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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 01:56
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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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E
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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 02:04
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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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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 02:06
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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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\((3^{101})(7^{103})\)

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

above can be written as

\((3^{1})(7^{3})\)
3*343
units digit = 9
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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 03:38
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[Math Revolution GMAT math practice question]

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

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
Show more

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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E
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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 07:40
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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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\(3^4\) = 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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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 08:48
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MathRevolution
[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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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 :grin: :lol:

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 :)
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What is the units digit of (3^{101})(7^{103})? 29 Aug 2018, 09:06
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dave13


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 :grin: :lol:

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

\(= (1^{25}∗3)(1^{25}∗3)\) { 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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What is the units digit of (3^{101})(7^{103})? 31 Aug 2018, 01:15
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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
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What is the units digit of (3^{101})(7^{103})? Updated on: 03 Oct 2018, 10:34
Originally posted by arjunnath on 31 Aug 2018, 10:00.
Last edited by arjunnath on 03 Oct 2018, 10:34, edited 1 time in total.
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MathRevolution
[Math Revolution GMAT math practice question]

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

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
Show more


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)
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What is the units digit of (3^{101})(7^{103})? 03 Sep 2018, 19:01
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MathRevolution
[Math Revolution GMAT math practice question]

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

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
Show more

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
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What is the units digit of (3^{101})(7^{103})? 03 Oct 2018, 08:31
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[Math Revolution GMAT math practice question]

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

\(A. 1\)
\(B. 3\)
\(C. 5\)
\(D. 7\)
\(E. 9\)
Show more

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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