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GMAT 1: 760 Q51 V42
Posts: 10,057
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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[Math Revolution GMAT math practice question]
What is the remainder when \(7^8\) is divided by \(100\)?
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
What is the remainder when \(7^8\) is divided by \(100\)?
A. \(1\)
B. \(2\)
C. \(3\)
D. \(4\)
E. \(5\)
Posts: 1,348
MathRevolution
When an integer is divided by 100, the remainder will have the same units digit as the integer.
Thus, to determine which answer choice represents the remainder when \(7^8\) is divided by 100, we need to know the units digit of \(7^8\).
When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE.
\(7^1\) --> units digit of 7.
\(7^2\) --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
\(7^3\) --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
\(7^4\) --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 7 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, \(7^8\) has a units digit of 1.
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General Discussion
Posts: 65
Typical example of big exponents, where we need to find the pattern for the last digit.
7^1=7
7^2=49
7^3 last digit will be last digit of 9x7=63 => 3
7^4 last digit will be last digit of 3x7=21 =>1
7^5 last digit will be last digit of 1x7=7 => 7
7^6 last digit will be last digit of 7x7=49 => 9
now the pattern repeats itself
7^7 last digit will be 3
7^8 last digit will be 1
Answer A
7^1=7
7^2=49
7^3 last digit will be last digit of 9x7=63 => 3
7^4 last digit will be last digit of 3x7=21 =>1
7^5 last digit will be last digit of 1x7=7 => 7
7^6 last digit will be last digit of 7x7=49 => 9
now the pattern repeats itself
7^7 last digit will be 3
7^8 last digit will be 1
Answer A
Posts: 6,727
MathRevolution
Let's examine 7^8 - 1
Why would I do this?
Well, I know that 7^2 + 1 = 50, which is a factor of 100.
So, perhaps it's the case that 7^8 - 1 is divisible by 100, in which case 7^8 will leave a remainder of 1 when divided by 100
7^8 - 1 is a difference of squares.
So, 7^8 - 1 = (7^4 + 1)(7^4 - 1)
= (7^4 + 1)(7^2 + 1)(7^2 - 1)
= (7^4 + 1)(7^2 + 1)(7 + 1)(7 - 1)
= (7^4 + 1)(50)(8)(6)
= (7^4 + 1)(2400)
= (7^4 + 1)(24)(100)
So, we can see that 7^8 - 1 is divisible by 100
7^8 is 1 greater than 7^8 - 1, so we must get a remainder of 1 when 7^8 is divided by 100
Answer: A
Cheers,
Brent
I did this in following way.
49^4*2^4/2^4*100
= 98^4/2^4*100
=(-2)^4/2^4
= 1
That's the answer.
Please appraise my approach and give kudos if I am correct on my way.
49^4*2^4/2^4*100
= 98^4/2^4*100
=(-2)^4/2^4
= 1
That's the answer.
Please appraise my approach and give kudos if I am correct on my way.
Abhishek009
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Joined: 11 Jun 2011
Last visit: 17 Dec 2025
Posts: 5,880
Status:QA & VA Forum Moderator
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Posts: 5,880
MathRevolution
\(\frac{7^4}{100}\) = Remainder 1
Thus, \(\frac{7^8}{100}\) = Remainder 1, Hence answer must be (A)
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pkd niks18 chetan2u KarishmaB gmatbusters
I could not recall such a rule under timing pressure. Although w/o clock I figured out that if I take random number
say 202 and divide it by 100 I get remainder as 2, which is same as unit digits of 202.
My mind intuitively went into a decimal form of integers under timed stress e.g. 202/100 = 2.02 and I could not relate cyclicity to a decimal form of a fraction. Any two cents here to relate divisibility by 100 with unit digits?
Quote:
I could not recall such a rule under timing pressure. Although w/o clock I figured out that if I take random number
say 202 and divide it by 100 I get remainder as 2, which is same as unit digits of 202.
My mind intuitively went into a decimal form of integers under timed stress e.g. 202/100 = 2.02 and I could not relate cyclicity to a decimal form of a fraction. Any two cents here to relate divisibility by 100 with unit digits?
adkikani
It is also true for divisibility by 10..
Reason is if the number has to be divisible by 10 or 100, it has to have 0 as units digit. So whatever on top of 0 will be units digit..
202/100= (200+2)/100=200/100+2/100..
So 200 is div and left is 2
_________________
adkikani
Hi adkikani
The basic problem that you discussed here is that you went on to think of a decimal number for a remainder question.
In most "Remainder" type question in GMAT you need not think of a decimal number. The options for these questions are also pretty straight forward integers. So as a thumb rule when you see a remainder question you should think of D=Q+R form, where R is the remainder.
and the cyclicity and properties of 10 and its powers are pretty simple to learn. You can practice few questions to internalize it
adkikani
Another thing - remainders and decimals are two different (but equivalent of course) ways in which you can show the output of a division.
e.g.13/5 => 2 quotient and 3 remainder
or 13/5 => 2.6 (the integer part corresponds to the quotient and 6/10 = 3/5 = remainder/divisor)
So when talking about remainders, we are looking at the result of division from the quotient/remainder perspective, not from the decimal perspective.
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GMAT 1: 760 Q51 V42
Posts: 10,057
=>
The remainder when \(7^8\) is divided by \(100\) is equal to the final two digits of \(7^8\).
Now, \(7^1 = 7, 7^2 = 49,\) \(7^3 = 343\), and \(7^4 = 2401\).
So, the final two digits of \(7^n\) have period \(4\):
The tens digits are \(0 -> 4 -> 4 -> 0\)
and the units digits are \(7 -> 9 -> 3 -> 1.\)
It follows that the tens and units digits of \(7^8\) are \(0\) and \(1\), respectively.
Therefore, the remainder when \(7^8\) is divided by \(100\) is \(1\).
Therefore, the answer is A.
Answer : A
The remainder when \(7^8\) is divided by \(100\) is equal to the final two digits of \(7^8\).
Now, \(7^1 = 7, 7^2 = 49,\) \(7^3 = 343\), and \(7^4 = 2401\).
So, the final two digits of \(7^n\) have period \(4\):
The tens digits are \(0 -> 4 -> 4 -> 0\)
and the units digits are \(7 -> 9 -> 3 -> 1.\)
It follows that the tens and units digits of \(7^8\) are \(0\) and \(1\), respectively.
Therefore, the remainder when \(7^8\) is divided by \(100\) is \(1\).
Therefore, the answer is A.
Answer : A
MathRevolution
using 7,9,3,1 cycle, 7^8/100 remainder=7^4/100 remainder
7^4=2401
2401/100 gives remainder of 1
1
A
Abhishek009
Board of Directors
Joined: 11 Jun 2011
Last visit: 17 Dec 2025
Posts: 5,880
Status:QA & VA Forum Moderator
Location: India
GPA: 3.5
WE:Business Development (Commercial Banking)
Posts: 5,880
MathRevolution
\(\frac{7^4}{100} =\) Rem \(1\)
Thus, \(\frac{7^{4*2}}{100}\) , hence Answer must be (A) 1
_________________
Remainder of \(7^8\) is divided by \(100\) = Last two digits of \(7^8\)
Last Two digits of 7 follow following pattern
=> We have a cycle of 4
=> last two digits of 7^8 = Last two digits of 7^4 = 01
So, Answer will be A
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
_________________
Last Two digits of 7 follow following pattern
- Last Two digits of 7^1 = 07
- Last Two digits of 7^2 = 49
- Last Two digits of 7^3 = 49*7 = 43
- Last Two digits of 7^4 = 43*7 = 01
- Last Two digits of 7^5 = 01*7 = 07
- Last Two digits of 7^6 = 07*7 = 49
=> We have a cycle of 4
=> last two digits of 7^8 = Last two digits of 7^4 = 01
So, Answer will be A
Hope it helps!
Link to Theory for Last Two digits of exponents here.
Link to Theory for Units' digit of exponents here.
_________________
How to Solve :
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
All Quant Topics Theory || Last Two Digits of Exponents Ending with 1, 2, 3, 4, 5, 6, 7, 8, 9 || Units' Digit Problems || Even and Odd || Divisibility Rules || Positive and Negative || Absolute Value Basics || Inequalities || Absolute Value + Inequality || Functions and Custom Characters || LCM & GCD || Remainders || Factors and Multiples || Arithmetic and Geometric Progression || Co-ordinate Geometry || Circles || Probability: Coin Toss || Work Rate || Statistics || Stem & Leaf Plot || Box And Whisker Plot
GMAT Club Topic Wise Problems
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