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# What is the remainder when (2^16)(3^16)(7^16) is divided by 10?

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What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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06 Feb 2020, 07:47
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25% (medium)

Question Stats:

76% (01:13) correct 24% (01:31) wrong based on 54 sessions

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What is the remainder when $$(2^{16})(3^{16})(7^{16})$$ is divided by 10?

A. 0
B. 2
C. 4
D. 6
E. 8

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Joined: 11 Dec 2013
Posts: 121
Location: India
GMAT Date: 03-15-2015
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Re: What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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06 Feb 2020, 07:51
Finding the remainder when divided by 10 is equivalent to finding the last digit .

Last digit of 2^16 = 6
Last digit of 3^16 = 1
Last digit of 7^16 = 1

Hence the last digit will be (6)(1)(1) = 6
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Re: What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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06 Feb 2020, 07:55
What is the remainder when (2^16)(3^16)(7^16) is divided by 10?

A. 0
B. 2
C. 4
D. 6
E. 8

Remainder when divided by 10 is the units digit and remainders can be multiplied (when adjusted for excess)

Cyclicity of 2 -> 2,4,8,6 -> exponent/cyclicity - remainder 0 -> Units digit of 2 is 6 -> Remainder = 6
Cyclicity of 3 -> 3,9,7,1 -> exponent/cyclicity - remainder 0 -> Units digit of 3 is 1 -> Remainder = 1
Cyclicity of 7 -> 7,9,3,1 -> exponent/cyclicity - remainder 0 -> Units digit of 2 is 1 -> Remainder = 1

Therefore remainder = 6*1*1 = 6 -> Answer - D
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Re: What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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06 Feb 2020, 08:14
1
What is the remainder when $$(2^{16})(3^{16})(7^{16})$$ is divided by 10?

A. 0
B. 2
C. 4
D. 6 --> correct
E. 8

Solution:
N = $$(2^{16})(3^{16})(7^{16})$$
= $$(2*3*7)^{16}$$
= $$(42)^{16}$$
= $$(40+2)^{16}$$
Reminder of $$(40+2)^{4*4}$$ divided by 10 = Reminder of $$2^{16}$$ divided by 10

2^1 = 2, reminder = 2
2^2=4, reminder = 4
2^3=8, reminder = 8
2^4=16, reminder = 6 --> cyclicity of 4,
2^5 = 32, reminder = 2
Reminder of $$2^{16}$$ divided by 10 = 6
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What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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06 Feb 2020, 08:16
2
Top Contributor
Bunuel wrote:
What is the remainder when $$(2^{16})(3^{16})(7^{16})$$ is divided by 10?

A. 0
B. 2
C. 4
D. 6
E. 8

Key concept: $$(x^n)(y^n)(z^n) = (xyz)^n$$

So, we can write: $$(2^{16})(3^{16})(7^{16}) = (2 \times 3 \times 7)^{16}$$

$$= (42)^{16}$$

At this point we need only find the unit's digit of $$(42)^{16}$$

42¹ = 42
42² = ---4
42³ = ---8
42⁴ = ---6
42⁵ = ---2

42 has a cycle of 4 (the units digit repeats every 4 powers)

This means 42^8 = ---6, 42^12 = ---6, 42^16 = ---6, etc

Since 42^16 = ---6, we can conclude that 42^16, when divided by 10, will leave a remainder of 6

Cheers,
Brent
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Re: What is the remainder when (2^16)(3^16)(7^16) is divided by 10?  [#permalink]

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08 Feb 2020, 12:32
Bunuel wrote:
What is the remainder when $$(2^{16})(3^{16})(7^{16})$$ is divided by 10?

A. 0
B. 2
C. 4
D. 6
E. 8

Solution:

Let’s look at some quick examples: If we divide 76 by 10, the quotient is 7, remainder 6. If we divide 23 by 10, the quotient is 2, remainder 3. Thus, we see that when we divide a number by 10, the remainder is always equal to the units digit.

Thus, to find the remainder when (2^16)(3^16)(7^16) is divided by 10, we simply have to find the units digit of (2^16)(3^16)(7^16). Now, let’s look at the units digit of each factor.

The pattern of units digits for a base of 2 is 2-4-8-6.

So, 2^16 has a units digit of 6.

The pattern of units digits for a base of 3 is 3-9-7-1.

So, 3^16 has a units digit of 1.

The pattern of units digits for a base of 7 is 7-9-3-1.

So, 7^16 has a units digit of 1.

Thus, the remainder of 2^16 x 3^16 x 7^16 when divided by 10 is 6 x 1 x 1 = 6.

Alternate solution:

Again, to find the remainder when (2^16)(3^16)(7^16) is divided by 10 is to find the units digit of (2^16)(3^16)(7^16). However, we can see that (2^16)(3^16)(7^16) = (2 x 3 x 7)^16 = 42^16 and 42^16 has the same units digit as 2^16. Since the pattern of units digits for a base of 2 is 2-4-8-6,
2^16 has a units digit of 6.

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Re: What is the remainder when (2^16)(3^16)(7^16) is divided by 10?   [#permalink] 08 Feb 2020, 12:32
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