If a, b, c, d, e, f, g, h, i and j are distinct positive integers, wha

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.

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 = 19,200,000

Small Base Large Exponent:
1 * 500 * 6,500 = 3,250,000
multiplied by
15,600 * 16,000 = 249,000,000

The 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

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.

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


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 =
7,742,895,390,720

And \(1^{10} * 2^9 * 3^8 * 4^7 * 5^6\) =
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:
__7,742,895,390,720

There 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?

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 =
7,742,895,390,720

And \(1^1{10} * 2^9 * 3^8 * 4^7 * 5^6\) =
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:
__7,742,895,390,720

The 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!

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!

Generally speaking, given two different numbers, the smaller number in the exponent results in a lower number, and we're trying to find the lowest overall number first before determining its units digit

So we want 1-5 as the exponents.

It then makes sense to assign the numbers 6-10 to the highest down to lowest exponents, so:

9^2 8^3 7^4 6^5

But, not 10^1, since clearly 1^10 is preferred.

So the units digit of 1^10 is

1

9^2 = 81, units digit of

1

8^3 = 64*8 Clearly a digit of

2

7^4 = 49^2 Clearly a digit of

1

6^5 = always a

6

So the units digits multiplied are:

1*1*2*1*6 = 2 units digit

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