Find the units digit of (1998^1999) - (1999^1998).?

Answer 1

You don't have to actually take the exponents to do this. Instead, let's look at the units digit of exponents of 8:

8^1 = 8
8^2 = 4 (looking only at the units digit)
8^3 = 2
8^4 = 6
8^5 = 8 (notice we are at 8 again, so we will just repeat our pattern from here on out)

So the digits unit of 8^n repeats every 4 digits, so we look at the remainder of n/4 to determine our units digit.

Since n = 1999 in our case, and 1999/4 = 499 remainder 3, we can know therefore that 1998^1999 has the same units digit as 8^3, namely 2.

Similarly for 1999^1998, let's look at the units digits for exponents of 9:

9^1 = 9
9^2 = 1 (looking only at units digit)
9^3 = 9 (notice we are at 9 again, so we will just repeat our pattern from here on out)

So here our pattern is pretty easy -- for 9^n, the units digit is 1 if n is odd, 2 if n is even.

So, 1999^1998 will have a 1 as the units digit (since 1998 is even)

Therefore the units digit of (1998^1999) - (1999^1998) =
2-1 = 1

Pretty neat way to solve it, huh? :-)

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Answer 2

1999^1998 = k(1998^1999),
k =1/1998((1999/1998)^1998).
1998(1999^1998) = ((1999/1998)^1998)(1998^1999),
1998(1999^1998)=1999^1999 - 1999^1998
1999^1999 - 1999^1998 =
((1999/1998)^1998)(1998^1999).
1999^1999-1999^1998=1998(1999^1998)
1999^1999=1999^1998+1998(1999^1998),

cant help ya treehugger!

Answer 3

The first answerer has the best answer. Just give her 10 points. What else can I say about the question? I'd just be repeating her answer no matter how I try it!