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Re M2602 [#permalink]
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16 Sep 2014, 01:24
Official Solution:What is the units digit of \((17^3)^41973^{3^2}\)? 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 \(1113=2\), which gives the final answer that the units digit of \({(17^3)}^41973^{3^2}\) is 2. Answer: B
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Re: M2602 [#permalink]
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01 Nov 2014, 06:59
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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Re: M2602 [#permalink]
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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 = 13,841,267,51 8. 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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Re: M2602 [#permalink]
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10 Jan 2015, 05:55
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 = 13,841,267,51 8. 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: M2602 [#permalink]
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04 Jul 2015, 05:26
Bunuel wrote: Official Solution:
What is the units digit of \((17^3)^41973^{3^2}\)?
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 \(1113=2\), which gives the final answer that the units digit of \({(17^3)}^41973^{3^2}\) 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 13 = 2, but the answer is 2. Please if you can help. Cheers



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Re: M2602 [#permalink]
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05 Jul 2015, 08:36
anewbeginning wrote: Bunuel wrote: Official Solution:
What is the units digit of \((17^3)^41973^{3^2}\)?
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 \(1113=2\), which gives the final answer that the units digit of \({(17^3)}^41973^{3^2}\) 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 13 = 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: M2602 [#permalink]
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05 Jul 2015, 13:02
Bunuel wrote: anewbeginning wrote: Bunuel wrote: Official Solution:
What is the units digit of \((17^3)^41973^{3^2}\)?
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 \(1113=2\), which gives the final answer that the units digit of \({(17^3)}^41973^{3^2}\) 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 13 = 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 113 =8 why is that not possible



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Re: M2602 [#permalink]
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05 Jul 2015, 13:10
anewbeginning wrote: Bunuel wrote: Bunuel wrote: Official Solution:
What is the units digit of \((17^3)^41973^{3^2}\)?
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 \(1113=2\), which gives the final answer that the units digit of \({(17^3)}^41973^{3^2}\) 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 113 =8 why is that not possible Please read the highlighted part and this post: m26184441.html#p1467411
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Re: M2602 [#permalink]
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04 Feb 2016, 05:05
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 1111973 then the results ends in 2. If 11111  1973 > 8.
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Re: M2602 [#permalink]
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27 May 2016, 21:24
Thanks Bunuel, this is an excellent question!
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Re M2602 [#permalink]
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15 Jan 2017, 02:21
I think this is a highquality 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 213== then units digit is 8



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Re: M2602 [#permalink]
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17 Dec 2017, 02:44
We can write the the equation as 17^121973^6 or 7^123^6 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 (119 =2)
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