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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
What is the units digit of (3^{101})(7^{103})?  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 70% (01:08) correct 30% (01:33) wrong based on 179 sessions

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[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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Manager  G
Joined: 14 Jun 2018
Posts: 211
Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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3^101 * 7^103 = 21^101 * 49
cycle of last digit of 21 is 1.
Ans E
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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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

101= 4k+1
103=4k+3

above can be written as

$$(3^{1})(7^{3})$$
3*343
units digit = 9
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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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.

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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

$$3^4$$ = 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)
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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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Abhishek009 wrote:
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

$$3^4$$ = 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 Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4834
Location: India
GPA: 3.5
Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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1
dave13 wrote:

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!!!

PS : Mr Bean is my fav character, and I really love your innocent looking DP, good evening friend, plz feel free to revert in case of any further doubt...
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Abhishek....

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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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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 $$3*3 = 9.$$

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

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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)

Originally posted by Natty97 on 31 Aug 2018, 10:00.
Last edited by Natty97 on 03 Oct 2018, 10:34, edited 1 time in total.
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Posts: 2809
Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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.

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Re: What is the units digit of (3^{101})(7^{103})?  [#permalink]

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Top Contributor
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

$$A. 1$$
$$B. 3$$
$$C. 5$$
$$D. 7$$
$$E. 9$$

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

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