent question!

Thanks ! This is one gem of a question
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I think this is a high-quality question and I agree with explanation. Deadly question

I think this is a high-quality question and I agree with explanation.

Bunuel GMATinsight chetan2u

https://gmatclub.com/forum/m35-311776.html
In this question 5-6 => 9

I saw 1 & 3 and did 1-3 => 8.

Can you please explain this?

(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, ...

(greater positive number with the units digit of 1) - (smaller positive number with the units digit of 3) = (a number with the units digit of 8). For example, 11 - 3 = 8, 21 - 3 = 18, 81 - 53 = 28, ...


In this question: https://gmatclub.com/forum/m35-311776.html The same logic.

(smaller positive number with the units digit of 6) - (greater positive number with the units digit of 5) = (a number with the units digit of 9). For example, 6 - 15 = -9, 16 - 35 = -19, 26 - 55 = -29, ...

(greater positive number with the units digit of 6) - (smaller positive number with the units digit of 5) = (a number with the units digit of 1). For example, 6 - 5 = 1, 16 - 5 = 11, 86 - 15 = 71, ...
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Hello, everyone. I solved the question using the same cyclisity (a new word to me) framework that appears in the official solution—7: 7, 9, 3, 1 and 3: 3, 9, 7, 1. But I did actually calculate one of the values:

\(17^3 = 289 * 17 = 4913\)

The difference was then easier to gauge:

\((4913)^4-1973^9\)

4913 is close enough to 4900, or 70 squared.

\((70^2)^4-1973^9\)

\(70^8-1973^9\)

The two values are not even close, and the second must be larger. Thus, whatever the actual values might be, it has to be the same as, say, 21 - 33, meaning the units digit has to be 2, (B).

I do not trust myself enough to intuit what will be larger when I am dealing with gigantic numbers, so I took a little comfort in approximation. I would rather solve the question in a less elegant way and walk away with a correct answer than stumble into an unfounded assumption, and I am guessing you would feel the same way. (Plus, the above process did not take too long.)

Good luck with your studies.

- Andrew
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I think this is a high-quality question and I agree with explanation.

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Why is 1973 to the power of 9? Shouldn’t it be to the power of 6 as 3*2=6?

Remember the rule for working with stacked exponents, which is to begin with the highest exponent and work our way down. For example, \(a^{m^n}\) means we first compute \(m^n\) and then use that result as the exponent for \(a\). Therefore, \(a^{m^n} = a^{(m^n)}\). On the other hand, if we have \((a^m)^n\), we compute \(a^m\) first and then raise the result to the power of \(n\), so \((a^m)^n=a^{mn}\).

Therefore, since \(a^{m^n} = a^{(m^n)}\), then \(1973^{3^2}=1973^{(3^2)} =1973^{9}\).

If it were \((1973^3)^2\), then because \((a^m)^n=a^{mn}\), it would have been equal to \(1973^{3*2}=1973^{6}\).

Hope it helps.
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Thank you for the clarification.

I'm super confused about this portion:


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


I don't get why [m]3^{3^2} is not equal to 3^6

I think this is explained above in this thread but here it is again

Remember the rule for working with stacked exponents, which is to begin with the highest exponent and work our way down. For example, \(a^{m^n}\) means we first compute \(m^n\) and then use that result as the exponent for \(a\). Therefore, \(a^{m^n} = a^{(m^n)}\).

So, \(3^{3^2}=3^{(3^2)}=3^9\).

On the other hand, if we have \((a^m)^n\), we compute \(a^m\) first and then raise the result to the power of \(n\), so \((a^m)^n=a^{mn}\).

So, if it were \((3^3)^2\), then it would be \((3^3)^2=3^{3*2}=3^6\).

Hope it's clear.
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Please tell me if my thinking is correct here:
I was confused about the second number being larger or not than the second number, so I found out the nearest perfect square which was 44*44=1936. For a second considering the second number to be 1936 instead of 1973, I went on to understand this as 44^18.
Thus, 17^12-44^18, hence clearly the second number is bigger. Bunuel