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What is the remainder, after division by 100, of 7^10 ?

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by GMATGuruNY » Mon Jun 18, 2012 3:13 am
gmatter2012 wrote:What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70
When an integer is divided by 100, the remainder = tens digit + units digit.
In other words, the integer formed by the last two digits.
125/100 = 1 R25.
3289/100 = 32 R89.

Since each answer choice here has a different units digit, all we have to determine is the units digit of 7^10.

7^1 = 7.
7² = 49.
Since the product of 7 and the units digit of 7² = 7*9 = 63, the units digit of 7³ is 3.
Since the product of 7 and the units digit of 7³ = 7*3 = 21, the units digit of 7^4 is 1.
From here, the units digits will repeat in a cycle of 4:
7,9,3,1,7,9,3,1,7,9...
The 10th value in the list above is 9, implying that the units digit of 7^10 is 9.

The correct answer is D.
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by dhonu121 » Mon Jun 18, 2012 4:19 am
gmatter2012 wrote:What is the remainder, after division by 100, of 7^10 ?

(A) 1
(B) 7
(C) 43
(D) 49
(E) 70
You just need to find the units digit of 7^10.
powers of 7 follow a cyclicity of 4.
7,9,3,1.
Hence 7^10 is effectively 7^(4+4+2). Forget about the 4's as its a cycle. It eventually boils down to 7^2 = 49.
Thus the last digit of 7^10 will be 9.
Only D has last digit as 9.
Hence D is the right answer.
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by Anurag@Gurome » Mon Jun 18, 2012 4:36 am
gmatter2012 wrote:What is the remainder, after division by 100, of 7^10 ?
7^10 = (7²)^5 = 49^5
Odd powers of 9 have 9 as the unit's digit and even powers of 9 has 1 as the unit's digit.

Now, unit's digit of 7^10 = unit's digit of 49^5 = 9

Hence, the unit's digit of the remainder when 7^10 s divided by 100 will be 9. Only such option is 49.

The correct answer is D.
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by aakgoel » Mon Jun 18, 2012 10:01 pm
Is there any way to find the digit in tens place also.
If exists, please tell.
Thank you.
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by Anurag@Gurome » Mon Jun 18, 2012 11:27 pm
aakgoel wrote:Is there any way to find the digit in tens place also.
If exists, please tell.
Thank you.
In general to find out the ten's digit of numbers of the form x^n you need to know binomial expansion.

We can expand (a + b)^n as follows,
Image

Now, the trick is to write x as sum or difference of the nearest multiple of 10. For example, say we need to find out the last two digits of 121^21. Then we will write 121 = (120 + 1)

So, 121^21 = (120 + 1)^21 = (1 + 120)^21
Now, we will consider a = 1, b = 120, and n = 21, and accordingly we will expand (1 + 120)^21. Note that, every term in the expansion after the second term will be a multiple of 100. Hence, we just need to determine the first two terms.

First term = a^n = (1)^21 = 1
Second term = n*[a^(n - 1)]*b = 21*[1^20]*120 = 21*120 = 2520

Hence, last two digits of 121^21 are last two digits of (2520 + 1) = 2521, i.e. 2 and 1.

This method works great when we can write x as sum or difference of nearest multiple of 10 with respect small numbers like 1 or 2. Even for 3, it gets complicated.

For example, in this case 7^10 = (-3 + 10)^10
First term = (-3)^10 = 3^10 ---> Needs lot of calculation.

This method will work for last three digits also. In that case, we need to write x as sum or difference of nearest multiple of 100.

Hope that helps.
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by sabeer6870 » Tue Jun 19, 2012 7:38 am
7^4=2401
2401/100 rem =1
so 7^8/100 rem =01
7^2 is left ...49/100 rem =49

D is the answer
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by gmatter2012 » Wed Jun 20, 2012 5:49 am
many thanks to both GmatguruNY and Anurag and others , this is indeed good information
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