Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 60460

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
20 Jul 2017, 22:06
Question Stats:
15% (02:17) correct 85% (01:48) wrong based on 130 sessions
HideShow timer Statistics
If a, b, c, d, e, f, g, h, i and j are distinct positive integers, what is the units digit of the lowest possible value of \(a^b * c^d * e^f * g^h * i^j\)? (A) 0 (B) 1 (C) 2 (D) 4 (E) 8
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3296
Location: India
GPA: 3.12

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
20 Jul 2017, 22:49
Given : a, b, c, d, e, f, g, h, i and j are distinct positive integers. Since we have been asked to find the lowest positive integer values of expression \(a^b * c^d * e^f * g^h * i^j\) Please find below the cyclicity of all numbers : Number^1^2^3^4Cyclicity 2 24864 3 39714 4 46462 5 55551 6 66661 7 79314 8 84264 9 91912The choice of numbers have to be such that each of the numbers is repeated once and the total value of the expression is the least. This is possible when the choice of numbers is a=1,b=10,c=9,d=2,e=8,f=3,g=7,h=4,i=6,j=5. The expression \(a^b * c^d * e^f * g^h * i^j\) becomes \(1^{10} * 9^2 * 8^3 * 7^4 * 6^5\) Since we are only concerned about the units digit of this expression \(1^{10} * 9^2 * 8^3 * 7^4 * 6^5\) will have units digits \(1 * 1 * 2 * 1 * 6 = 2 * 6 = 2\) Therefore, units digit of the expression will be 2(Option C).
_________________
You've got what it takes, but it will take everything you've got



Manager
Joined: 02 Mar 2017
Posts: 71
GMAT 1: 700 Q51 V34

Re: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
21 Jul 2017, 07:05
pushpitkc wrote: Given : a, b, c, d, e, f, g, h, i and j are distinct positive integers.
Since we have been asked to find the lowest positive integer values of expression \(a^b * c^d * e^f * g^h * i^j\)
Please find below the cyclicity of all numbers : Number^1^2^3^4Cyclicity 2 24864 3 39714 4 46462 5 55551 6 66661 7 79314 8 84264 9 91912
The choice of numbers have to be such that each of the numbers is repeated once and the total value of the expression is the least. This is possible when the choice of numbers is a=1,b=9,c=2,d=8,e=3,f=7,g=4,h=6,i=5,j=0.
The expression \(a^b * c^d * e^f * g^h * i^j\) becomes \(1^9 * 2^8 * 3^7 * 4^6 * 5^0\) Since we are only concerned about the units digit of this expression \(1^9 * 2^8 * 3^7 * 4^6 * 5^0\) will have units digits \(1 * 6 * 7 * 6 * 1 = 2 * 6 = 2\)
Therefore, units digit of the expression will be 2(Option C). J should not be 0 ( all are positive integer). Hence a=1, b=10, c=2, d=9, e=3, f=8, g=4, h=7, i=5, j=6. ( I hope our thinking that lowest base highest power is the correct way to go). using this and your cyclic table answer should be A=0



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3296
Location: India
GPA: 3.12

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
21 Jul 2017, 07:58
pulkitaggi wrote: pushpitkc wrote: Given : a, b, c, d, e, f, g, h, i and j are distinct positive integers.
Since we have been asked to find the lowest positive integer values of expression \(a^b * c^d * e^f * g^h * i^j\)
Please find below the cyclicity of all numbers : Number^1^2^3^4Cyclicity 2 24864 3 39714 4 46462 5 55551 6 66661 7 79314 8 84264 9 91912
The choice of numbers have to be such that each of the numbers is repeated once and the total value of the expression is the least. This is possible when the choice of numbers is a=1,b=9,c=2,d=8,e=3,f=7,g=4,h=6,i=5,j=0.
The expression \(a^b * c^d * e^f * g^h * i^j\) becomes \(1^9 * 2^8 * 3^7 * 4^6 * 5^0\) Since we are only concerned about the units digit of this expression \(1^9 * 2^8 * 3^7 * 4^6 * 5^0\) will have units digits \(1 * 6 * 7 * 6 * 1 = 2 * 6 = 2\)
Therefore, units digit of the expression will be 2(Option C). J should not be 0 ( all are positive integer). Hence a=1, b=10, c=2, d=9, e=3, f=8, g=4, h=7, i=5, j=6. ( I hope our thinking that lowest base highest power is the correct way to go). using this and your cyclic table answer should be A=0 I had missed that 0 is not a positive number. I have made the correction in the solution. One thing i noticed about numbers is that \(9^2 < 2^9, 8^3 < 3^8, 7^4 < 4^7, and 6^5 < 5^6\) So, I suppose that's something we had missed.
_________________
You've got what it takes, but it will take everything you've got



Senior SC Moderator
Joined: 22 May 2016
Posts: 3734

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
Updated on: 22 Jul 2017, 18:41
Bunuel wrote: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, what is the units digit of the lowest possible value of \(a^b * c^d * e^f * g^h * i^j\)?
(A) 0 (B) 1 (C) 2 (D) 4 (E) 8 I got Answer A. If we want the lowest possible value, pushpitkc , I think it's exactly the reverse of your method (which is where I started, and then changed my mind). Exponential growth is huge. So I paired largest base with smallest exponent, second largest base with second smallest exponent, and so on. We need the big factors to be multiplied the fewest times. Think of 10\(^2\) = 100 compared to 2\(^{10}\) = 1,024. We can only use the smallest first ten digits once each. So I got 10\(^1\) 9\(^2\) 8\(^3\) 7\(^4\) 6\(^5\) Where a=10, b=1, c=9, d=2, e=8, f=3, g=7, h=4, i=6, j=5 We can pay attention to just the units' digits of these factors of what I am hoping is the lowest value possible for those digits in that expression. 10\(^1\) = units digit 0 9\(^2\) = units digit 1 8\(^3\) = units digit 2 7\(^4\) = units digit 1 6\(^5\) = units digit 6 0*1*2*1*6 = 0 Let's compare the factors, rounded to the nearest hundred Large Base Small Exponent: 10\(^1\) = 10 9\(^2\) = 80 8\(^3\) = 500 7\(^4\) = 2,400 6\(^5\) = 8,000 Small Base Large Exponent: 1\(^{10}\) = 1 2\(^9\) = 500 3\(^8\) = 6500 4\(^7\) = 16,000 5\(^6\) = 15,600 The number of zeros in the second group already tells us that that number is larger. But take just the first three, then the last two, factors in each list and multiply Large Base Small Exponent: 10 * 80 * 500 = 400,000 multiplied by 2,400 * 8,000 = 19,200,000Small Base Large Exponent: 1 * 500 * 6,500 = 3,250,000multiplied by 15,600 * 16,000 = 249,000,000The first set of factors, multiplied, produces a much smaller value than the second set. There's no way to test those numbers on the exam. But I tested 2\(^{10}\) vs. 10\(^2\) and 3\(^4\) (81) vs. 4\(^3\) (64). They were all calculable. The numbers confirmed my instinct: to get a lower value, multiply the big values as few times as possible. Make the exponents small numbers. So 10\(^1\) * 9\(^2\) * 8\(^3\) * 7\(^4\) * 6\(^5\) yield units digit: 0*1*2*1*6 = 0 Answer A
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has  Margaret Mead
Originally posted by generis on 22 Jul 2017, 17:52.
Last edited by generis on 22 Jul 2017, 18:41, edited 1 time in total.



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3296
Location: India
GPA: 3.12

Re: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
22 Jul 2017, 18:18
I agree @genxer123 But that doesn't apply to each number combination. 1^10 is smaller than 10^1. So number got by multiplying X with 1^10 will definitely be smaller than X multiplied by 10^1. That is why the units digit must be 2, not 0.
_________________
You've got what it takes, but it will take everything you've got



Senior SC Moderator
Joined: 22 May 2016
Posts: 3734

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
22 Jul 2017, 19:42
pushpitkc wrote: I agree genxer123But that doesn't apply to each number combination. 1^10 is smaller than 10^1. So number got by multiplying X with 1^10 will definitely be smaller than X multiplied by 10^1. That is why the units digit must be 2, not 0. Hmm ... Aren't we looking for the product of all those x\(^y\) "number combinations"? I'm not following you. Perhaps this question is phrased awkwardly. With respect to the "number combinations" (and elsewhere where you say we want the "total value of the expression [that] is the least"), while it is true that 10X is much bigger than 1X, all the other number combinations / expressions are the other way: 2^9 > 9^2 3^8 > 8^3 4^7 > 7^4 5^6 > 6^5 So are you arguing that by virtue of 10 > 1, the rest of the number combinations don't matter? Bunuel wrote Quote: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, what is the units digit of the lowest possible value of \(a^b * c^d * e^f * g^h * i^j\) I interpret "what is the units digit of the lowest possible value of \(a^b * c^d ...\)]" to mean "what is the units digit of the smallest number you can derive from multiplying these factors"? Substitute Q, R, S, T and U for each term of the form \(a^b\). Put another way, I think the question asks, "What is the smallest number you can make by multiplying Q x R x S x T x U, and once you have found that number, what is its units' digit?" If that is true, then \(10^1 * 9^2 * 8^3 * 7^4 * 6^5\) = 10 x 81 x 512 x 2401 x 7776 = 7,742,895,390,720And \(1^{10} * 2^9 * 3^8 * 4^7 * 5^6\) = 1 x 512 x 6561 x 16384 x 15625 = 861,449,408,741,376. Here is the other number, and I will put two underlines before it to align the digits' columns: __ 7,742,895,390,720There are the two "values" as I understand the question to be asking for them. It seems as if you think the question asks something else. . . What am I missing here? (And why the h#ll am I debating a math brainiac? Let's go on over to the language and reading threads. . .) pushpitkc ,what do you think the question asks?
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has  Margaret Mead



Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3296
Location: India
GPA: 3.12

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
22 Jul 2017, 19:59
What I think the questions asks is to use distinct positive values for a,b,c,d,e,f,g,h,i,j and find the lowest possible value of expression \(a^b*c^d*e^f*g^h*i^j\) Going by your logic : I interpret "what is the units digit of the lowest possible value of ab∗cd...ab∗cd...]" to mean "what is the units digit of the smallest number you can derive from multiplying these factors"? Substitute Q, R, S, T and U for each term of the form abab. Put another way, I think the question asks, "What is the smallest number you can make by multiplying Q x R x S x T x U, and once you have found that number, what is its units' digit?" If that is true, then \(10^1 * 9^2 * 8^3 * 7^4 * 6^5\) = 10 x 81 x 512 x 2401 x 7776 = 7,742,895,390,720And \(1^1{10} * 2^9 * 3^8 * 4^7 * 5^6\) = 1 x 512 x 6561 x 16384 x 15625 = 861,449,408,741,376. Here is the other number, and I will put two underlines before it to align the digits' columns: __ 7,742,895,390,720The first number is definitely smaller than the second number. But if the combination were as follows: \(1^{10} * 9^2 * 8^3 * 7^4 * 6^5\) = 1 x 81 x 512 x 2401 x 7776 = 7,742,895,390,72 which is the least possible possible value for the expression \(a^b*c^d*e^f*g^h*i^j\) (using any distinct positive numbers for a,b,c,d,e,f,g,h,i,j) Here a=1,b=10,c=9,d=2,e=8,f=3,g=7,h=4,i=6,g=5 The reason I use these numbers is we need the minimum value of the expression. Hope this helps clear things!
_________________
You've got what it takes, but it will take everything you've got



Senior SC Moderator
Joined: 22 May 2016
Posts: 3734

If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
22 Jul 2017, 20:22
Quote: Hope this helps clear things! Completely. You used exactly four of the same combinations I did  and then switched \(10^1\) for \(1^{10}\), decreasing it by a factor of 10. Sorry for the fog brain here! (At least I had the "dumb idea to debate a math brainiac" part correct.) Thank you!
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has  Margaret Mead



NonHuman User
Joined: 09 Sep 2013
Posts: 13963

Re: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
Show Tags
30 May 2019, 19:02
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha
[#permalink]
30 May 2019, 19:02






