Could anyone answer this math question...It's really bugging me and I've tried everything?!!?

Question

It's:

Find the units digits of
(1998^1999)-(1999^1998)

And if you somehow my a miracle get the answer could you show me the work..This is intended to do without a calculator...

Thanks!!

Answer 1

You have to look for the repeating pattern of units digits for each problem.

1998^1 ends in 8
1998^2 ends in 4 (8X8=64)
1998^3 ends in 2
1998^4 ends in 6
1998^5 ends in 8 so it's starting to repeat. It repeats every 4 so 1998^1999 is the same as 1998^3, ends in a 2.

Now for 1999^1998.
1999^1 ends in a 9
1999^2 ends in a 1 9X9 = 81
1999^3 ends in 9 It's already repeating

So 1999^1998 ends just like 1999^2 did, with a 1

Finally subtract these two end digits since that's all you care about: 2 - 1 = 1

So it ends in a 1

Answer 2

Ignore all but the unit digits. So the units digit of 1998^1999 is the same as the units digit of 8^1999 (1998 = 1990 + 8, (1990 + 8)^2 = 1990^2 + 16*1990 + 64, note all but the last are divisible by 10)

8^1 has units digit 8 (8)
8^2 has units digit 4 (64)
8^3 has units digit 2 (512)
8^4 has units digit 6 (4096)
8^5 has units digit 8 (32768)
8^6 has units digit 4
...
1999 == 3 mod 4, so the units digit is the same as for 1995, 1991, ... 3, is 2.

Same thing for 9:

9^1 has units digit 9
9^2 has units digit 1
9^3 has units digit 9
...

This is even easier, odd has units digit 9, even has 1. Since 1998 is even 1999^1998 has units digit 1.

So the difference has units digit 2 - 1 = 1.



So

Answer 3

Since you are only looking for the units digit, you can do all your arithmetic mod 10.

(1998^1999)-(1999^1998) ≡ 8^1999 - 9^1998 mod 10.

Now we look for a pattern in the powers of 8 and 9 mod 10:

9≡-1 mod 10, so 9^n≡(-1)^n mod 10, so 9 to any even power is congruent to 1 mod 10

8 is only slightly more complicated:

8^1 ≡ 8 mod 10
8^2 ≡ 4 mod 10
8^3 ≡ 2 mod 10
8^4 ≡ 6 mod 10
8^5 ≡ 8 mod 10

And after that the cycle repeats. So the powers of 8 form a cycle of length four. 1999 has a remainder of 3 when divided by 4, so 8^1999 ≡ 8^3 ≡ 2 mod 10. So we have:

(1998^1999)-(1999^1998) ≡ 2-1 mod 10
And 2-1 is 1. Thus the units digit of (1998^1999)-(1999^1998) is 1.

Answer 4

Def.: the units digits(the number of unis's place?) of a^b = x(a,b), m=0,1,2,3,…

x(8,1)=8
x(8,2)=4
x(8,3)=2
x(8,4)=6
x(8,5)=8
x(8,6)=4
x(8,7)=2
x(8,9)=6
…
x(8,4m+1)=8
x(8,4m+2)=4
x(8,4m+3)=2
x(8,4m)=6


since 1999=4*499+3 then
x(8,1999)=2


In same way
x(9,2m+1)=9
x(9,2m)=1

since 1998=2*999 then
x(9,1998)=1

x(8,1999)-x(9,1998)=2-1=1

The answer may be one.

Answer 5

how the hell do you find a units digits