What is the units digit of 248^20?

D

factoring, you get 248 = 2 x 2 x 2 x 31 or

(2^3 * 31)^20

This equals \(2^{60} * 31^{20}\)

We individually calculate units digits of the two factors.

For 2:
\(2^1 = 2\)
\(2^2 = 4\)
\(2^3 = 8\)
\(2^4 = 16\)
\(2^5 = 2\)

and the pattern of unit digits 2,4,8 and 6 repeats after every multiplication of 2. so \(2^{60}\) has the units digit the same as \(2^4\) which is 6.

For 31:
Units digit of 31x31 is 1 and it is again multiplied by 31 to result in units digit of 1 and the pattern continues.

so essentially we end up with units digits of 6 and 1 for expressions of 2 and 31. Multiplying both results in 6 which would be the units digit of \(248^{20}\)

When numbers have last digit of 2, 3, 7, 8, they have a cyclicity of 4. When 20 is divided by 4, the remainder is 0. So, 8^4 has unit digit of 6. HENCE, ANS IS D

\(248^{20}\)

We just need to check the unit digit of \(8\).

\(8^1 = 8\) ---------(Unit digit is \(8\))

\(8^2 = 64\) -------- (Unit digit is \(4\))

\(8^3 = 64 * 8 = 4*8 = 32\) ----- (we just need to multiply the unit digits to get the unit digit value, hence \(4*8\)) ----- (Unit digit is \(2\))

\(8^4 = 32 * 8 = 2*8 = 16\) ----- (we just need to multiply the unit digits to get the unit digit value, hence \(2*8\)) ----- (Unit digit is \(6\))

\(8^5 = 16 * 8 = 6*8 = 48\) ----- (we just need to multiply the unit digits to get the unit digit value, hence \(6*8\)) ------(Unit digit is \(8\))

\(8\) has cyclicity of \(4\). Hence \(248\) will also have the cyclicity of \(4\).

\(20\) is divisible by \(4\), hence the remainder would be \(0\).

Hence the unit digit of \(248 = 6\)

Answer (D)...

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Theory:
The cyclicity of 2,3,7,8 is 4.
When Power is divided by cyclicity; Remainder is zero, then unit digit^cyclicity
When power is divided by cyclicity; Remainder is Non zero, then unit digit ^Remainder

20=4x5 + 0
Thus, remainder =0
so, last digit of 248^cyclicity(4)
=8^4
=6
Ans is D

==>You get \(~8^1=~8, ~8^2=~4, ~8^3=~2, ~8^4=~6,\) so the units digit becomes 8-->4-->2-->6-->8, and the index has a period of 4.
Thus, you get \(248^{20}=(248^4)^5=(~6)^5=~6.\)

The answer is D.
Answer: D

Since we only care about units digits, we can rewrite the expression as:

8^20

Let’s start by evaluating the pattern of the units digits of 8^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 8. When writing out the pattern, notice that we are ONLY concerned with the units digit of 8 raised to each power.

8^1 = 8

8^2 = 4

8^3 = 2

8^4 = 6

8^5 = 8

The pattern of the units digit of powers of 8 repeats every 4 exponents. The pattern is 8–4–2–6. In this pattern, all positive exponents that are multiples of 4 will produce an 8 as the units digit. Thus:

8^20 has a units digit of 6.

Answer: D
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Look for a pattern

248^1 = 248
248^2 = (248)(248) = ---4 [aside: we need not determine the other digits. All we care about is the units digit]
248^3 = (248)(248^2) = (248)(---4) = ----2
248^4 = (248)(248^3) = (248)(---2) = ----6
248^5 = (248)(248^4) = (248)(---6) = ----8

NOTICE that we're back to where we started.
248^5 has units digit 8, and 248^1 has units digit 8
So, at this point, our pattern of units digits keep repeating 8, 4, 2, 6, 8, 4, 2, 6, 8,...
We say that we have a "cycle" of 4, which means the digits repeat every 4 powers.

So, we get:
248^1 = --8
248^2 = ---4
248^3 = ----2
248^4 = ----6
248^5 = ----8
248^6 = ---4
248^7 = ----2
248^8 = ----6
248^9 = ----8
248^10 = ----4
etc.

Notice that when the exponent is a MULTIPLE of 4 (4, 8, 12, 16, ...), the units digit will be 6
Since 20 is a MULTIPLE of 4, we know that the units digit of 248^20 will be 6
Answer:

Here's an article I wrote on this topic (with additional practice questions): https://www.gmatprepnow.com/articles/un ... big-powers

Cheers,
Brent

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