What is the tens digit of 7^241?

We can extract the tens digit by finding the last two digits of \(7^{241}\)
The last two digits of \(7^{241}\) = Remainder when \(7^{241}\) is divided by 100.

Thus, we need to find the Remainder of \(\frac{7^{241}}{100}\)

Finding the remainder using Euler’s Theorem:
We will convert \(7^{241}\) into a smaller and “similar” number which will give the same remainder as \(7^{241}\) when divided by 100.

Converting \(7^{241}\) into a smaller and similar number:
For that, we will first find Euler’s Number of the denominator-
Mathematically, the Euler number of a number z denoted by the symbol E(z) is calculated as explained below.
\(E(z) = z * [1 – \frac{1}{P}]*[1 – \frac{1}{Q}]\)
where P, and Q are the different prime factors of z.

Therefore, \(E(100) = 100 * [1 – \frac{1}{2}] * [1 – \frac{1}{5}] = 100 * \frac{1}{2} * \frac{4}{5} = 40\)-> (a)

The smaller and similar number will have a base equal to Remainder(\(\frac{7}{100}\)) = 7 -> (b)
And it will be raised to the power of Remainder(\(\frac{241}{Euler’s number of 100}\)) = Remainder of \(\frac{241}{40}\) = 1 -> (c)

Therefore from (b) and (c), the smaller and “similar” number which will give the same remainder as \(7^{241}\) when divided by 100 is \(7^1\) = 7 -> (d)

Therefore the last two digits of \(7^1\) = Remainder of \(\frac{7^1}{100}\) = Last two digits of \(7^{241}\) = 07 -> (e)

Hence from (e), the tens digit of \(7^{241}\) is 0

Answer A

What is the tens digit of \(7^2^4^1\)?

A. 0
B. 2
C. 4
D. 6
E. 7

There are several ways to deal with this problems some easier some harder, but almost all of them are based on the pattern recognition.

The tens digit of 7 in integer power starting from 2 (6^1 has no tens digit) repeats in pattern of 4: {4, 4, 0, 0 }:
The tens digit of 7^2=49 is 4;
The tens digit of 7^3=343 is 4;
The tens digit of 7^4=...01 is 0 (how to calculate: multiply 43 by 7 to get ...01 as the last two digits);
The tens digit of 7^5=...07 is 0 (how to calculate: multiply 01 by 7 to get ...07 as the last two digit);
The tens digit of 7^6=...49 is 4 (how to calculate: multiply 07 by 7 to get ...49 as the last two digits);
The tens digit of 7^7=...43 is 4 (how to calculate: multiply 49 by 7 to get ...43 as the last two digits).
The tens digit of 7^8=...01 is 4 (how to calculate: multiply 43 by 7 to get ...01 as the last two digits).
The tens digit of 7^9=...07 is 4 (how to calculate: multiply 01 by 7 to get ...07 as the last two digits).


In general, the tens digit of \(7^(^4^x^+^1^) = 0\)
241 = 4*60 + 1
Tenth digit of \(7^2^4^1\) = 0

IMO A

Now GMAT obviously does not expect you to work out how much 7^241 is to see what the tens digit might be. With questions like these, there will ALWAYS be a pattern that needs finding. So let us get on with it:

\(7^1=7\) - tens digit: 0
\(7^2=49\) - tens digit: 4
\(7^3=343\) - tens digit: 4
\(7^4=xx01\) - tens digit: 0 Note: no need to compute the whole value, we are just interested in the last two digits of the number
\(7^5=xxx07\) - tens digit: 0
\(7^6=xxxx49\) - tens digit: 4
\(7^7=xxxxx43\) - tens digit: 4

Now, do you see it? the tens digit alternates between 0 and 4 with both 0 and 4 repeated twice, so there is a set of 4, two numbers with 0 in the tens digit and 2 powers with a zero. If the pattern continues, the 240th power of 7 will be zero, as \(\frac{240}{6}=60\), the 241st power of 7 will also be a zero as the pattern shows (see \(7^4\) and \(7^5\)).

The answer is A 0.

What is the tens digit of 7^241?
Solution:
7th power -> last 2 digit
7^1->07
7^2->49
7^3->49*7->43
7^4->43*7->01
7^5->01*7->07
so this will repeat after 4 powers like 07-49-43-01-07
241 = 4*60+1 --> so last two digits of 7^241 are 07 i.e. the tens digit of 7^241 is 0.

A. 0 --> correct
B. 2
C. 4
D. 6
E. 7

What is the tens digit of \(7^{241}\)?

Let’s find the periodicity of \(7^x\):

\(7^1=07\)
\(7^2=49\)
\(7^3=..43\)
\(7^4=..01\)
\(7^5=..07\)

We can see that the periodicity of \(7^x\) is \(4\) and the tens digit will repeat as follows:

1. \(..0..\)
2. \(..4..\)
3. \(..4..\)
4. \(..0..\)
5. \(..0..\)
6. \(..4..\)
7. etc.

When the power \(241\) of \(7\) is divided by its periodicity \(4\), we have a remainder \(1\). That means that tens digit of \(7^{241}\) is the same the tens digit of \(7^1\) or \(7^5\) or \(7^9\) or in other words equals \(0\).

Hence \(A\)

What is the tens digit of 7^241?

We need to establish the pattern of the tens digit of the powers of 7.

This way, we can write the following equation:

241=kn + R
Where k is the cyclicity and n is a positive integer and R is the remainder when 241 is divided by the cyclicity of the tens digit of the power of 7. The Rth term in the cyclicity of the tens digit is the tens digit of 7^241.

7^1=07 tens digit is 0
7^2=49 tens digit is 4
7^3=343 tens digit is 4
7^4=2401 tens digit is 0
7^5=16807 tens digit is 0
7^6=117649 tens digit is 4
7^7=823543 tens digit is 4

We can see from above that the cyclicity of the tens digit of the powers of 7 is 4, in the order 0440

Hence 242 divided by 4 leaves a remainder of 1.
Therefore 241=4(60)+1
R is 1, hence the cyclicity of 7^241 = 0.

Answer is therefore A.

Posted from my mobile device

Divide by 100 ——> the remainder after dividing by 100 will give the last 2 digits of the number

(1st way) we can use the Chinese Remainder Theorem and break the DEN into co-prime divisors of (25) and (4) [note: 25 * 4= 100 and 25 and 4 so not share any prime factors]

and then find the General form of the number “(7)^241 when divided by 100” by comparing the 2 results


Part 1:
(7)^41 / 25 ——-> remainder of = ?

(7)^2 = 49 ———> which is (-1) away from a multiple of 25

[ (49)^20 * (7) ] / 25 ———-> Rem of = ?

Excess remainder = (-1)^20 * (7) ———> after multiplying remainders, yields a remainder of 7

General Form 1:

N = 25a + 7 (eq 1)


Part 2:
(7)^41 / 4 ——-> remainder of = ?

(7)^2 = 49 ———> is (+1) more than a multiple of 4

[ (49)^20 * (7) ] / 4 ——-> excess remainder of = (1)^20 * 7 = 7

(7) / 4 ———> Rem of = 3

General form part 2:
N = 4b + 3 (eq 2)


Combining the 2 parts to find the General Form of “the dividend (7)^41 when divided by 100”:

N = 25a + 7 = 4b + 3

Where a and b are non negative integer quotients

4b = 25a + 4

b = (25a + 4) / 4

The first non negative value of a that will make (25a + 4) divisible by 4 ———> a = 0

B = (25* 0 + 4) / 4 = 1 ———-> so b = 1

N = 25a + 7 ——-> (25) (0) + 7 = 7

N = 4b + 3 ———-> (4) (1) + 3 = 7


Thus: the General form of “(7)^41 when divided by 100” will take the general form for any number positive integer N where:

N = 100k + 7


(7)^41 is just one of many N values that have the same remainder properties when divided by 100

Thus: (N / 100) ———-> (100k + 7) / 100 >>>>> yields a remainder of = 07

Thus the Tens Digit of the Term (7)^41 is

0 (A)


(Quicker method)
From memory, (7)^3 = 343

Thus, (7)^4 = (7) * (7)^3 = (7) * (343) = (7) (300 + 40 + 3) = (2100 + 280 + 21) =

2,401 ———-> which is (+1) more than a multiple of 100. Again, when we divide any dividend by 100, the remainder will give use the last 2 digits of the NUM

(7)^41 ———> (2,401)^ 10 * (7)

Dividing by 100, the excess remainder will be:

(1)^10 * (7) = 07

The tens digit or (7)^41 will be 0

(A) 0

Edit: is this an official question?

Posted from my mobile device

is this an official GMAT question ? can we expect questions like these ?

No, it's not, and I'm not sure why it's tagged as such. This is not like a real GMAT question. If it asked about units digits, then it might be realistic, though even then, the GMAT prefers to provide less awkward numbers (I would never expect to see an exponent like '241' on a similar official question). So if you're preparing for the GMAT, then if you haven't already, I'd suggest instead looking at official questions testing units digits and exponents, to get a clear idea of what you might be asked on test day.

Posted from my mobile device

You can expect questions that can be solved in under 2 minutes by engaging in manual labor, like several awesome replies here - those that simply wrote out the tens and units digit for the first several iterations of 7^x, finding the patterns as a result.

Of course, this raises the question about when to engage in such manual labor. This might sound superficial, but it’s very real: when you have no idea what else to do.

This clearly would not work for all math. But it’s not a math test. It’s quantitative problem-solving...

Posted from my mobile device

Thanks IanStewart and AlexTheTrainer for such quick responses.

The last 2 digits for powers of 7 follow the pattern:
07, 49, 43 and 01 (a cycle of 4 places)

Thus, 7^240 would complete the cycle and have 01 in the last 2 places (since 240 is a multiple of 4, the cycle)
Hence, 7^241 would be the next in the pattern, i.e. 07

The tens digit is therefore 0.

Answer A

Posted from my mobile device