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MathRevolution
[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. \(5\)

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,
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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.
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MathRevolution
[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. \(5\)

\(\frac{7^4}{100}\) = Remainder 1

Thus, \(\frac{7^8}{100}\) = Remainder 1, Hence answer must be (A)
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pkd niks18 chetan2u KarishmaB gmatbusters

Quote:
When an integer is divided by 100, the remainder will have the same units digit as the integer.

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?
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adkikani
pkd niks18 chetan2u KarishmaB gmatbusters

Quote:
When an integer is divided by 100, the remainder will have the same units digit as the integer.

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?

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
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adkikani
pkd niks18 chetan2u KarishmaB gmatbusters

Quote:
When an integer is divided by 100, the remainder will have the same units digit as the integer.

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?

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
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adkikani
pkd niks18 chetan2u KarishmaB gmatbusters

Quote:
When an integer is divided by 100, the remainder will have the same units digit as the integer.

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?

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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=>

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
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MathRevolution
[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. \(5\)

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
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MathRevolution
[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. \(5\)
\(\frac{7^4}{100} =\) Rem \(1\)

Thus, \(\frac{7^{4*2}}{100}\) , hence Answer must be (A) 1
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Remainder of \(7^8\) is divided by \(100\) = Last two digits of \(7^8\)

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.
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