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S98-12

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S98-12  [#permalink]

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New post 16 Sep 2014, 00:52
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A
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D
E

Difficulty:

  35% (medium)

Question Stats:

64% (01:03) correct 36% (01:15) wrong based on 69 sessions

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Re S98-12  [#permalink]

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New post 16 Sep 2014, 00:52
1
1
Official Solution:

What is the remainder, after division by 100, of \(7^{10}\)?

A. 1
B. 7
C. 43
D. 49
E. 70


First, we should understand what the question is asking for. What is "the remainder, after division by 100" of a large integer? Test some numbers, if necessary:

321 divided by 100 leaves a remainder of 21.

432 divided by 100 leaves a remainder of 32.

Thus, we can see that the remainder, after division by 100, of a large integer is just the two-digit number formed by the last two digits of the integer (the tens digit and the units digit).

Now, we turn our attention to the actual integer in question, \(7^{10}\). This is a power of 7, so we need to look for any pattern in the last two digits of the powers of 7. (We can assume that there must be such a pattern; otherwise, this question would not be realistically solvable on the GMAT.)
\(7^1 = 7\)
\(7^2 = 49\)

\(7^3 = 49 \times 7 = 343\). Note that we only need to pay attention to the last 2 digits (43), so we will write ...43.
\(7^4 = ...43 \times 7 = ...01\)
\(7^5 = ...01 \times 7 = ...07\)

At this point, we see that the cycle is starting to repeat. The next power (\(7^6\)) will end in ...49, and so forth. Since we only have to go to \(7^{10}\), we may as well just keep going:
\(7^7 = ...43\)
\(7^8 = ...01\)
\(7^9 = ...07\)
\(7^{10} = ...49\)


Answer: D
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Re: S98-12  [#permalink]

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New post 09 Jul 2015, 03:01
Hello Bunuel,

Could we also do this by only paying attention to the last digit of the powers of 7?

In other words, we know that 7^10 will end in 9. So, we could just pick a number that ends in 9 and is higher than 10 and divide those?
For example, 19/10 leaves a remainder of 9. So, we would pick D for this reason. I only choose 19, because it needed to be higher than 10, so I just went for the first possivle number that ends in 9 and is higher than 10.

The same happens if you divide 109/100.It leaves a remainder of 9. So, I went for D.
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Re: S98-12  [#permalink]

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New post 09 Jul 2015, 03:38
pacifist85 wrote:
Hello Bunuel,

Could we also do this by only paying attention to the last digit of the powers of 7?

In other words, we know that 7^10 will end in 9. So, we could just pick a number that ends in 9 and is higher than 10 and divide those?
For example, 19/10 leaves a remainder of 9. So, we would pick D for this reason. I only choose 19, because it needed to be higher than 10, so I just went for the first possivle number that ends in 9 and is higher than 10.

The same happens if you divide 109/100.It leaves a remainder of 9. So, I went for D.


Yes. Jist notice that we are dividing by 100 not by 10.

Check my solution here: what-is-the-remainder-after-division-by-100-of-121666.html

Hope it helps.
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Re: S98-12  [#permalink]

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New post 30 Jul 2015, 19:59
Bunuel is correct to the extent of finding rem when divided by 100. even i followed the same "cyclicity" of remainders approach for ten's and unit's digit.
But for this particular problem, since all answer choices have different unit's digit, we can reduce our labor to finding just the unit's digit.
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Re: S98-12  [#permalink]

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New post 01 Feb 2018, 04:59
7^4 = 2401. which will leave a remainder 1 when divided with 100. Therefore, 7^4n (7^8) will also leave a remainder 1. 7^10=7^8.7^2 so the remainder will be 7^2 or 49.
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Re: S98-12  [#permalink]

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New post 02 Jul 2018, 00:54
Did it by approximation:
we have --> 7^10/10^2
--> we can say --> 49 x 7^8/ 10^2 --> 0.49 x 7^8 --> we know the number for 7^ 8 will end in 1. So shift decimal two spaces left , which is 1, then 3, leaving us with 9. So the product will have number ending 9 in it. Answer = 49
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Re: S98-12 &nbs [#permalink] 02 Jul 2018, 00:54
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