What is the units digit of (3^{101})(7^{103})?

When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE.

\(3^{101}\):
3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3⁴ --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, \(3^{100}\)has a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus, \(3^{101}\) has a units digit of 3.

7¹⁰³:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7⁴ --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 7 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, \(7^{100}\) has a units digit of 1.
From here, the cycle of units digits will repeat: 7, 9, 3, 1...
\(7^{101}\)--> units digit of 7.
\(7^{102}\) --> units digit of 9.
\(7^{103}\)--> units digit of 3.

Result:
\(3^{101}7^{103}\) = (integer with a units digit of 3)(integer with a units digit of 3) = integer with a units digit of 9.

\(3^4\) = Units digit 1
\(7^4\) = Units digit 1

\((3^{101})(7^{103})\)

= \((3^{4*25}*3^1)(7^{4*25}*7^3)\)

3^1 will have units digit 1 and 7^3 will have units digit 3

So, The units digit of the expression will be \(1*3*1*3 = 9\) , Answer must be (E)

hey there Abhishek009 hope your solo guitar career is thriving let me know when you are gving your next rock concert , i will buy tickets

if 3^1 will have units digit 1 and 7^3 will have units digit 3

So we have unit digit 1 and unit digit 3 hence 1*3 = 3 ?

where from did you get so many numbers \(1*3*1*3 = 9\)


have a great evening

\(= (3^{4∗25}∗3^1)(7^{4∗25}∗7^3)\)

\(= (1^{25}∗3)(1^{25}∗3)\) { Units digit of 3^4 = 1 and Units digit of 7^4 = 3 }

\(= 1* 3 * 1 * 3\)

Hope this helps!!!

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The units digit is the remainder when \((3^{101})(7^{103})\) is divided by \(10\).

The remainders when powers of \(3\) are divided by \(10\) are
\(3^1: 3,\)
\(3^2: 9,\)
\(3^3: 7,\)
\(3^4: 1,\)
\(3^5: 3,\)
…
So, the units digits of \(3^n\) have period \(4\):
They form the cycle \(3 -> 9 -> 7 -> 1.\)
Thus, \(3^n\) has the units digit of \(3\) if \(n\) has a remainder of \(1\) when it is divided by \(4\).
The remainder when \(101\) is divided by \(4\) is \(1\), so the units digit of \(3^{101}\) is \(3\).

The remainders when powers of \(7\) are divided by \(10\) are
\(7^1: 7,\)
\(7^2: 9,\)
\(7^3: 3,\)
\(7^4: 1,\)
\(7^5: 7,\)
…
So, the units digits of \(7^n\) have period \(4\):
They form the cycle \(7 -> 9 -> 3 -> 1\).
Thus, \(7^n\) has the units digit of \(3\) if \(n\) has a remainder of \(3\) when it is divided by \(4\).
The remainder when \(103\) is divided by \(4\) is \(3\), so the units digit of \(7^{103}\) is \(3\).

Thus, the units digit of \((3^{101})(7^{103})\) is \(3*3 = 9.\)

Therefore, the answer is E.
Answer: E

3^101 * 7^103

cycle of 3 is 3,9,7,1
101st through cyclicity will be 3
cycle of 7 is 7,9,3,1
103rd through cyclicity will be 3

Units digit i.e. 3*3 = 9

Ans (E)

Let’s start by evaluating the pattern of the units digits of 3^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 3. When writing out the pattern, notice that we are concerned ONLY with the units digit of 3 raised to each power.

3^1 = 3

3^2 = 9

3^3 = 7

3^4 = 1

3^5 = 3

The pattern of the units digit of powers of 3 repeats every 4 exponents. The pattern is 3–9–7–1. In this pattern, all positive exponents that are multiples of 4 will produce 1 as its units digit. Thus:

3^100 has a units digit of 1, and so 3^101 has a units digit of 3.

Next, we can evaluate the pattern of the units digits of 7^n for positive integer values of n. That is, let’s look at the pattern of the units digits of powers of 7. When writing out the pattern, notice that we are concerned ONLY with the units digit of 7 raised to each power.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of the units digit of powers of 7 repeats every 4 exponents. The pattern is 7–9–3–1. In this pattern, all positive exponents that are multiples of 4 will produce 1 as its units digit. Thus:

7^104 has a units digit of 1, and so 7^103 has a units digit of 3.

Thus, the units digit of 3^101 x 7^103 is 3 x 3 = 9.

Answer: E

This is a great question for applying the following property: (x^n)(y^n) = (xy)^n
For example, (3^7)(5^7) = 15^7

(3^101)(7^103) = (3^101)(7^101)(7^2) [rewrote 7^103 as the product of 7^101 and 7^2]
= (3^101)(7^101)(49) [evaluated 7^2]
= (21^101)(49) [applied above property]

Notice that 21^n will have units digit 1 for all positive integer values of n
So, = 21^101 = --------1 [some big number ending in 1]
So, we get:
= (--------1)(49)
= --------9

Answer: E

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