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We have such a pattern 7 7*7=49 7*7*7=343 7*7*7*7=...1 7*7*7*7*7=7 So, the last number repeats every 5th time. 10/4 = 2, and remainder is 2. we choose 7*7 it means that 7*7*7*7*7*7*7*7*7*7=.......49 We divide by 100, it means .....,49 where 49 is a remainder.

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

1 7 43 49 70

PLZ EXPLAIN THE TRICK IN SOLVING THIS TYPE REMAINDER PROBLEM

Using Binomial theorem, last two digits of an exponent can be found as 7^(10)=7^(2*5)=49^5=(-1+50)^5=(-1)^5+5*(-1)^4*50=-1+50(Just considered last 2-digit of the product)=49

Re: PS: Remainder of 7^(10) divided by 100 [#permalink]
14 Dec 2012, 01:21

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thangvietnam wrote:

if we do not hav formular, how do I do?

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

(A) 1 (B) 7 (C) 43 (D) 49 (E) 70

The remainder when 7^10 is divided by 100 will be the last two digits of 7^10 (for example 123 divided by 100 yields the remainder of 23, 345 divided by 100 yields the remainder of 45).

\(7^{10}=(7^2)^5=49^5\) --> the units digit of 49^5 will be 9 (the units digit of 9^even is 1 and the units digit of 9^odd is 9).

So, we have that \(7^{10}=49^5\) has the units digit of 9, thus the units digit of the remainder must also be 9. Only answer D fits.

Re: What is the remainder, after division by 100, of 7^10 ? [#permalink]
16 Dec 2012, 22:39

1

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see wrote:

Hey Bunnel,

this rule is general ? like if XXXXXX9^even the unit digit of the remainder is always 1 and XXXXX9^odd the unit digit of the remainder is always 9 ??

Thanks for your help

The units digit of 9^even is 1 and the units digit of 9^odd is 9.

If the units digit of a number is 1, then the remainder when this number will be divided by 100 will have the units digit of 1, for example 231 divided by 100 gives the reminder of 31.

If the units digit of a number is 9, then the remainder when this number will be divided by 100 will have the units digit of 9, for example 239 divided by 100 gives the reminder of 39. _________________

Re: What is the remainder, after division by 100, of 7^10 ? [#permalink]
18 Dec 2012, 00:38

Ans:

7^10 can be written as 49^5 which can be written as (49^2)^2. 49 when divided by 100 it will give a remainder of (1)^2.49=49 answer (D). _________________

Re: What is the remainder, after division by 100, of 7^10 ? [#permalink]
17 Sep 2014, 10:13

kkalyan wrote:

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

(A) 1 (B) 7 (C) 43 (D) 49 (E) 70

sol:

7=7 7^2=..9 7^3=..3 7^4=1

now 10/4= 2 i.e. second from top of the pattern...which is 9 since we are dividing the number by 100 last number will be reminder check the answers....D is the only choice

Re: What is the remainder, after division by 100, of 7^10 ? [#permalink]
10 May 2015, 00:53

Bunuel wrote:

thangvietnam wrote:

if we do not hav formular, how do I do?

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

(A) 1 (B) 7 (C) 43 (D) 49 (E) 70

The remainder when 7^10 is divided by 100 will be the last two digits of 7^10 (for example 123 divided by 100 yields the remainder of 23, 345 divided by 100 yields the remainder of 45).

\(7^{10}=(7^2)^5=49^5\) --> the units digit of 49^5 will be 9 (the units digit of 9^even is 1 and the units digit of 9^odd is 9).

So, we have that \(7^{10}=49^5\) has the units digit of 9, thus the units digit of the remainder must also be 9. Only answer D fits.

Answer: D.

excellent. I can not say a word for this wonderful explanation. thank you Buuney

gmatclubot

Re: What is the remainder, after division by 100, of 7^10 ?
[#permalink]
10 May 2015, 00:53