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What is the remainder if 7^10 is divided by 100?

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What is the remainder if 7^10 is divided by 100?

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What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49

OA E

Source: Manhattan Prep

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BTGmoderatorDC wrote:
What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49
The remainder when 7^10 is divided by 100 is equal to the last two digits of the expansion of 7^10.

Notice that 7^2 = 49, or 50 - 1. So 7^4 = (50 - 1)^2 = 2500 - 100 + 1 = 2401. The last two digits of 7^2 is 49 and those of 7^4 is 01 or simply 1. Since 7^10 = 7^2 x 7^4 x 7^4, the last two digits of 7^10 is

49 x 1 x 1 = 49

Answer: E

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BTGmoderatorDC wrote:
What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49
Source: Manhattan Prep
First note that:

> 1 is the remainder of 101 (=1*100+1) divided by 100
> 32 is the remainder of 532 (=5*100+32) divided by 100
> 47 is the remainder of 7847 (=78*100+47) divided by 100
$${7^{10}} = K \cdot 100 + R{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} {\mkern 1mu} \left( {K\,\,{\mathop{\rm int}} \,\,,\,\,\,0 \le R \le 99\,\,{\mathop{\rm int}} } \right){\mkern 1mu} $$
$$? = R$$
$${7^{10}} = {\left( {{7^2}} \right)^5} = {49^5}$$
$${49^2} = {\left( {50 - 1} \right)^2} = {5^2} \cdot {10^2} - 100 + 1 = M \cdot 100 + 1\,\,\,,\,\,\,M\,\,{\mathop{\rm int}} \ge 1\,\,\,\,\,\,\,\,\,\,\left( {M = {5^2} - 1} \right)$$
$${49^4} = {\left( {M \cdot 100 + 1} \right)^2} = {M^2} \cdot {10^4} + M \cdot 200 + 1 = N \cdot 100 + 1\,\,\,,\,\,\,\,N\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\,\,\,\,\,\left( {N = {M^2} \cdot {{10}^2} + 2M} \right)$$
$${49^5} = \left( {N \cdot 100 + 1} \right) \cdot 49 = K \cdot 100 + 49\,\,\,,\,\,\,\,K\,\,{\mathop{\rm int}} \,\, \ge 1\,\,\,\left( {K = 49N} \right)$$
$$? = 49$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Last edited by fskilnik@GMATH on Sun Oct 14, 2018 12:03 pm; edited 1 time in total

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Junior | Next Rank: 30 Posts Default Avatar
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If we are asked to calculate the remainder when 6*5 is divided by 4, we observe that we can divide each of the numbers 6 and 5 and multiply their remainders to get the required answer.
For example , $$\frac{6}{4}$$ yields a remainder of 2 whereas $$\frac{5}{4}$$ gives 1.
Multiplying 2 and 1 is 2 - the same as the remainder when 30 is divided by 4.

We use this same principle in the above problem. We break the given product into small numbers whose remainder we can easily find. In this case we can write it as

$$7^9\cdot7$$ = $$343^3\cdot7$$
When 343 is divided by 100, the remainder is 43.
Our required answer is
(43*43)*(43*7)
=1849 * 301
The remainders when 1849 and 301 are divided by 100 are 49 and 1.
The product of 49 and 1 is 49 ... which is the required answer since 49 divided by 100 would continue to yield a remainder of 49.

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GMAT/MBA Expert

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BTGmoderatorDC wrote:
What is the remainder if 7^10 is divided by 100?

A] 1
B] 43
C] 19
D] 70
E] 49
When a positive integer is divided by 100, the remainder is yielded by the last two digits:
123/100 = 1 R23
548/100 = 5 R48
692/100 = 6 R92

Thus:
The remainder when 7¹⁰ is divided by 100 is equal to the last two digits of 7¹⁰.

Calculate the last two digits for consecutive powers of 7 and look for a pattern:
7¹ --> 07
7² --> 49
7³ --> 43
7⁴ --> 01
7⁵ --> 07

The last two digits appear in a CYCLE OF 4:
07, 49, 43, 01...07, 49, 43, 01...
Implication:
When 7 is raised to a power that is a MULTIPLE OF 4 -- constituting the end of a cycle -- the last two digits will be 01.
From there, the cycle will repeat:
07, 49, 43, 01...

Since 8 is a multiple of 4, the last two digits for 7⁸ are 01.
The cycle then repeats:
7⁹ ---> 07
7¹⁰ --> 49

The correct answer is E.

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Sweeeeeeeeeeeeet solution, Scott!!

Cheers,
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Brent@GMATPrepNow wrote:
Sweeeeeeeeeeeeet solution, Scott!!

Cheers,
Brent
Thanks Brent!

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